Question

Exercise 5.6. Suppose that X is a random variable which has geometric distribution with parameter p, for some pe (0,1). Compute E[9(X)], where So if t = 0, g(t) = if t +0. 11/t

Answer 1

Assiming that is the geometric distribution counting failures before a first success.

Χε {1, 2, ...)

P(X = 1) = p(1-p-1 forr = 1,2,3,....

$E(g(x))=\sum_{x=0}^{\infty }g(x)P(X=x)=\sum_{k=1}^{\infty }\frac{p(1-p)^{k-1}}{k}$

(d = (1 - p)k-1 k k=1

Let us find the formula for the sum of the series

-1 k k a k k=1 k=1

when 0<a<1. Note that this series indeed converges since it's dominated by a convergent geometric series.

apply the Integration Theorem for power series to an appropriate series (that is a series that produces the series under scrutiny after integrating term by term).

Let's consider the series

$\sum_{k=1}^{\infty }a^{k-1}$

We will think of this series as a power series in the variable a. For a fixed a∈(0,1), term by term integration of this series gives

$\int_{0}^{a}\sum_{k=1}^{\infty }t^{k-1}dt=\sum_{k=1}^{\infty }\int_{0}^{a}t^{k-1}dt=\sum_{k=1}^{\infty }\frac{a^{k}}{k}$   

But, for

$0<t\leq a$, $\sum_{k=1}^{\infty }t^{k-1}=\frac{1}{1-t}$

so the integral on the left hand side of above equation is

$\int_{0}^{a}\frac{1}{1-t}dt=-ln(1-a)$

and so, with a=1-p<1

$E(g(x))=\sum_{k=1}^{\infty }\frac{p(1-p)^{k-1}}{k}=\frac{-pln(p)}{1-p}$