Practice Problem Set 7 – a discrete joint distribution

The practice problems presented here deal with a discrete joint distribution that is defined by multiplying a marginal distribution and a conditional distribution – similar to the joint distribution found here and here. Thus this post provides additional practice opportunities.

Practice Problems

Let X be the value of a roll of a fair die. For X=x, suppose that Y \lvert X=x has a binomial distribution with n=4 and p=x / 10.

Practice Problem 7-A
Compute the conditional binomial distributions Y \lvert X=x where x=1,2,3,4,5,6.

Practice Problem 7-B
Calculate the joint probability function P[X=x,Y=y] for x=1,2,3,4,5,6 and y=0,1,2,3,4.

Practice Problem 7-C
Determine the probability function for the marginal distribution of Y. Calculate the mean and variance of Y.

Practice Problem 7-D
Calculate the backward conditional probabilities P[X=x \lvert Y=y] for all applicable x and y.

Problems 7-A to 7-D are similar to the ones in this previous post.

Practice Problem 7-E
Calculate the mean and variance of X.

Practice Problem 7-F
Calculate the mean and variance of Y (use the methods discussed here).

Practice Problem 7-G
Calculate the covariance \text{Cov}(X,Y) and the correlation coefficient \rho.

Problems 7-E to 7-G are similar to the ones in this previous post.

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Answers

Practice Problem 7-A

    \displaystyle \begin{aligned} &P[Y=0 \lvert X=1]=0.6561 \\&P[Y=1 \lvert X=1]=0.2916 \\&P[Y=2 \lvert X=1]=0.0486 \\&P[Y=3 \lvert X=1]=0.0036 \\&P[Y=4 \lvert X=1]=0.0001 \end{aligned}

    \displaystyle \begin{aligned} &P[Y=0 \lvert X=2]=0.4096 \\&P[Y=1 \lvert X=2]=0.4096 \\&P[Y=2 \lvert X=2]=0.1536 \\&P[Y=3 \lvert X=2]=0.0256 \\&P[Y=4 \lvert X=2]=0.0016 \end{aligned}

    \displaystyle \begin{aligned} &P[Y=0 \lvert X=3]=0.2401 \\&P[Y=1 \lvert X=3]=0.4116 \\&P[Y=2 \lvert X=3]=0.2646 \\&P[Y=3 \lvert X=3]=0.0756 \\&P[Y=4 \lvert X=3]=0.0081 \end{aligned}

    \displaystyle \begin{aligned} &P[Y=0 \lvert X=4]=0.1296 \\&P[Y=1 \lvert X=4]=0.3456 \\&P[Y=2 \lvert X=4]=0.3456 \\&P[Y=3 \lvert X=4]=0.1536 \\&P[Y=4 \lvert X=4]=0.0256 \end{aligned}

    \displaystyle \begin{aligned} &P[Y=0 \lvert X=5]=0.0625 \\&P[Y=1 \lvert X=5]=0.25 \\&P[Y=2 \lvert X=5]=0.375 \\&P[Y=3 \lvert X=5]=0.25 \\&P[Y=4 \lvert X=5]=0.0625 \end{aligned}

    \displaystyle \begin{aligned} &P[Y=0 \lvert X=6]=0.0256 \\&P[Y=1 \lvert X=6]=0.1536 \\&P[Y=2 \lvert X=6]=0.3456 \\&P[Y=3 \lvert X=6]=0.3456 \\&P[Y=4 \lvert X=6]=0.1296 \end{aligned}

Practice Problem 7-B

    \displaystyle \begin{aligned} &P[Y=4,X=1]=\frac{0.0001}{6} \\&P[Y=4,X=2]=\frac{0.0016}{6} \\&P[Y=4,X=3]=\frac{0.0081}{6} \\&P[Y=4,X=4]=\frac{0.0256}{6} \\&P[Y=4,X=5]=\frac{0.0625}{6} \\&P[Y=4,X=6]=\frac{0.1296}{6} \end{aligned}

    \displaystyle \begin{aligned} &P[Y=3,X=1]=\frac{0.0036}{6} \\&P[Y=3,X=2]=\frac{0.0256}{6} \\&P[Y=3,X=3]=\frac{0.0756}{6} \\&P[Y=3,X=4]=\frac{0.1536}{6} \\&P[Y=3,X=5]=\frac{0.25}{6} \\&P[Y=3,X=6]=\frac{0.3456}{6} \end{aligned}

    \displaystyle \begin{aligned} &P[Y=2,X=1]=\frac{0.0486}{6} \\&P[Y=2,X=2]=\frac{0.1536}{6} \\&P[Y=2,X=3]=\frac{0.2646}{6} \\&P[Y=2,X=4]=\frac{0.3456}{6} \\&P[Y=2,X=5]=\frac{0.375}{6} \\&P[Y=2,X=6]=\frac{0.3456}{6} \end{aligned}

    \displaystyle \begin{aligned} &P[Y=1,X=1]=\frac{0.2916}{6} \\&P[Y=1,X=2]=\frac{0.4096}{6} \\&P[Y=1,X=3]=\frac{0.4116}{6} \\&P[Y=1,X=4]=\frac{0.3456}{6} \\&P[Y=1,X=5]=\frac{0.25}{6} \\&P[Y=1,X=6]=\frac{0.1536}{6} \end{aligned}

    \displaystyle \begin{aligned} &P[Y=0,X=1]=\frac{0.6561}{6} \\&P[Y=0,X=2]=\frac{0.4096}{6} \\&P[Y=0,X=3]=\frac{0.2401}{6} \\&P[Y=0,X=4]=\frac{0.1296}{6} \\&P[Y=0,X=5]=\frac{0.0625}{6} \\&P[Y=0,X=6]=\frac{0.0256}{6} \end{aligned}

Practice Problem 7-C

    \displaystyle \begin{aligned} &P[Y=4]=\frac{0.2275}{6} \\&P[Y=3]=\frac{0.854}{6} \\&P[Y=2]=\frac{1.533}{6} \\&P[Y=1]=\frac{1.862}{6} \\&P[Y=0]=\frac{1.5235}{6} \end{aligned}

    \displaystyle E[Y]=1.4

    \displaystyle E[Y^2]=3.22

    \displaystyle Var[Y]=1.26

Practice Problem 7-D

    \displaystyle \begin{aligned} &P[X=1 \lvert Y=0]=\frac{0.6561}{1.5235}=0.4307 \\&P[X=2 \lvert Y=0]=\frac{0.4096}{1.5235}=0.2689 \\&P[X=3 \lvert Y=0]=\frac{0.2401}{1.5235}=0.1576 \\&P[X=4 \lvert Y=0]=\frac{0.1296}{1.5235}=0.0851 \\&P[X=5 \lvert Y=0]=\frac{0.0625}{1.5235}=0.0410 \\&P[X=6 \lvert Y=0]=\frac{0.0256}{1.5235}=0.0168 \end{aligned}

