Question

Show that if X is a geometric random variable with parameter p, then $E[1/X]=\frac{-p\log \left(p\right)}{1-p}$Hint: You will need to evaluate an expression of the form$\sum _{i=1}^{\infty }{a}^i/i$

to do so, write$a^i/i={\int }_0^a{x}^{i-1}dx$

and then interchange the sum and the integral.

Short answer

In the given information the answer is$E(1/X)=\frac{p}{1-p}·-\log \left(p\right)$

Step-by-step solution

:Given Information

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

$E(1/X)=\sum _{k=1}^{\infty }\frac{1}{k}·P(X=k)=\sum _{k=1}^{\infty }\frac{1}{k}(1-p{)}^{k-1}p$

$=\frac{p}{1-p}\sum _{k=1}^{\infty }\frac{(1-p{)}^k}{k}$

Calculation

$\sum _{k=1}^{\infty }\frac{(1-p{)}^k}{k}=\sum _{k=1}^{\infty }{\int }_0^{1-p}{x}^{k-1}dx={\int }_0^{1-p}\left(\sum _{k=1}^{\infty }{x}^{k-1}\right)dx={\int }_0^{1-p}\left(\sum _{k=0}^{\infty }{x}^k\right)dx$

This sum and integral can be written as

${\int }_0^{1-p}\left(\sum _{k=0}^{\infty }{x}^k\right)dx={\int }_0^{1-p}\frac{1}{1-x}dx$

making substitution $y=1-x$we get that

${\int }_0^{1-p}\frac{1}{1-x}dx={\int }_1^p\frac{1}{y}(-dy)={\int }_p^1\frac{1}{y}dy=\log \left(1\right)-\log \left(p\right)=-\log \left(p\right)$

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

$E(1/X)=\frac{p}{1-p}·-\log \left(p\right)$

Final Answer

The answer is $E(1/X)=\frac{p}{1-p}·-\log \left(p\right)$