Monthly Archives: June 2013

Practice Problems for Conditional Distributions, Part 2

The following are practice problems on conditional distributions. The thought process of how to work with these practice problems can be found in the blog post Conditionals Distribution, Part 2.

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Practice Problems

Practice Problem 1

Suppose that X is the lifetime (in years) of a brand new machine of a certain type. The following is the density function.

    \displaystyle f(x)=\frac{1}{8 \sqrt{x}}, \ \ \ \ \ \ \ \ \ 1<x<25

You just purchase a 9-year old machine of this type that is in good working condition. Compute the following:

  • What is the expected lifetime of this 9-year old machine?
  • What is the expected remaining life of this 9-year old machine?

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Practice Problem 2

Suppose that X is the total amount of damages (in millions of dollars) resulting from the occurrence of a severe wind storm in a certain city. The following is the density function of X.

    \displaystyle f(x)=\frac{81}{(x+3)^4}, \ \ \ \ \ \ \ \ \ 0<x<\infty

Suppose that the next storm is expected to cause damages exceeding one million dollars. Compute the following:

  • What is the expected total amount of damages for the next storm given that it will exceeds one million dollars?
  • The city has a reserve fund of one million dollars to cover the damages from the next storm. Given the amount of damages for the next storm will exceeds one million dollars, what is the expected total amount of damages in excess of the amount in the reserve fund?

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Answers

The thought process of how to work with these practice problems can be found in the blog post Conditionals Distribution, Part 2.

Practice Problem 1

    \displaystyle E(X \lvert X>9)=\frac{49}{3}=16.33 \text{ years}

    \displaystyle E(X-9 \lvert X>9)=\frac{22}{3}=7.33 \text{ years}

Practice Problem 2

    \displaystyle E(X \lvert X>1)=3 \text{ millions}

    \displaystyle E(X-1 \lvert X>1)=2 \text{ millions}

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\copyright \ 2013 \text{ by Dan Ma}

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Practice Problems for Conditional Distributions, Part 1

The following are practice problems on conditional distributions. The thought process of how to work with these practice problems can be found in the blog post Conditionals Distribution, Part 1.

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Description of Problems

Suppose X and Y are independent binomial distributions with the following parameters.

    For X, number of trials n=5, success probability \displaystyle p=\frac{1}{2}

    For Y, number of trials n=5, success probability \displaystyle p=\frac{3}{4}

We can think of these random variables as the results of two students taking a multiple choice test with 5 questions. For example, let X be the number of correct answers for one student and Y be the number of correct answers for the other student. For the practice problems below, passing the test means having 3 or more correct answers.

Suppose we have some new information about the results of the test. The problems below are to derive the conditional distributions of X or Y based on the new information and to compare the conditional distributions with the unconditional distributions.

Practice Problem 1

  • New information: X<Y.
  • Derive the conditional distribution for X \lvert X<Y.
  • Derive the conditional distribution for Y \lvert X<Y.
  • Compare these conditional distributions with the unconditional ones with respect to mean and probability of passing.
  • What is the effect of the new information on the test performance of each of the students?
  • Explain why the new information has the effect on the test performance?

Practice Problem 2

  • New information: X>Y.
  • Derive the conditional distribution for X \lvert X>Y.
  • Derive the conditional distribution for Y \lvert X>Y.
  • Compare these conditional distributions with the unconditional ones with respect to mean and probability of passing.
  • What is the effect of the new information on the test performance of each of the students?
  • Explain why the new information has the effect on the test performance?

Practice Problem 3

  • New information: Y=X+1.
  • Derive the conditional distribution for X \lvert Y=X+1.
  • Derive the conditional distribution for Y \lvert Y=X+1.
  • Compare these conditional distributions with the unconditional ones with respect to mean and probability of passing.
  • What is the effect of the new information on the test performance of each of the students?
  • Explain why the new information has the effect on the test performance?

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Partial Answers

To let you know that you are on the right track, the conditional distributions are given below.

The thought process of how to work with these practice problems can be found in the blog post Conditional Distributions, Part 1.

Practice Problem 1

    \displaystyle P(X=0 \lvert X<Y)=\frac{1023}{22938}=0.0446

    \displaystyle P(X=1 \lvert X<Y)=\frac{5040}{22938}=0.2197

    \displaystyle P(X=2 \lvert X<Y)=\frac{9180}{22938}=0.4

    \displaystyle P(X=3 \lvert X<Y)=\frac{6480}{22938}=0.2825

    \displaystyle P(X=4 \lvert X<Y)=\frac{1215}{22938}=0.053

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    \displaystyle P(Y=1 \lvert X<Y)=\frac{10}{22933}=0.0004

    \displaystyle P(Y=2 \lvert X<Y)=\frac{540}{22933}=0.0235

    \displaystyle P(Y=3 \lvert X<Y)=\frac{4320}{22933}=0.188

    \displaystyle P(Y=4 \lvert X<Y)=\frac{10530}{22933}=0.459

    \displaystyle P(Y=5 \lvert X<Y)=\frac{7533}{22933}=0.328

Practice Problem 2

    \displaystyle P(X=1 \lvert X>Y)=\frac{5}{3386}=0.0013

    \displaystyle P(X=2 \lvert X>Y)=\frac{160}{3386}=0.04

    \displaystyle P(X=3 \lvert X>Y)=\frac{1060}{3386}=0.2728

    \displaystyle P(X=4 \lvert X>Y)=\frac{1880}{3386}=0.4838

    \displaystyle P(X=5 \lvert X>Y)=\frac{781}{3386}=0.2

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    \displaystyle P(Y=0 \lvert X>Y)=\frac{31}{3386}=0.008

    \displaystyle P(Y=1 \lvert X>Y)=\frac{390}{3386}=0.1

    \displaystyle P(Y=2 \lvert X>Y)=\frac{1440}{3386}=0.37

    \displaystyle P(Y=3 \lvert X>Y)=\frac{1620}{3386}=0.417

    \displaystyle P(Y=4 \lvert X>Y)=\frac{405}{3386}=0.104

Practice Problem 3

    \displaystyle P(X=0 \lvert Y=X+1)=\frac{15}{8430}=0.002

    \displaystyle P(X=1 \lvert Y=X+1)=\frac{450}{8430}=0.053

    \displaystyle P(X=2 \lvert Y=X+1)=\frac{2700}{8430}=0.32

    \displaystyle P(X=3 \lvert Y=X+1)=\frac{4050}{8430}=0.48

    \displaystyle P(X=4 \lvert Y=X+1)=\frac{1215}{8430}=0.144

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    \displaystyle P(Y=1 \lvert Y=X+1)=\frac{15}{8430}=0.002

    \displaystyle P(Y=2 \lvert Y=X+1)=\frac{450}{8430}=0.053

    \displaystyle P(Y=3 \lvert Y=X+1)=\frac{2700}{8430}=0.32

    \displaystyle P(Y=4 \lvert Y=X+1)=\frac{4050}{8430}=0.48

    \displaystyle P(Y=5 \lvert Y=X+1)=\frac{1215}{8430}=0.144

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\copyright \ 2013 \text{ by Dan Ma}