    \displaystyle \begin{aligned} &P[X=1 \lvert Y=1]=\frac{0.2916}{1.862}=0.1566 \\&P[X=2 \lvert Y=1]=\frac{0.4096}{1.862}=0.2200 \\&P[X=3 \lvert Y=1]=\frac{0.4116}{1.862}=0.2211 \\&P[X=4 \lvert Y=1]=\frac{0.3456}{1.862}=0.1856 \\&P[X=5 \lvert Y=1]=\frac{0.25}{1.862}=0.1343 \\&P[X=6 \lvert Y=1]=\frac{0.1536}{1.862}=0.0825 \end{aligned}

    \displaystyle \begin{aligned} &P[X=1 \lvert Y=2]=\frac{0.0486}{1.533}=0.0317 \\&P[X=2 \lvert Y=2]=\frac{0.1536}{1.533}=0.1002 \\&P[X=3 \lvert Y=2]=\frac{0.2646}{1.533}=0.1726 \\&P[X=4 \lvert Y=2]=\frac{0.3456}{1.533}=0.2254 \\&P[X=5 \lvert Y=2]=\frac{0.375}{1.533}=0.2446 \\&P[X=6 \lvert Y=2]=\frac{0.3456}{1.533}=0.2254 \end{aligned}

    \displaystyle \begin{aligned} &P[X=1 \lvert Y=3]=\frac{0.0036}{0.854}=0.0042 \\&P[X=2 \lvert Y=3]=\frac{0.0256}{0.854}=0.0300 \\&P[X=3 \lvert Y=3]=\frac{0.0756}{0.854}=0.0885 \\&P[X=4 \lvert Y=3]=\frac{0.1536}{0.854}=0.1799 \\&P[X=5 \lvert Y=3]=\frac{0.25}{0.854}=0.2927 \\&P[X=6 \lvert Y=3]=\frac{0.3456}{0.854}=0.4047 \end{aligned}

    \displaystyle \begin{aligned} &P[X=1 \lvert Y=4]=\frac{0.0001}{0.2275}=0.0004 \\&P[X=2 \lvert Y=4]=\frac{0.0016}{0.2275}=0.0070 \\&P[X=3 \lvert Y=4]=\frac{0.0081}{0.2275}=0.0356 \\&P[X=4 \lvert Y=4]=\frac{0.0256}{0.2275}=0.1125 \\&P[X=5 \lvert Y=4]=\frac{0.0625}{0.2275}=0.2747 \\&P[X=6 \lvert Y=4]=\frac{0.1296}{0.2275}=0.5697 \end{aligned}

Practice Problem 7-E

    \displaystyle E[X]=\frac{7}{2}=3.5

    \displaystyle E[X^2]=\frac{91}{6}

    \displaystyle Var[X]=\frac{35}{12}

Practice Problem 7-F

    \displaystyle E[Y]=1.4

    \displaystyle E[Y^2]=3.22

    \displaystyle Var[Y]=1.26

Practice Problem 7-G

    \displaystyle \text{Cov}(X,Y)=\frac{7}{6}

    \displaystyle \rho=\frac{7}{6 \sqrt{3.675}}=0.60858

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Practice Problem Set 6 – transformation of univariate random variables

This post provides practice problems to reinforce the concept discussed in this post on transformation of univariate distributions.

In each problem in this post, a pdf for the random variable X is given and a transformation Y=g(X) is given where g(x) is a one-to-one function. The problem is to obtain the pdf of the transformed variable Y as well as to calculate probabilities regarding Y.

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Practice Problem 6-A

Let Y=2 X+5 where the pdf of the random variable X is given by:

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  \frac{1}{50} \ (10-x) &\ \ \ \ \ \ 0 < x < 10 \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

  • Find the pdf of Y.
  • Determine the mean and variance of Y.
  • Compute the probability P[10<Y<20]
  • Determine the 75th percentile of Y.

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Practice Problem 6-B

Let Y=\sqrt{X} where the pdf of the random variable X is given by:

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  \frac{1}{1024} \ (-63+61 x+3 x^2-x^3) &\ \ \ \ \ \ 1 < x < 9 \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

  • Find the pdf of Y.
  • Compute the probability P[1<Y<2]

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Practice Problem 6-C

Suppose that the random variable X follows an exponential distribution with the following pdf.

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  \frac{1}{4} \ e^{-x / 4} &\ \ \ \ \ \ 0 < x < \infty \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

Let Y=3 X+2.

  • Find the pdf of Y.
  • Determine the mean and variance of Y.
  • Compute the probability P[6<Y<18]
  • Determine the 90th percentile of Y.

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Practice Problem 6-D
The random variable X has a pdf that is given below.

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  \frac{1}{4} \ x (4-x^2) &\ \ \ \ \ \ 0 < x < 2 \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

Let Y=5-\frac{X}{2}.

  • Find the pdf of Y.
  • Determine the mean and variance of Y.

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Practice Problem 6-E

Suppose that the random variable X follows an exponential distribution with the following pdf.

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  \frac{1}{4} \ e^{-x / 4} &\ \ \ \ \ \ 0 < x < \infty \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

Let Y=X^2.

  • Determine of pdf of Y.
  • Determine of CDF of Y.
  • Calculate the probabilities P(Y > y) where y=4,8,16,32,64,128,256.

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Practice Problem 6-F
Suppose that the random variable X follows an exponential distribution with the following pdf.

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  \frac{1}{4} \ e^{-x / 4} &\ \ \ \ \ \ 0 < x < \infty \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

Let Y=\sqrt{X}.

  • Determine of pdf of Y.
  • Determine of CDF of Y.
  • Calculate the probabilities P(Y > y) where y=4,8,16,32.

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Practice Problem 6-G

Suppose that X follows a uniform distribution whose pdf is given by the following.

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  1 &\ \ \ \ \ \ 0 < x < 1 \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

Let Y=-\theta \ln(X) where \theta is a positive constant.

  • Determine the pdf of Y.
  • What is this distribution?

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Practice Problem 6-H

Consider the random variable X whose pdf is given below.

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  \frac{3}{125} x^2 &\ \ \ \ \ \ 0 < x < 5 \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

Let Y=X^3.

