Q:

Uniqueness of distribution and Laplace transform

Let $X$ and $Y$ be positive random variable. 
Show that the following are equivalent:
a) $X$ and $Y$ have the same distribution
b)$Ee^{-rX}=Ee^{-rY}$ for every $r$ in $R^+$.
Could anyone kindly help me with this?
I tried to use the definition, but I don't know how to connect these two.
Thanks!

A:

I'm going to put this in more analytic terms than probabilistic terms; there's one bit I don't quite see how to state probabilistically. Requires a little background in a few areas:
Consider the distributions of $X$ and $Y$. We have two probability measures $\mu$ and $\nu$ on $[0,\infty)$, and $$\int_{[0,\infty)}e^{-rt}\,d\mu(t) =
\int_{[0,\infty)}e^{-rt}\,d\nu(t)\quad(r>0).$$
For complex numbers $z=r+is$ with $r>0$ define $$f(r+is)=
\int_{[0,\infty)}e^{-(r+is)t}\,d\mu(t) -
\int_{[0,\infty)}e^{-(r+is)t}\,d\nu(t).$$The function $f$ is holomorphic (analytic) in the right half-plane; this follows from Fubini's Theorem plus Morera's Theorem, or in various other ways. We have $f(r)=0$ for $r>0$; now a basic uniqueness result for holomorphic functions shows that $f=0$ everywhere in the right half-plane. In particular $$\int_{[0,\infty)}e^{-(1+is)t}\,d\mu(t) =
\int_{[0,\infty)}e^{-(1+is)t}\,d\nu(t)\quad(s\in\mathbb R).$$
Define a (finite) real-valued measure $\lambda$ by $$\int_{[0,1)}\phi(t)\,d\lambda(t)=\int_{[0,\infty)}\phi(t)e^{-t}\,d\mu(t)
-\int_{[0,\infty)}\phi(t)e^{-t}\,d\nu(t).$$The previous display says exactly that $$\hat\lambda=0,$$where $\hat\lambda$ is the Fourier Transform. So uniqueness for the Fourier transform (for complex measures) shows that $\lambda=0$, hence $\mu=\nu$.

