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Show that if X is a geometric random variable with parameter p, then E[1/X]=-plog(p)1-pHint: You will need to evaluate an expression of the form∑i=1∞ai/i

to do so, writeai/i=∫0axi-1dx

and then interchange the sum and the integral.

Short Answer

Expert verified

In the given information the answer isE(1/X)=p1-p·-log(p)

Step by step solution

01

:Given Information

We have that X~Geom(p)by the theorem about the mean of a function of random variable, we have that

E(1/X)=∑k=1∞1k·P(X=k)=∑k=1∞1k(1-p)k-1p

=p1-p∑k=1∞(1-p)kk

02

Calculation

∑k=1∞(1-p)kk=∑k=1∞∫01-pxk-1dx=∫01-p∑k=1∞xk-1dx=∫01-p∑k=0∞xkdx

This sum and integral can be written as

∫01-p∑k=0∞xkdx=∫01-p11-xdx

making substitution y=1-xwe get that

∫01-p11-xdx=∫1p1y(-dy)=∫p11ydy=log(1)-log(p)=-log(p)

plug these calculations back in (2) and we get that

E(1/X)=p1-p·-log(p)

03

Final Answer

The answer is E(1/X)=p1-p·-log(p)

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