  • Determine the pdf of Y.
  • What is this distribution?

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Practice Problem 6-I

Suppose that X has the following density function.

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  6 \ x \ e^{-3 \ x^2} &\ \ \ \ \ \ 0 < x < \infty \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

Let Y=X^2.

  • Determine the pdf of Y.
  • What is this distribution?

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Practice Problem 6-J

Suppose that the pdf of the random variable X is given below.

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  \frac{3}{x^4} &\ \ \ \ \ \ 1 < x < \infty \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

Let Y=\ln(X).

  • Determine the pdf of Y.
  • What is this distribution?

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Problem ………..Answer
6-A
  • \displaystyle f_Y(y)=\frac{1}{200} (25-y) \ \ \ \ \ \ 5<y<25
  • \displaystyle E[Y]=\frac{35}{3}, \displaystyle Var[Y]=\frac{200}{9}
  • \displaystyle P[10<Y<20]=\frac{1}{2}
  • 75th percentile = 15
6-B
  • \displaystyle f_Y(y)=\frac{1}{512} (-63y+61y^3+3y^5-y^7) \ \ \ \ \ \ 1<y<3
  • \displaystyle P[1<Y<2]=\frac{1071}{4096}
6-C
  • \displaystyle f_Y(y)=\frac{1}{12} e^{-\frac{1}{12} (y-2) } \ \ \ \ \ \ 2<y<\infty
  • \displaystyle E[Y]=14, \displaystyle Var[Y]=144
  • \displaystyle P[16<Y<18]=e^{-4/12}-e^{-16/12}=0.4529341725
  • 90th percentile = 29.63102112
6-D
  • \displaystyle f_Y(y)=4 (-120+74y-15y^2+y^3) \ \ \ \ \ \ 4<y<5
  • \displaystyle E[Y]=\frac{67}{15}, \displaystyle Var[Y]=\frac{11}{225}
6-E
  • \displaystyle f_Y(y)=\frac{1}{8} \ \frac{1}{\sqrt{y}} \ e^{-(y/16)^{0.5}} \ \ \ \ \ \ 0<y<\infty
  • \displaystyle F_Y(y)=1- e^{-(y/16)^{0.5}} \ \ \ \ \ \ 0<y<\infty
  • \displaystyle 1-F_Y(4)=e^{-0.5}=0.6065306597
  • \displaystyle 1-F_Y(8)=e^{-1/ \sqrt{2}}=0.4930686914
  • \displaystyle 1-F_Y(16)=e^{-1}=0.3678794412
  • \displaystyle 1-F_Y(32)=e^{- \sqrt{2}}=0.2431167344
  • \displaystyle 1-F_Y(64)=e^{-2}=0.1353352832
  • \displaystyle 1-F_Y(128)=e^{- \sqrt{8}}=0.0591057466
  • \displaystyle 1-F_Y(256)=e^{-4}=0.0183156389
6-F
  • \displaystyle f_Y(y)=\frac{1}{2} \ y \ e^{-(y/2)^{2}} \ \ \ \ \ \ 0<y<\infty
  • \displaystyle F_Y(y)=1- e^{-(y/2)^{2}} \ \ \ \ \ \ 0<y<\infty
  • \displaystyle 1-F_Y(4)=e^{-4}=0.0183156389
  • \displaystyle 1-F_Y(8)=e^{-16}=1.12535 \cdot 10^{-7}
  • \displaystyle 1-F_Y(16)=e^{-64}=1.6038 \cdot 10^{-28}
  • \displaystyle 1-F_Y(32)=e^{-256}=6.6163 \cdot 10^{-112}
6-G
  • \displaystyle f_Y(y)=\frac{1}{\theta} \ e^{-y / \theta} \ \ \ \ \ \ 0<y<\infty
  • Exponential distribution
6-H
  • \displaystyle f_Y(y)=\frac{1}{125} \ \ \ \ \ \ 0<y<125
  • Uniform distribution
6-I
  • \displaystyle f_Y(y)=3 \ e^{-3 y} \ \ \ \ \ \ 0<y<\infty
  • Exponential distribution with mean 1/3
6-J
  • \displaystyle f_Y(y)=3 \ e^{-3 y} \ \ \ \ \ \ 0<y<\infty
  • Exponential distribution with mean 1/3

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Transformation of univariate random variables

Creating new probability distributions from old ones (or existing ones) is a familiar theme in the study pf probability. This post shows how to generate a distribution under a transformation. The process is illustrated with examples.

Practice problems are found in the next post.

Examples

The starting point is the random variable X whose probability density function (pdf) is given by the following.

    \displaystyle  f(x) = \left\{ \begin{array}{ll}                   \displaystyle  \frac{1}{8} \ (4-x) &\ \ \ \ \ \ 0 < x < 4 \\           \text{ } & \text{ } \\          \displaystyle  0 &\ \ \ \ \ \ \text{otherwise} \\                              \end{array} \right.

The given X is transformed in four different ways as follows:

    Y_1=X^2

    Y_2=16-X^2

    Y_3=\sqrt{X}

    Y_4=\sqrt{4-X}

We demonstrate how to derive the pdfs of these four new random variables based on the pdf given at the beginning. Note that the support of X is the interval (0, 4). Because of the transformations, the supports of the Y variables are different. The support of Y_1 is the interval (0, 16). The support of Y_2 is also (0, 16). The support of Y_3 is (0, 2). The support of Y_4 is also (0, 2).

There are two ways to derive the pdfs of Y_i, i=1,2,3,4. One way is the CDF method: to find the CDF of the new variable and then take the derivative to get the pdf. Another way is the method of transformation, which is the focus here. We show how to use CDF method on Y_1 in order to draw out the idea of the method of transformation.

The f_{Y_1}(y) and F_{Y_1}(y) be the pdf and CDF of Y_1. Let F(x) be the CDF of X. The following derives F_{Y_1}(y).

    \displaystyle \begin{aligned} F_{Y_1}(y)&=P(Y_1 \le y) \\&=P(X^2 \le y) \\&=P(X \le \sqrt{y}) \\&=F(\sqrt{y})  \end{aligned}

Thus the CDF F_{Y_1}(y) is the CDF F(x) evaluated at \sqrt{y}. Since pdf is the derivative of the CDF, the pdf f_{Y_1}(y) is obtained by taking derivative of F(\sqrt{y}).

