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Show how to compute Cov(X,Y) from the joint moment generating function ofX and Y.

Short Answer

Expert verified

The ComputeCov(X,Y)from the joint moment generating function value areCov(X,Y)=∂2∂t1∂t2MX,Y-∂∂t1MX,Y∂∂t2MX,Y(0,0).

Step by step solution

01

Given Information

The joint moment generating function of XandY.

02

Given Information

We have that the joint moment generating function of Xand Yis,

MX,Yt1,t2=Eet1X+t2Y

Observe that,

∂∂t1MX,Yt1,t2=EXet1X+t2Y

Which implies

E(X)=∂∂t1MX,Y(0,0)and

E(Y)=∂∂t2MX,Y(0,0).

03

Explanation

Now, consider what happens if we partially differentiateMX,Yt1,t2respective to t1and then to t2. We end up with

∂2∂t1∂t2MX,Yt1,t2=EXYet1X+t2Y

Which implies,

∂2∂t1∂t2MX,Y(0,0)=E(XY)

Finally we have that, Cov(X,Y)=E(XY)-E(X)E(Y)

=∂2∂t1∂t2MX,Y-∂∂t1MX,Y∂∂t2MX,Y(0,0).

04

Final answer

TheCov(X,Y) joint moment generating function value areCov(X,Y)=∂2∂t1∂t2MX,Y-∂∂t1MX,Y∂∂t2MX,Y(0,0).

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Most popular questions from this chapter

Let ϕbe the standard normal distribution function, and let X be a normal random variable with mean μ and variance 1. We want to find E[ ϕ(X)]. To do so, let Z be a standard normal random variable that is independent of X, and let

I=1, â¶Ä…â¶Ä…â¶Ä…ifZ<X0, â¶Ä…â¶Ä…â¶Ä…ifZ≥X

(a) Show that E[I∣X=x]=Φ(x).

(b) Show that E[Φ(X)]=P{Z<X}.

(c) Show that E[Φ(X)]=Φμ2.

Hint: What is the distribution of X-Z?

The preceding comes up in statistics. Suppose you are about to observe the value of a random variable X that is normally distributed with an unknown mean μ and variance 1, and suppose that you want to test the hypothesis that the mean μ is greater than or equal to 0. Clearly you would want to reject this hypothesis if X is sufficiently small. If it results that X = x, then the p-value of the hypothesis that the mean is greater than or equal to 0 is defined to be the probability that X would be as small as x if μ were equal to 0 (its smallest possible value if the hypothesis were true). (A small p-value is taken as an indication that the hypothesis is probably false.) Because X has a standard normal distribution when μ = 0, the p-value that results when X = x is ϕ (x). Therefore, the preceding shows that the expected p-value that results when the true mean is μ is φμ2 .

Show that Xis stochastically larger than Yif and only ifE[f(X)]≥E[f(Y)]

for all increasing functions f..

Hint: Show that X≥stY, then E[f(X)]≥E[f(Y)]by showing that f(X)≥stf(Y)and then using Theoretical Exercise 7.7. To show that if E[f(X)]≥E[f(Y)]for all increasing functions f, then P{X>t}≥P{Y>t}, define an appropriate increasing function f.

A deck of n cards numbered 1 through n is thoroughly shuffled so that all possible n! orderings can be assumed to be equally likely. Suppose you are to make n guesses sequentially, where the ith one is a guess of the card in position i. Let N denote the number of correct guesses.

(a) If you are not given any information about your earlier guesses, show that for any strategy, E[N]=1.

(b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that under this strategy

E[N]=1n+1n−1+⋯+1≈∫1n1xdx=logn

(c) Supposethatyouaretoldaftereachguesswhetheryou are right or wrong. In this case, it can be shown that the strategy that maximizes E[N] is one that keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy, show that

E[N]=1+12!+13!+⋯+1n!≈e−1

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Compute EX3∣Y=y.

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