4

I am trying to answer the question $F(x)$ is distribution with $F(0)=0$ and $F(x)<1$ for some $x>0$. Show $F(x)$ is the distribution function of an exponential random variable iff $$F(x+y)-F(y)=F(x)\left(1-F(y)\right).$$

I started with the CDF of the exponential function which is $F(x)=1-e^{-kx}$. I then took this function and plugged it into the inequality $F(x+y)-F(y)=F(x)\left(1-F(y)\right)$ and got the let side equal to the right side.

However where I am stuck is I am unsure how to prove the reverse.

Tunk-Fey
  • 24,849
user130614
  • 79
  • 3
    you know the solutions to the functional equation $f(x+y)=f(x)f(y)$ are given by the exponential function. Rearranging your equation combined with the intial conditions should give you the result – aflous Apr 17 '14 at 14:05

1 Answers1

2

If we assume that the random variable has the density function $f$, i.e. assuming that the CDF is differentiable, we can write: $$ \frac{F(x+y)-F(y)}x=\frac{F(x)}{x}\left(1-F(y)\right). $$ Now let $x\to 0$, we get the following differential equation ($F'(0)=\lambda$): $$ F'(y)=\lambda\left(1-F(y)\right)\implies \left(e^{\lambda y}F(y)\right)'=\lambda e^{\lambda y}\implies\\ e^{\lambda y}F(y)=e^{\lambda y}+C\implies F(y)=1+Ce^{-\lambda y}. $$ Using $F(0)=0$, it turns out that $C=-1$.

However differentiability is not part of your assumptions, so you have to go for functional equation type solution, which is mentioned in comments. Choose $G(x)=1-F(x)$ and you get $G(0)=1$: $$ G(x+y)=G(x)G(y), $$ whose solution can be found in a straight forward way (look at comments).

Arash
  • 11,131