    \displaystyle \begin{aligned} f_{Y_1}(y)&=\frac{d}{dy} F_{Y_1}(y) \\&=\frac{d}{dy} F(\sqrt{y}) \\&=f(\sqrt{y}) \ \frac{d}{dy} \sqrt{y} \ \ \ \ \ *\\&=\frac{1}{8} \biggl(4-\sqrt{y} \biggr) \frac{1}{2 \sqrt{y}} \\&=\frac{1}{16} \biggl(\frac{4}{\sqrt{y}}-1 \biggr) \ \ \ \ \ \ \ \ \ \ 0<y<16 \end{aligned}

The step that is labeled with * is the key step in the derivative and will be discussed in further details.

The Method of Transformation

Let’s describe the method demonstrated in the above derivation. There is a starting probability distribution represented by the random variable X. Its pdf is f(x) whose support is a subset of the x-axis, likely an interval (of finite or infinite length). Let’s call the support S. The support of a pdf is the set of all x such that f(x)>0. We have a differentiable function g(x) defined on S. This function g(x) is a one-to-one function over the support S. The function g(x) does not have to be a one-to-one function over all of the x-axis. It just has to be one-to-one over the support S. As a result, the function g(x) is either an increasing function or a decreasing function over the support.

Since y=g(x) is a one-to-one function, it has an inverse x=g^{-1}(y). The inverse g^{-1}(y) is defined over the set g(S).

Consider the new random variable Y=g(X). The following gives the pdf of Y.

(1)……\displaystyle f_Y(y)=f(g^{-1}(y)) \ \biggl \lvert \frac{d}{dy} \ g^{-1}(y)  \biggr \lvert

The formula (1) gives the method of transformation and is illustrated by the step labeled with * above. With Y_1=X^2, the transformation is the function g(x)=x^2. It is not a one-to-one function over the entire x-axis but it is a one-to-one function on the support (0, 4). In fact, g(x)=x^2 is an increasing function over (0, 4). The inverse function is then g^{-1}(y)=\sqrt{y}. Applying (1) gives the pdf f_Y(y).

One thing to keep in mind is that the method works only if the transformation g(x) is a one-to-one function over the support of the original pdf f(x) (either g(x) is increasing or decreasing). If not, the method will produce a wrong answer. Another thing to keep in mind is that when the transformation g(x) is a decreasing function, its inverse g^{-1}(y) is also a decreasing function. Then its derivative would be negative. In (1), we use the absolute value of the derivative.

Examples Continued

Using the method of transformation, the following shows the pdfs of Y_i, i=1,2,3,4.

    \displaystyle f_{Y_1}(y)=\frac{1}{16} \biggl(\frac{4}{\sqrt{y}}-1 \biggr) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0<y<16

    \displaystyle f_{Y_2}(y)=\frac{1}{16} \biggl(\frac{4}{\sqrt{16-y}}-1 \biggr) \ \ \ \ \ \ \ \ \ \ 0<y<16

    \displaystyle f_{Y_3}(y)=\frac{1}{4} \ (4 y-y^3 ) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0<y<2

    \displaystyle f_{Y_4}(y)=\frac{1}{4} \ y^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0<y<2

It will be instructive to examine the graphs of the pdfs. The following is the graph of the pdf of X.

Figure 1

The starting pdf (Figure 1) is a straight line with negative slope. In this distribution, more probabilities are found near zero. For example P(X \le 1)=0.4375. About 44% of the values from this distribution are expected to be less than 1. The following is the graph of the pdf of Y_1=X^2.

Figure 2

Figure 2 shows that the effect of the transformation y=x^2 is to push the probabilities further to zero. The 44% that is less than 1 in Figure 1 is further pushed toward zero. Hence the graph in Figure 2 is extremely positively skewed (or skewed to the right since the right tail is longer). The following is the graph of the pdf of Y_2=16-X^2.

Figure 3

Figure 3 shows that the effect of the transformation y=16-x^2 is to push the probabilities in the opposite direction toward 4. The hence the distribution of Y_2 is extremely negative skewed (or left skewed since the left tail is longer).

Remarks

The method of transformation is a great tool making it possible to create new distributions with desired characteristics from old ones.

Some named distributions are generated from transformation. For example, the lognormal distribution is a transformation from the normal distribution where the transformation is an exponential function. More specifically, if X has a normal distribution with mean \mu and variance \sigma^2, then Y=e^X has a lognormal distribution and parameters \mu and \sigma^2. The transformation goes the other way too. If Y has a lognormal distribution and parameters \mu and \sigma^2, then X=\ln(Y) has a normal distribution with mean \mu and variance \sigma^2. See here for a discussion of lognormal distribution. Practice problems on lognormal distribution are found here.

Another named distribution that is generated from a transformation is the Weibull distribution. It is generated by raising an exponential random variable to a power (discussed here). The topic of raising an exponential distribution to a power is further discussed here. For more distributions created by transformation, explore to the site in the given links.

Practice problems are found in the next post.

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Practice Problem Set 5 – bivariate normal distribution

This post provides practice problems to reinforce the concept of bivariate normal distribution discussed in two posts – one is a detailed introduction to bivariate normal distribution and the other is a further discussion that brings out more mathematical properties of the bivariate normal distribution. The properties discussed in these two posts form the basis for the calculation behind the practice problems presented here.

The practice problems presented here are mostly on calculating probabilities. The normal probabilities can be obtained using a normal table or a calculator that has a function for normal distribution (such as TI84+). The answers for normal probabilities given at the end of the post have two versions – one using a normal table (found here) and the other one using TI84+.

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Practice Problem 5-A

Suppose that X and Y follow a bivariate normal distribution with parameters \mu_X=15, \sigma_X=4, \mu_Y=20, \sigma_Y=5 and \rho=-0.7.

Determine the following.

  • Compute the probability P[12<Y<28]
  • For X=20, determine the mean and standard deviation of the conditional distribution of Y given X=20.
  • Determine P[12<Y<28 \lvert X=20], the probability that 12<Y<28 given X=20.

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Practice Problem 5-B

Suppose that X and Y follow a bivariate normal distribution with parameters \mu_X=6, \sigma_X=1.6, \mu_Y=4, \sigma_Y=1.2 and \rho=0.8.

Determine the following.

  1. Compute the probability P[3<Y<5]
  2. Determine E[Y \lvert X=x], the mean of the conditional distribution of Y given X=x.
  3. Determine \sigma_{Y \lvert x}^2=Var[Y \lvert X=x] and \sigma_{Y \lvert x}, the variance and the standard deviation of the conditional distribution of Y given X=x.
  4. For each of the x values 6, 8, 10 and 12, determine the 99.7% interval (a,b) for the conditional distribution of Y given x, i.e. a is three standard deviations below the mean and b is 3 standard deviations above the mean.
  5. For each of the x values 6, 8, 10 and 12, determine P[3<Y<5 \ \lvert X=x]. Explain the magnitude of each of these probabilities based on the intervals in 6.

