Category Archives: Probability

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 1 – Order Statistics

This post presents exercises on order statistics, reinforcing the concepts discussed in two blog posts in a companion blog on mathematical statistics.

The first blog post from the companion blog is an introduction to order statistics. That post presents the probability distributions of the order statistics, both individually and jointly. The second post presents basic examples illustrating how to calculate the order statistics.

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Practice Problem 1-A
A random sample of size 4 is drawn from a population that has a uniform distribution on the interval (0,5). The resulting order statistics are X_{(1)}, X_{(2)}, X_{(3)} and X_{(4)}.

Determine the cumulative distribution function (CDF) of the 3rd order statistic X_{(3)}. Evaluate the probability P(X_{(3)}>2).

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Practice Problem 1-B

As in Problem 1-A, a random sample of size 4 is drawn from a population that has a uniform distribution on the interval (0,5). The resulting order statistics are X_{(1)}, X_{(2)}, X_{(3)} and X_{(4)}.

Evaluate the conditional probability P(X_{(4)}>4 \lvert X_{(3)} >2).

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Practice Problem 1-C

The random sample X_1,X_2,\cdots,X_9 of size 9 is drawn from a population that has a uniform distribution on the interval (0,10).

Evaluate the mean and variance of the 7th order statistic X_{(7)}.

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Practice Problem 1-D

Suppose that the random sample X_1,X_2 of size 2 is drawn from a population that has an exponential distribution with mean 10. Let X_{(1)} be the sample minimum and let X_{(2)} be the sample maximum.

Evaluate the conditional probability P(X_{(1)}<5 \lvert X_{(2)} <10).

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Practice Problem 1-E

Suppose that X_1,X_2,X_3 is a random sample drawn from an exponential distribution with mean 10. The sample median here is the 2nd order statistic X_{(2)}.

Evaluate the probability that the sample median is between 5 and 10.

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Practice Problem 1-F
Suppose that X_1,X_2,X_3 is a random sample drawn from a uniform distribution on the interval (0,2). Let R=X_{(3)}-X_{(1)} be the sample range.

  • Determine the CDF of the sample range R.
  • Evaluate the mean and variance of the sample range R.

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Practice Problem 1-G
As in Problem 1-F, X_1,X_2,X_3 is a random sample drawn from a uniform distribution on the interval (0,2). Let R=X_{(3)}-X_{(1)} be the sample range. The following relationship relates the variance of the sample range R with the variances and covariance of X_{(1)} and X_{(3)}.

    Var(R)=Var(X_{(1)})+Var(X_{(3)})-2 \ Cov(X_{(1)},X_{(3)})

  • Evaluate Cov(X_{(1)},X_{(3)}) using the above relationship.
  • Evaluate the correlation coefficient \rho of X_{(1)} and X_{(3)}.

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Practice Problem 1-H
Suppose that X_1,X_2,X_3 is a random sample drawn from a uniform distribution on the interval (0,5). Let X_{(1)},X_{(2)} and X_{(3)} be the resulting order statistics.

  • Determine the conditional density function of X_{(3)} given that X_{(1)}=x, X_{(2)}=y for all 0<x<y<5.
  • What is the distribution indicated by the conditional density function?
  • Evaluate the condition mean E(X_{(3)} \lvert  X_{(1)}=x, X_{(2)}=y) and the conditional variance Var(X_{(3)} \lvert  X_{(1)}=x, X_{(2)}=y).

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Practice Problem 1-I
Suppose that X_1,X_2,X_3 is a random sample drawn from an exponential distribution with mean \theta. Let X_{(1)},X_{(2)} and X_{(3)} be the resulting order statistics.

  • Determine the conditional density function of X_{(3)} given that X_{(1)}=x, X_{(2)}=y for all 0<x<y<\infty.
  • What is the distribution indicated by the conditional density function?
  • Evaluate the condition mean E(X_{(3)} \lvert  X_{(1)}=x, X_{(2)}=y) and the conditional variance Var(X_{(3)} \lvert  X_{(1)}=x, X_{(2)}=y).

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Practice Problem 1-J

Suppose that X_1,X_2,X_3 is a random sample drawn from a continuous distribution with density function f(x)=2x where 0<x<1. Let the resulting order statistics be X_{(1)}, X_{(2)} and X_{(3)} where X_{(1)} is the sample minimum, X_{(2)} is the sample median and X_{(3)} is the sample maximum.

  • Evaluate the mean and variance of the sample minimum X_{(1)}.
  • Evaluate the mean and variance of the sample maximum X_{(3)}.

