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

pls help

Answer 1

We have the distribution of X here as:

$X \sim Geometric(p)$

As the value of X can never be 0, therefore the function g(X) here is given as:

G(x) = 1/x

P(X = 1) = p
P(X = 2) = (1-p)p
P(X = 3) = (1-p)²p and so on.....

Therefore,
P(G(x) = 1) = p
P(G(x) = 1/2) = (1-p)p
P(G(x) = 1/3) = (1-p)²p and so on...

Therefore the expected value of g(X) here is computed as:

E(G(x)) = P(G(x) = t)

$E(G(x)) = p + \frac{(1-p)p}{2} + \frac{(1-p)^2p}{3} + ... \infty$

$E(G(x)) = p(1+ \frac{(1-p)}{2} + \frac{(1-p)^2}{3} + ... \infty )$

Now we will use the taylor expansion of Ln(1 + x) here which is given as:

$Ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ....\infty$

$Ln(1 - x) = - x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} + ....\infty$

Putting x = (1-p) here, we get:

$Ln(p) = - ((1 -p) + \frac{(1-p)^2}{2} + \frac{(1-p)^3}{3} + \frac{(1-p)^4}{4} + ....\infty )$

$Ln(p) = -(1-p) (1 + \frac{(1-p)}{2} + \frac{(1-p)^2}{3} + \frac{(1-p)^3}{4} + ....\infty )$

$E(g(x)) = \frac{Ln(p)}{p-1}$

This is the required expected value of g(x) here.