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Practice Problem 5-C

Let X and Y have a bivariate normal distribution with parameters \mu_X=50, \sigma_X=10, \mu_Y=60, \sigma_Y=5 and \rho=0.6. Determine the following.

  • Calculate P[100<X+Y<140]
  • Determine the 5 parameters of the bivariate normal random variables L=X+Y and M=X-Y.
  • Calculate P[100<X+Y<140 \ \biggl \lvert \ X-Y=5]

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Practice Problem 5-D

Suppose X is the height (in inches) and Y is the weight (in pounds) of a male student in a large university. Furthermore suppose that X and Y follow a bivariate normal distribution with parameters \mu_X=69, \mu_Y=155, \sigma_X=2.5, \sigma_Y=20 and \rho=0.55.

  • What is the distribution of the weights of all male students what are 5 feet 11 inches tall (71 inches)?
  • For a randomly chosen male student who is 71 inches tall, what is the probability that his weight is between 170 and 200 pounds?
  • For male students who are 71 inches tall, what is the 90th percentile of weight?

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Practice Problem 5-E
Suppose that X and Y have a bivariate normal distribution with parameters \mu_X=70, \mu_Y=70, \sigma_X=5, \sigma_Y=10 and \rho>0.

Further suppose that P[58.24<Y<81.76 \lvert X=70]=0.95. Determine \rho.

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Practice Problem 5-F

Suppose that X and Y have a bivariate normal distribution with parameters \mu_X=70, \mu_Y=60, \sigma_X=10, \sigma_Y=12 and \rho=0.8.

  • Compute P[45<Y<55 \lvert X=60]
  • When X=60, 4 values of Y are observed. Compute P[45<\overline{Y}<55 \lvert X=60] where \overline{Y} is the mean of the sample of size 4.

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Practice Problem 5-G
Let X and Y have a bivariate normal distribution with parameters \mu_X=70, \mu_Y=50, \sigma_X=10, \sigma_Y=12 and \rho=-0.65. Determine the following.

  • P[X-Y<50]
  • \displaystyle P[55<\frac{X+Y}{2}<65]

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Practice Problem 5-H
Let X and Y have a bivariate normal distribution with parameters \mu_X=70, \sigma_X=5, \mu_Y=50, \sigma_Y=10 and \rho=0.75. Determine the following probabilities.

  • P \biggl[ \frac{X+Y}{2}<68 \biggr]
  • P \biggl[ \frac{X+Y}{2}<68 \ \biggl \lvert  Y=60 \biggr]

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Practice Problem 5-I

For a couple from a large population of married couples, let X be the height (in inches) of the husband and let Y be the height (in inches) of the wife. Suppose that X and Y have a bivriate normal distribution with parameters \mu_X=68, \mu_Y=65, \sigma_X=2.2, \sigma_Y=2.5 and \rho=0.5.

  • For a randomly selected wife from this population, determine the probability that her height is between 68 inches and 72 inches.
  • For a randomly selected wife from this population whose husband is 72 inches tall, determine the probability that her height is between 68 inches and 72 inches.
  • For a randomly selected couple from this population, determine the probability that the wife is taller than the husband.

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Practice Problem 5-J

The annual revenues of Company X and Company Y are positively correlated since the correlation coefficient between the two revenues is 0.65. The annual revenue of Company X is, on average, 4,500 with standard deviation 1,500. The annual revenue of Company Y is, on average, 5,500 with standard deviation 2,000.

  • Calculate the probability that annual revenue of Company X is less than 6,800 given that the annual revenue of Company Y is 6,800.
  • Calculate the probability that the annual revenue of Company X is greater than that of Company Y given that their total revenue is 12,000.

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Problem ………..Answer
5-A
  1. P[12<Y<28]=0.8904 (table), 0.8904014212 (TI84+)
  2. E[Y \lvert X=20]=15.625, Var[Y \lvert X=20]=12.75
  3. P[12<Y<28 \lvert X=20]=0.8458 (table), 0.8447309876 (TI84+)
5-B
  • P[3<Y<5]=0.5934 (table),0.5953433508 (TI84+)
  • \displaystyle E[Y \lvert x]=0.4+0.6 \ x
  • \displaystyle Var[Y \lvert x]=0.5184, standard deviation = 0.72
    • For x = 6, (1.84, 6.16)
    • For x = 8, (3.04, 7.36)
    • For x = 10, (4.24, 8.56)
    • For x = 12, (5.44, 9.76)
    • P[3<Y<5 \lvert X=6]=0.8354 (table), 0.8351333522 (TI84+)
    • P[3<Y<5 \lvert X=8]=0.3886 (table), 0.3894682472 (TI84+)
    • P[3<Y<5 \lvert X=10]=0.0262 (table), 0.025919702 (TI84+)
    • P[3<Y<5 \lvert X=12]=0.0002 (table), 0.0001524802646 (TI84+)
5-C
  • P[100<X+Y<140]=0.7568 (table), 0.7551912515 (TI84+)
  • \displaystyle \mu_L=110 \ \ \ \sigma_L=\sqrt{185} \ \ \ \mu_M=-10 \ \ \ \sigma_M=\sqrt{65} \ \ \ \rho_{L,M}=\frac{75}{\sqrt{185} \sqrt{65}}
  • P[100<X+Y<140 \lvert X-Y=5]=0.8967 (table), 0.8966089617 (TI84+)
5-D
  • Normal with mean 163.8 and standard deviation \sqrt{279}.
  • P[170<Y<200 \lvert X=71]=0.3407 (table), 0.3401418637 (TI84+)
  • 90th percentile = 185.18 (table), 185.2061314 (TI84+)
5-E
  • 0.8
5-F
  • P[45<Y<55 \lvert X=60]=0.5123 (table), 0.5119251771 (TI84+)
  • P[45<\overline{Y}<55 \lvert X=60]=0.8329 (table), 0.8325288097 (TI84+)
5-G
  • P[X-Y<50]=0.9332 (table), 0.9331927713 (TI84+)
  • \displaystyle P[55<\frac{X+Y}{2}<65]=0.7154 (table), 0.7135779177 (TI84+)
5-H
  • P \biggl[ \frac{X+Y}{2}<68 \biggr]=0.8708 (table), 0.8710504336 (TI84+)
  • P \biggl[ \frac{X+Y}{2}<68 \ \biggl \lvert  Y=60 \biggr]=0.7517 (table), 0.7518542213 (TI84+)
5-I
  • 0.1125 (table), 0.1125145409 (TI84+)
  • 0.3523 (table), 0.3539664536 (TI84+)
  • 0.1020 (table), 0.1022447094 (TI84+)
5-J
  • 0.9279 (table), 0.9280950079 (TI84+)
  • 0.1736 (table), 0.1736950626 (TI84+)

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Calculating bivariate normal probabilities

This post extends the discussion of the bivariate normal distribution started in this post from a companion blog. Practice problems are given in the next post.