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Practice Problem 1-K

As in Problem 1-J, suppose that X_1,X_2,X_3 is a random sample drawn from a continuous distribution with density function f(x)=2x where 0<x<1. Let the resulting order statistics be X_{(1)}, X_{(2)} and X_{(3)} where X_{(1)} is the sample minimum, X_{(2)} is the sample median and X_{(3)} is the sample maximum.

  • Evaluate the covariance between X_{(1)} and X_{(3)}.
  • Evaluate the correlation between X_{(1)} and X_{(3)}.

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Problem ………..Answer
1-A
  • \displaystyle F_{X_{(3)}}(x)=\frac{20}{625} \ x^3-\frac{3}{625} \ x^4
  • \displaystyle \frac{513}{625}=0.8208
1-B
  • \displaystyle \frac{337}{513}=0.65692
1-C
  • E(X_{(7)}=7
  • \displaystyle Var(X_{(7)}=\frac{21}{11}=1.9
1-D
  • \displaystyle P(X_{(2)}<10)=(1-e^{-1})^2
  • \displaystyle P(X_{(1)}<5 \lvert X_{(2)}<10)=\frac{1-3 e^{-1}+2 e^{-1.5}}{(1-e^{-1})^2}=0.8575
1-E
  • \displaystyle 3 \ (e^{-1}-e^{-2})-2 \ (e^{-1.5}-e^{-3})=0.3509
1-F
  • \displaystyle F_R(r)=\frac{3}{4} \ r^2-\frac{1}{4} \ r^3
  • \displaystyle E(R)=1
  • \displaystyle Var(R)=\frac{1}{5}
1-G
  • \displaystyle Cov(X_{(1)},X_{(3)})=\frac{1}{20}
  • \displaystyle \rho=\frac{1}{3}
1-H
  • For 0<x<y<5, \displaystyle f_{X_{(3)} \lvert X_{(1)}=x,X_{(2)}=y}(z \lvert x, y)=\frac{1}{5-y} \ \ \ \ \ y<z<5
  • This is a uniform distribution on (y,5)
  • \displaystyle E(X_{(3)} \lvert X_{(1)}=x,X_{(2)}=y)=\frac{1}{2} \ (y+5)
  • \displaystyle Var(X_{(3)} \lvert X_{(1)}=x,X_{(2)}=y)=\frac{(5-y)^2}{12}
1-I
  • For 0<x<y<\infty, \displaystyle f_{X_{(3)} \lvert X_{(1)}=x,X_{(2)}=y}(z \lvert x, y)=\frac{\frac{1}{\theta} e^{-z/\theta}}{e^{-y/\theta}} \ \ \ \ \ y<z<\infty
  • This is an exponential distribution conditional on z>y
  • \displaystyle E(X_{(3)} \lvert X_{(1)}=x,X_{(2)}=y)=y+\theta
  • \displaystyle Var(X_{(3)} \lvert X_{(1)}=x,X_{(2)}=y)=\theta^2
1-J
  • \displaystyle E(X_{(1)})=\frac{16}{35}
  • \displaystyle Var(X_{(1)})=\frac{201}{4900}
  • \displaystyle E(X_{(3)})=\frac{6}{7}
  • \displaystyle Var(X_{(3)})=\frac{3}{196}
1-K
  • \displaystyle Cov(X_{(1)},X_{(3)})=\frac{2}{245}
  • \displaystyle \rho=\frac{8}{\sqrt{603}}=0.326

Daniel Ma mathematics

Dan Ma math

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Dan Ma probability

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Dan Ma statistics

\copyright 2018 – Dan Ma

Basic exercises for lognormal distribution

This post presents exercises on the lognormal distribution. These exercises are to reinforce the basic properties discussed in this companion blog post.

Additional resources: another discussion of lognormal, a concise summary of lognormal and a problem set on lognormal.

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Exercises

Exercise 1
Let X be a normal random variable with mean 6.5 and standard deviation 0.8. Consider the random variable Y=e^X. what is the probability P(800 \le Y \le 1000)?

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Exercise 2
Suppose Y follows a lognormal distribution with parameters \mu=1 and \sigma=1. Let Y_1=1.25 Y. Determine the following:

  • The probability that Y_1 exceed 1.
  • The 40th percentile of Y_1.
  • The 80th percentile of Y_1.

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Exercise 3
Let Y follows a lognormal distribution with parameters \mu=4 and \sigma=0.9. Compute the mean, second moment, variance, third moment and the fourth moment.

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Exercise 4
Let Y be the same lognormal distribution as in Exercise 3. Use the results in Exercise 3 to compute the coefficient of variation, coefficient of skewness and the kurtosis.

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Exercise 5
Given the following facts about a lognormal distribution:

  • The lower quartile (i.e. 25% percentile) is 1000.
  • The upper quartile (i.e. 75% percentile) is 4000.

