Question

Problem 1.33. Let X be an exponential random variable with unit rate Fix two positive numbers x and y. Prove that P(X > x+91X > x) P(X > y). This shows that conditioning the exponential clock on not having rung by time r and then restarting the count at that point gives statistically the same exponential clock! This is called the memoriless property of the exponential distribution. The same holds for the geometric distribution.

Answer 1

Let $X\sim exp(\lambda)$

So, $f(x)=\lambda e^{-\lambda x},F(x)=P(X<x)=1- e^{-\lambda x}$

Now, $P(X>x&plus;y|X>x)=\frac{P(X>x&plus;y\cap X>x)}{P(X>x)}=\frac{P(X>x&plus;y)}{P(X>x)}$

$=\frac{1-[1- e^{-\lambda (x&plus;y)}]}{1-[1- e^{-\lambda x}]}=\frac{e^{-\lambda (x&plus;y)}}{e^{-\lambda x}}=e^{-\lambda y}=1-[1-e^{-\lambda y}]=1-P(X<y)$

$=P(X>y)$