Suppose that the continuous random variables X and Y follow a bivariate normal distribution with parameters \mu_X, \sigma_X, \mu_Y, \sigma_Y and \rho. What to make of these five parameters? According to the previous post, we know that

  • \mu_X and \sigma_X are the mean and standard deviation of the marginal distribution of X,
  • \mu_Y and \sigma_Y are the mean and standard deviation of the marginal distribution of Y,
  • and finally \rho is the correlation coefficient of X and Y.

So the five parameters of a bivariate normal distribution are the means and standard deviations of the two marginal distributions and the fifth parameter is the correlation coefficient that serves to connect X and Y. If \rho=0, then X and Y are simply two independent normal distributions.

When calculating probabilities involving a bivariate normal distribution, keep in mind that both marginal distributions are normal. Furthermore, the conditional distribution of one variable given a value of the other is also normal. Much more can be said about the conditional distributions.

The conditional distribution of Y given X=x is usually denoted by Y \lvert X=x or Y \lvert x. In additional to being a normal distribution, it has a mean that is a linear function of x and has a variance that is constant (it does not matter what x is, the variance is always the same). The linear conditional mean and constant variance are given by the following:

    \displaystyle E[Y \lvert X=x]=\mu_Y+\rho \ \frac{\sigma_Y}{\sigma_X} \ (x-\mu_X)

    \displaystyle Var[Y \lvert X=x]=\sigma_Y^2 \ (1-\rho^2)

Similarly, the conditional distribution of X given Y=y is usually denoted by X \lvert Y=y or X \lvert y. In additional to being a normal distribution, it has a mean that is a linear function of x and has a variance that is constant. The linear conditional mean and constant variance are given by the following:

    \displaystyle E[X \lvert Y=y]=\mu_X+\rho \ \frac{\sigma_X}{\sigma_Y} \ (y-\mu_Y)

    \displaystyle Var[X \lvert Y=y]=\sigma_X^2 \ (1-\rho^2)

The information about the conditional distribution of Y on X=x is identical to the information about the conditional distribution of X on Y=y, except for the switching of X and Y. An example is helpful.

Example 1
Suppose that the continuous random variables X and Y follow a bivariate normal distribution with parameters \mu_X=10, \sigma_X=10, \mu_Y=20, \sigma_Y=5 and \rho=0.6. The first two parameters are the mean and standard deviation of the marginal distribution of X. The next two parameters are the mean and standard deviation of the marginal distribution of Y. The parameter \rho is the correlation coefficient of X and Y. Both marginal distributions are normal.

Let’s focus on the conditional distribution of Y given X=x. It is normally distributed. Its mean and variance are:

    \displaystyle \begin{aligned} E[Y \lvert X=x]&=\mu_Y+\rho \ \frac{\sigma_Y}{\sigma_X} \ (x-\mu_X) \\&=20+0.6 \ \frac{5}{10} \ (x-10) \\&=20+0.3 \ (x-10) \\&=17+0.3 \ x  \end{aligned}

    \displaystyle \sigma_{Y \lvert x}^2=Var[Y \lvert X=x]=\sigma_Y^2 (1-\rho^2)=25 \ (1-0.6^2)=16

    \displaystyle \sigma_{Y \lvert x}=4

The line y=17+0.3 \ x is also called the least squares regression line. It gives the mean of the conditional distribution of Y given x. Because X and Y are positively correlated, the least squares line has positive slope. In this case, the larger the x, the larger is the mean of Y. The standard deviation of Y given x is constant across all possible x values.

With mean and standard deviation known, we can now compute normal probabilities. Suppose the realized value of X is 25. Then the mean of Y \lvert 25 is E[Y \lvert 25]=24.5. The standard deviation, as indicated above, is 4. In fact, for any other x, the standard deviation of Y \lvert x is also 4. Now calculate the probability P[20<Y<30 \lvert X=25]. We first calculate it using a normal table found here.

    \displaystyle \begin{aligned} P[20<Y<30 \lvert X=25]&=P\bigg[\frac{20-24.5}{4}<Z<\frac{30-24.5}{4} \biggr] \\&=P[-1.13<Z<1.38] \\&=0.9162-(1-0.8708) \\&=0.7870  \end{aligned}

Using a TI84+ calculator, P[20<Y<30 \lvert X=25]=0.7851396569. In contrast, the probability P[20<Y<30] is (using the table found here):

    \displaystyle \begin{aligned} P[20<Y<30]&=P\bigg[\frac{20-20}{5}<Z<\frac{30-20}{4} \biggr] \\&=P[0<Z<2] \\&=0.9772-0.5 \\&=0.4772  \end{aligned}

Using a TI84+ calculator, P[20<Y<30]=0.4772499375. Note that P[20<Y<30] is for the marginal distribution of Y. It is not conditioned on any realized value of X.

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Practice Problem Set 4 – Correlation Coefficient

This post provides practice problems to reinforce the concept of correlation coefficient discussed in this
post in a companion blog. The post in the companion blog shows how to evaluate the covariance \text{Cov}(X,Y) and the correlation coefficient \rho of two continuous random variables X and Y. It also discusses the connection between \rho and the regression curve E[Y \lvert X=x] and the least squares regression line.

The structure of the practice problems found here is quite simple. Given a joint density function for a pair of random variables X and Y (with an appropriate region in the xy-plane as support), determine the following four pieces of information.

  • The covariance \text{Cov}(X,Y)
  • The correlation coefficient \rho
  • The regression curve E[Y \lvert X=x]
  • The least squares regression line y=a+b x

The least squares regression line y=a+bx whose slope b and y-intercept a are given by:

    \displaystyle b=\rho \ \frac{\sigma_Y}{\sigma_X}

    \displaystyle a=\mu_Y-b \ \mu_X

where \mu_X=E[X], \sigma_X^2=Var[X], \mu_Y=E[Y] and \sigma_Y^2=Var[Y].