Determine the mean and variance of the given lognormal distribution.

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Exercise 6
Suppose that a random variable Y follows a lognormal distribution with mean 149.157 and variance 223.5945. Determine the probability P(Y>150).

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Exercise 7
Suppose that a random variable Y follows a lognormal distribution with mean 1200 and median 1000. Determine the probability P(Y>1300).

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Exercise 8
Customers of a very popular restaurant usually have to wait in line for a table. Suppose that the wait time Y (in minutes) for a table follows a lognormal distribution with parameters \mu=3.5 and \sigma=0.10. Concerned about long wait time, the restaurant owner improves the wait time by expanding the facility and hiring more staff. As a result, the wait time for a table is cut by half. After the restaurant expansion,

  • what is the probability distribution of the wait time for a table?
  • what is the probability that a customer will have to wait more than 20 minutes for a table?

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Answers

Exercise 1

  • 0.1040

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Exercise 2

  • 0.8888
  • 1.4669
  • 7.8707

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Exercise 3

  • E(Y)=e^{4.405}
  • E(Y^2)=e^{9.62}
  • E(Y^3)=e^{15.645}
  • E(Y^4)=e^{22.48}
  • Var(Y)=(e^{0.81}-1) \ e^{8.81}

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Exercise 4

  • \displaystyle \text{CV}= 1.117098
  • \displaystyle \gamma_1= 4.74533
  • \displaystyle \beta_2= 60.41075686

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Exercise 5

  • \displaystyle E(Y)= 3415.391045
  • \displaystyle E(Y^2)= 34017449.61
  • \displaystyle Var(Y)= 22352553.62

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Exercise 6

  • 0.4562

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Exercise 7

  • 0.3336

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Exercise 8

  • longnormal with \mu=3.5+\log(0.5) and \sigma=0.1 where \log is the natural logarithm.
  • 0.0294

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\copyright \ 2015 \text{ by Dan Ma}
Revised October 18, 2018

Calculating the skewness of a probability distribution

This post presents exercises on calculating the moment coefficient of skewness. These exercises are to reinforce the calculation demonstrated in this companion blog post.

For a given random variable X, the Pearson’s moment coefficient of skewness (or the coefficient of skewness) is denoted by \gamma_1 and is defined as follows:

    \displaystyle \begin{aligned} \gamma_1&=\frac{E[ (X-\mu)^3 ]}{\sigma^3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\&=\frac{E(X^3)-3 \mu E(X^2)+3 \mu^2 E(X)-\mu^3}{\sigma^3} \\&=\frac{E(X^3)-3 \mu [E(X^2)+\mu E(X)]-\mu^3}{\sigma^3} \\&=\frac{E(X^3)-3 \mu \sigma^2-\mu^3}{\sigma^3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\&=\frac{E(X^3)-3 \mu \sigma^2-\mu^3}{(\sigma^2)^{\frac{3}{2}}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \end{aligned}

(1) is the definition which is the ratio of the third central moment to the cube of the standard deviation. (2) and (3) are forms that may be easier to calculate. Essentially, if the first three raw moments E(X), E(X^2) and E(X^3) are calculated, then the skewness coefficient can be derived via (3). For a more detailed discussion, see the companion blog post.

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Practice Problems

Practice Problems 1
Let X be a random variable with density function f(x)=10 x^9 where 0<x<1. This is a beta distribution. Calculate the moment coefficient of skewness in two ways. One is to use formula (3) above. The other is to use the following formula for the skewness coefficient for beta distribution.

    \displaystyle \gamma_1=\frac{2(\beta-\alpha) \ \sqrt{\alpha+\beta+1}}{(\alpha+\beta+2) \ \sqrt{\alpha \ \beta}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)

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Practice Problems 2
Calculate the moment coefficient of skewness for Y=X^2 where X is as in Practice Problem 1. It will be helpful to first calculate a formula for the raw moments E(X^k) of X.

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Practice Problems 3
Let X be a random variable with density function f(x)=8 (1-x)^7 where 0<x<1. This is a beta distribution. Calculate the moment coefficient of skewness using (4).

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Practice Problems 4
Suppose that X follows a gamma distribution with PDF f(x)=4 x e^{-2x} where x>0.

  • Show that E(X)=1, E(X^2)=\frac{3}{2} and E(X^3)=3.
  • Use the first three raw moments to calculate the moment coefficient of skewness.

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Practice Problems 5
Calculate the moment coefficient of skewness for Y=X^2 where X is as in Practice Problem 4. It will be helpful to first calculate a formula for the raw moments E(X^k) of X.

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Practice Problems 6
Verify the calculation of \gamma_1 and the associated calculation of Example 6 in this companion blog post.

