In Example 4f, we showed that the covariance of the multinomial random variables Niand Njis equal to -mPiPjby expressing Niand Njas the sum of indicator variables. We could also have obtained that result by using the formula VarNi+Nj=VarNi+VarNj+2CovNi,Nj

(a) What is the distribution of Ni+Nj?

(b) Use the preceding identity to show thatCovNi,Nj=-mPi,Pj

Short Answer

Expert verified

a) The distribution of Ni+Njis Pi+Pj.

b) It has been shown that the covariance of the multinomial random variables Niand Njis equal to -mPiPj.

Step by step solution

01

Given Information (Part a)

Using formula: VarNi+Nj=VarNi+VarNj+2CovNi,Nj

The covariance of the multinomial random variables localid="1647410996201" Ni+Nj=-mPiPj

Sum of indicator variables=Ni+Nj

The distribution ofNi+Nj=?

02

Explanation (Part a) 

Find the distribution of Ni+Nj

Since the sum of the indicator variables Niand Njfollows a Binomial with parameters mandPi+Pj

03

Final Answer (Part a) 

Therefore, the distribution ofNi+Nj isPi+Pj.

04

Given Information (Part b) 

Using formula: VarNj+Nj=VarNi+VarNj+2CovNi,Nj

The covariance of the multinomial random variables Ni+Nj=-mPiPj

Sum of indicator variables=Ni+Nj

05

Explanation (Part b)  

The total number of independent trails is m

The probability of success for the sum of the indicator variables is Pi+Pj

The probability of failure for the sum of the indicator variables is 1-(Pi+Pj)

The formula for the variance of the binomial distribution is,

V(X)=np(1-p)

06

Explanation 

Find the variance for the sum of the indicator variables,

VarNi+Nj=mPi+Pj1Pi+Pj

=mPi+Pj1PiPj

=mPi+mPj1PiPj

=mPimPi2mPiPj+mPjmPiPjmPj2

role="math" localid="1647526313298" =mPi1Pi+mPj1Pj2mPiPj

=VarNi+VarNj2CovNi,Nj

Therefore, the variance for the sum of the indicator variables is,

VarNi+Nj=VarNi+VarNj2CovNi,Nj(1)

07

Final Answer 

From the equation (1), the covariance term is,

CovNi,Nj=mPiPj

So, the covariance of the multinomial random variablesNi+Nj=Pi+Pj

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Most popular questions from this chapter

A fair die is successively rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a 6 and a 5. Find

(a) E[X];

(b) E[XY=1];

(c) E[XY=5];

The random variables X and Y have a joint density function is given by

f(x,y)={2e2x/x0x<,0yx0otherwise

ComputeCov(X,Y)

Two envelopes, each containing a check, are placed in front of you. You are to choose one of the envelopes, open it, and see the amount of the check. At this point, either you can accept that amount or you can exchange it for the check in the unopened envelope. What should you do? Is it possible to devise a strategy that does better than just accepting the first envelope? Let Aand B, A<B, denote the (unknown) amounts of the checks and note that the strategy that randomly selects an envelope and always accepts its check has an expected return of (A+B)/2. Consider the following strategy: Let F(·)be any strictly increasing (that is, continuous) distribution function. Choose an envelope randomly and open it. If the discovered check has the value x, then accept it with probability F(x)and exchange it with probability 1F(x).

(a) Show that if you employ the latter strategy, then your expected return is greater than (A+B)/2. Hint: Condition on whether the first envelope has the value Aor B. Now consider the strategy that fixes a value x and then accepts the first check if its value is greater than x and exchanges it otherwise. (b) Show that for any x, the expected return under thex-strategy is always at least (A+B)/2and that it is strictly larger than (A+B)/2if xlies between Aand B.

(c) Let X be a continuous random variable on the whole line, and consider the following strategy: Generate the value ofX, and if X=x, then employ the x-strategy of part (b). Show that the expected return under this strategy is greater than (A+B)/2.

The game of Clue involves 6 suspects, 6 weapons, and 9 rooms. One of each is randomly chosen and the object of the game is to guess the chosen three.

(a) How many solutions are possible? In one version of the game, the selection is made and then each of the players is randomly given three of the remaining cards. Let S, W, and R be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player. Also, let X denote the number of solutions that are possible after that player observes his or her three cards.

(b) Express X in terms of S, W, and R.

(c) Find E[X]

For an event A, let IA equal 1 if A occurs and let it equal 0 if A does not occur. For a random variable X, show that E[X|A] = E[XIA] P(A

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