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For some of the problems, the regression curves E[Y \lvert X=x] coincide with the least squares regression lines. When the regression curve is in a linear form, it coincides with the least squares regression line.

As mentioned, the practice problems are to reinforce the concepts discussed in this post.

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Practice Problem 4-A
    \displaystyle f(x,y)=\frac{3}{4} \ (2-y) \ \ \ \ \ \ \ 0<x<y<2

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Practice Problem 4-B
    \displaystyle f(x,y)=\frac{1}{2} \ \ \ \ \ \ \ \ \ \ \ \ 0<x<y<2

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Practice Problem 4-C
    \displaystyle f(x,y)=\frac{1}{8} \ (x+y) \ \ \ \ \ \ \ \ \ 0<x<2, \ 0<y<2

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Practice Problem 4-D
    \displaystyle f(x,y)=\frac{1}{2 \ x^2} \ \ \ \  \ \ \ \ \ \ \ \ \ 0<x<2, \ 0<y<x^2

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Practice Problem 4-E
    \displaystyle f(x,y)=\frac{1}{2} \ (x+y) \ e^{-x-y} \ \ \ \  \ \ \ \ \ \ \ \ \ x>0, \ y>0

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Practice Problem 4-F
    \displaystyle f(x,y)=\frac{3}{8} \ x \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 0<y<x<2

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Practice Problem 4-G
    \displaystyle f(x,y)=\frac{1}{2} \ xy \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 0<y<x<2

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Practice Problem 4-H
    \displaystyle f(x,y)=\frac{3}{14} \ (xy +x) \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 0<y<x<2

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Practice Problem 4-I
    \displaystyle f(x,y)=\frac{3}{32} \ (x+y) \ xy \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ 0<x<2, \ 0<y<2

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Practice Problem 4-J
    \displaystyle f(x,y)=\frac{3y}{(x+1)^6} \ \ e^{-y/(x+1)} \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ x>0, \ y>0

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Practice Problem 4-K
    \displaystyle f(x,y)=\frac{y}{(x+1)^4} \ \ e^{-y/(x+1)} \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ x>0, \ y>0

For this problem, only work on the regression curve E[Y \lvert X=x]. Note that E[X] and Var[X] do not exist.

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Problem ………..Answer
4-A
  • \displaystyle \text{Cov}(X,Y)=\frac{1}{10}
  • \displaystyle \rho=\sqrt{\frac{1}{3}}=0.57735
  • \displaystyle E[Y \lvert X=x]=\frac{2 (4-3 x^2+x^3)}{3 (4- 4x+x^2)}=\frac{2 (2+x-x^2)}{3 (2-x)} \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{2}{3} \ (x+1)
4-B
  • \displaystyle \text{Cov}(X,Y)=\frac{1}{9}
  • \displaystyle \rho=\frac{1}{2}
  • \displaystyle E[Y \lvert X=x]=1+\frac{1}{2} x \ \ \ \ \ 0<x<2
  • \displaystyle y=1+\frac{1}{2} x
4-C
  • \displaystyle \text{Cov}(X,Y)=-\frac{1}{36}
  • \displaystyle \rho=-\frac{1}{11}
  • \displaystyle E[Y \lvert X=x]=\frac{x+\frac{4}{3}}{x+1} \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{14}{11}-\frac{1}{11} x
4-D
  • \displaystyle \text{Cov}(X,Y)=\frac{1}{3}
  • \displaystyle \rho=\frac{1}{2} \ \sqrt{\frac{15}{7}}=0.7319
  • \displaystyle E[Y \lvert X=x]=\frac{x^2}{2} \ \ \ \ \ 0<x<2
  • \displaystyle y=-\frac{1}{3}+ x
4-E
  • \displaystyle \text{Cov}(X,Y)=-\frac{1}{4}
  • \displaystyle \rho=-\frac{1}{7}=-0.1429
  • \displaystyle E[Y \lvert X=x]=\frac{x+2}{x+1} \ \ \ \ \ x>0
  • \displaystyle y=\frac{12}{7}-\frac{1}{7} x
4-F
  • \displaystyle \text{Cov}(X,Y)=-\frac{3}{40}
  • \displaystyle \rho=\frac{3}{\sqrt{19}}=0.3974
  • \displaystyle E[Y \lvert X=x]=\frac{x}{2} \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{x}{2}
4-G
  • \displaystyle \text{Cov}(X,Y)=\frac{16}{225}
  • \displaystyle \rho=\frac{4}{\sqrt{66}}=0.4924
  • \displaystyle E[Y \lvert X=x]=\frac{2}{3} x \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{2}{3} x
4-H
  • \displaystyle \text{Cov}(X,Y)=\frac{298}{3675}
  • \displaystyle \rho=\frac{149}{3 \sqrt{12259}}=0.4486
  • \displaystyle E[Y \lvert X=x]=\frac{x (2x+3)}{3x+6}  \ \ \ \ \ 0<x<2
  • \displaystyle y=-\frac{2}{41}+\frac{149}{246} x
4-I
  • \displaystyle \text{Cov}(X,Y)=-\frac{1}{144}
  • \displaystyle \rho=-\frac{5}{139}=-0.03597
  • \displaystyle E[Y \lvert X=x]=\frac{4x+6}{3x+4}  \ \ \ \ \ 0<x<2
  • \displaystyle y=\frac{204}{139}-\frac{5}{139} x
4-J
  • \displaystyle \text{Cov}(X,Y)=\frac{3}{2}
  • \displaystyle \rho=\frac{1}{\sqrt{3}}=0.57735
  • \displaystyle E[Y \lvert X=x]=2 (x+1) \ \ \ \ \ x>0
  • \displaystyle y=2 (x+1)
4-K
  • \displaystyle E[Y \lvert X=x]=2 (x+1) \ \ \ \ \ x>0

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Practice Problem Set 3 – The Big 3 Discrete Distributions

This post presents exercises on the big 3 discrete distributions – binomial, Poisson and negative binomial, reinforcing the concepts discussed in several blog posts (here and here).

A previous problem set on Poisson and gamma is found here.

A previous problem set on Poisson distribution is found here.

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Practice Problem 3-A

The amount of damage from an auto collision accident is modeled by an exponential distribution with mean 5. Ten unrelated auto collision claims are examined by an insurance adjuster.

  • What is the probability that five of the claims will have damages exceeding the mean damage amount?
  • What is the probability that at most two of the claims will have damages exceeding the mean damage amount?
  • What is the expected number of claims with damages exceeding the mean?

