Show $X$~Exponential distribution given $P(X > nx) = (P( X > x ))^n$

Question

Show $X$~Exponential distribution given $P(X > nx) = (P( X > x ))^n$

Can someone give some hint ? The only thing I can think of is to write it as CDF format and say according to observation there must be an exponential term...

Many thanks!

Answer 1

This is a well-known property that only the exponential function can satisfy. To show it, you may want to proceed as follows.

1. Suppose $F:\mathbb{R}\to \mathbb{R}_+$ is a continuous real function such that $F(nx)=F(x)^n$ for all $n\in \mathbb{N}_+$. Let $\alpha = F(1)$, and define $G(x) = \log_\alpha F(x)$. We want to determine what $G(x)$ is.
2. Show that if $x\in \mathbb{N}_+\cup\{0\}$, then $G(x) = x$.
3. Let $x = p/q\in \mathbb{Q}_+$ with $p\geq 0$ and $q> 0$, be a rational number. Using the previous part, prove that $G(x)=x$.
4. Let $x\in \mathbb{R}_+$. We know that given any real number there exists a sequences of rational numbers convergent to $x$ (whether $x\in \mathbb{Q}$, or not). Without loss of generality we can assume all of the elements of this sequence are positive. Use continuity of $G$ to prove $G(x) = x$.

This should show you that $F(x) = \alpha^x = \exp (x\ln F(1))$ when $x\geq 0$. Similar arguments show that $G(x) = \exp(-x\log F(-1))$ when $x\leq 0$. Now if furthermore $F(x)$ is differentiable, then $F'(0)=\log F(1) = - \log F(-1)$, so that $F(x) = \exp(x\log F(1))$ throughout.

In your case, $F(x) = P(X>x)$ is indeed differentiable. The rest should be easy.