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Practice Problems 7
Verify the calculation of \gamma_1 and the associated calculation of Example 7 in this companion blog post.

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Practice Problems 8
Verify the calculation of \gamma_1 and the associated calculation of Example 8 in this companion blog post.

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Answers

Practice Problems 1

  • \displaystyle \gamma_1=\frac{-36 \sqrt{3}}{13 \sqrt{10}}=-1.516770159

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Practice Problems 2

  • \displaystyle \gamma_1=\frac{- \sqrt{7}}{\sqrt{5}}=-1.183215957

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Practice Problems 3

  • \displaystyle \gamma_1=\frac{7 \sqrt{10}}{11 \sqrt{2}}=1.422952349

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Practice Problems 4

  • \displaystyle \gamma_1=\sqrt{2}

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Practice Problems 5

  • \displaystyle \gamma_1=\frac{138}{7 \sqrt{21}}=4.302009836

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\copyright \ 2015 \text{ by Dan Ma}

Practice problems for order statistics and multinomial probabilities

This post presents exercises on calculating order statistics using multinomial probabilities. These exercises are to reinforce the calculation demonstrated in this blog post.

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Practice Problems

Practice Problems 1
Draw a random sample X_1,X_2,\cdots,X_{11} of size 11 from the uniform distribution U(0,4). Calculate the following:

  • P(Y_4<2<Y_5<Y_7<4<Y_8)
  • P(Y_4<2<Y_6<Y_7<4<Y_8)

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Practice Problems 2
Draw a random sample X_1,X_2,\cdots,X_7 of size 7 from the uniform distribution U(0,5). Calculate the probability P(Y_4<2<4<Y_7).

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Practice Problems 3
Same setting as in Practice Problem 2. Calculate P(Y_7>4 \ | \ Y_4<2) and P(Y_7>4). Compare the conditional probability with the unconditional probability. Does the answer for P(Y_7>4 \ | \ Y_4<2) make sense in relation to P(Y_7>4)?

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Practice Problems 4
Same setting as in Practice Problem 2. Calculate the following:

  • P(Y_4<2<Y_7<4)
  • P(2<Y_7<4 \ | \ Y_4<2)
  • P(2<Y_7<4)
  • Does the answer for P(2<Y_7<4 \ | \ Y_4<2) make sense in relation to P(2<Y_7<4)?

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Practice Problems 5
Draw a random sample X_1,X_2,\cdots,X_6 of size 6 from the uniform distribution U(0,4). Consider the conditional distribution Y_3 \ | \ Y_5<2. Calculate the following:

  • P(Y_3 \le t \ | \ Y_5<2)
  • f_{Y_3}(t \ | \ Y_5<2)
  • E(Y_3 \ | \ Y_5<2)
  • E(Y_3)

where 0<t<2. Compare E(Y_3) and E(Y_3 \ | \ Y_5<2). Does the answer for the conditional mean make sense?

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Practice Problems 6
Draw a random sample X_1,X_2,\cdots,X_7 of size 7 from the uniform distribution U(0,5). Calculate the following:

  • P(Y_4 > 4 \ | \ Y_2>2)
  • P(Y_4 > 4)
  • Compare the two probabilities. Does the answer for the conditional probability make sense?

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Answers

Practice Problems 1

  • \displaystyle \frac{11550}{177147}
  • \displaystyle \frac{18480}{177147}

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Practice Problems 2

  • \displaystyle \frac{11088}{78125}

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Practice Problems 3

  • \displaystyle P(Y_7>4 \ | \ Y_4<2)=\frac{11088}{22640}
  • \displaystyle P(Y_7>4)=\frac{61741}{78125}

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Practice Problems 4

  • \displaystyle P(Y_4<2<Y_7<4)=\frac{8064}{78125}
  • \displaystyle P(2<Y_7<4 \ | \ Y_4<2)=\frac{8064}{22640}
  • \displaystyle P(2<Y_7<4)=\frac{16256}{78125}

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Practice Problems 5

  • \displaystyle P(Y_3 \le t \ | \ Y_5<2)=\frac{-10t^6+84t^5-300t^4+400t^3}{448}
  • \displaystyle f_{Y_3}(t \ | \ Y_5<2)=\frac{-60t^5+420t^4-1200t^3+1200t^2}{448}
  • \displaystyle E(Y_3 \ | \ Y_5<2)=\frac{55}{49}
  • \displaystyle E(Y_3)=\frac{84}{49}

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Practice Problems 6

  • \displaystyle \frac{3641}{12393}
  • \displaystyle \frac{2605}{78125}

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\copyright \ 2015 \text{ by Dan Ma}