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Practice Problem 3-B

The jackpot of the Powerball lottery can sometimes be in the hundreds of millions dollars. The odds of winning the jackpot are one in 292 million. However, there are prizes other than the jackpot (some of the lesser prizes are $100 and $7). The odds of winning a prize in Powerball are one in 24.87. A Powerball player buys one ticket every month for a year.

  • What is the probability of winning at least one prize?
  • What is the probability of winning at least two prizes?
  • What is the probability of winning at least three prizes?
  • What is the probability of winning at least four prizes?

See here for the calculation of Powerball winning odds.

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Practice Problem 3-C

According to a poll conducted by AAA, 94% of teen drivers acknowledge the dangers of texting and driving but 35% admitted to doing it anyway. In a random sample of 20 teen drivers,

  • what is the probability that exactly five of the teen drivers do texting while driving?
  • what is the probability that more than five of the teen drivers do texting while driving?

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Practice Problem 3-D

According to aviation statistics in the commercial airline industry, approximately one in 225 bags or luggage that are checked is lost. A business executive will be flying frequently next year and will be checking 100 bags or luggage during that one year.

  • Determine the probability that the business executive will not lose any bags or luggage during his travel.
  • Determine the probability that the business executive will lose one or two bags or luggage during his travel.

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Practice Problem 3-E

A large group of insured drivers are classified as high risk and low risk. About 10% of the drivers in this group are considered high risk while the remaining 90% are considered low risk drivers. The number of auto accidents in a year for a high risk driver in this group is modeled by a binomial distribution with mean 0.8 and variance 0.64. The number of auto accidents in a year for a low risk driver is modeled by a binomial distribution with mean 0.4 and variance 0.36. Suppose that an insured driver is randomly selected from this group.

  • What is the probability that the randomly selected insured driver will have no auto accident in the next policy year?
  • What is the probability that the randomly selected insured driver will have more than 1 auto accident in the next policy year?
  • What is the variance of the number of auto accidents for the randomly selected insured drivers in the next policy year?

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Practice Problem 3-F
The number of TV sets of a particular brand sold in a given week at an electronic store has a Poisson distribution with mean 4.

  • Determine the probability that the store will sell more than 4 TV sets next week.
  • Determine the minimum number of TV sets that the manager should order for the next week so that the probability of having more sales than available TV sets is less than 0.10.

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Practice Problem 3-G

The number of vacant rooms in a given night in a certain hotel follows a Poisson distribution with mean 1.75. Three travelers without reservation walk into the hotel one night. Assume that they do not know each other.

  • Determine the probability that rooms are available for all three travelers.
  • Given that rooms are available for all three travelers, determine the probability that the hotel will still be able to accommodate three more travelers without reservation who also do not know each other.

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Practice Problem 3-H

Cars running the red light arrive at a busy intersection according to a Poisson process with the rate of 0.5 per hour.

  • What is the probability that there will be at most 4 cars running the red light in a 5-hour period?
  • After a period of having no activities in running red light, what is the probability that it will take more than 90 minutes to see another car running the red light?
  • After a period of having no activities in running red light, what is the probability that it will take more than 90 minutes to see two cars running the red light?

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Practice Problem 3-I
Consider a roulette wheel consisting of 38 numbers – 1 through 36, and 0 and 00. A player always makes bets on one of the numbers 1 through 12.

  • Determine the probability that the player will lose his first 5 bets.
  • Determine the probability that the first win of the player will occur on the 5th bet.
  • Determine the probability that the first win of the player will occur no later than the 5th bet.

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Practice Problem 3-J
Suppose that roughly 10% of the adult population have type II diabetes. A researcher wishes to find 3 adult patients who are diabetic. Suppose that the researcher evaluate one patient at a time until finding three diabetic patients.

  • What is the probability that the third diabetic patient is found after evaluating 10 or 11 patients?

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Practice Problem 3-K

For any high risk insured driver, the number of auto accidents in a year has a negative binomial distribution with mean 1.6 and variance 2.88. One such insured driver is selected at random and observed for one year. What is the probability that the insured driver will have more than one accident?

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Practice Problem 3-L

A discrete probability distribution has the following probability function.

    \displaystyle P(X=k)=\frac{(k+1) (k+2)}{2} \ \biggl(\frac{4}{9} \biggr)^3 \ \biggl(\frac{5}{9} \biggr)^k \ \ \ \ \ k=0,1,2,3,\cdots

Determine the mean and variance of X.

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Practice Problem 3-M

A large pool of insureds is made up of two subgroups – low risk (75% of the pool) and high risk (25% of the pool). The number of claims in a year for each insured can be any non-negative integer 0, 1, 2, 3, … The number of claims in a year for each insured in the low risk group has a negative binomial distribution with mean 0.5 and variance 0.625. The number of claims in a year for each insured in the high risk group has a negative binomial distribution with mean 0.75 and variance 0.9375.

If a randomly selected insured from the pool is observed to have one claim in a given year, what is the probability that the insured is a high risk insured?

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Practice Problem 3-N

An American roulette wheel has 38 areas – numbers 1 through 36 and 0 and 00. A player bets on odd numbers (1, 3, 5, 7, …, 35). He leaves the game when he wins 5 bets.

  • What is the expected number of bets the player will lose before winning 5 bets?
  • What is the probability that the player will lose 5 bets before leaving the game?

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Problem ………..Answer
3-A
  • 0.171367
  • 0.2247123
  • 10 e^{-1}=3.67879
3-B
  • 0.388889698
  • 0.081670443
  • 0.010882596
  • 0.000997406
3-C
  • 0.127199186
  • 0.754604255
3-D
  • 0.640545556
  • 0.348149413
3-E
  • 0.43425
  • 0.16795
  • 0.6264
3-F
  • \displaystyle 1-\frac{103}{3} e^{-4}=0.371163065
  • min is 7 since P(X>6)=0.11 and P(X>7)=0.0511
3-G
  • \displaystyle 1-4.28125 e^{-1.75}=0.256030305
  • 0.035673762
3-H
  • 0.891178019
  • \displaystyle e^{-0.75}=0.472366553
  • \displaystyle 1.75 e^{-0.75}=0.826641467
3-I
  • \displaystyle (13/19)^5=0.1499507895
  • \displaystyle (6/19) (13/19)^4=0.0692
  • \displaystyle 1-(13/19)^5=0.85
3-J
  • 0.036589713
3-K
  • \displaystyle \frac{304}{729}=0.417
3-L
  • 3.75
  • 8.4375
3-M
  • \displaystyle \frac{0.0768}{0.2688}=0.2857
3-M
  • \displaystyle \frac{50}{9}=5.56
  • 0.1213520403

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