$X,Y\geq 0$ are R.V.s s.t., $\mathbb{E}[e^{-tX}] = \mathbb{E}[e^{-tY}]$ for all $t>0$. Show X=Y in distribution Having issues getting started with this one. 
Let $X,Y\geq 0$ be random variables such that $\mathbb{E}[e^{-tX}] = \mathbb{E}[e^{-tY}]$ for all $t>0$. Show X=Y in distribution.
Any hints?
 A: Step 1. We claim the following:

Claim. Let $\varphi \in C([0, 1])$ be such that $\varphi = 0$ near $x = 0$. Then
  $$\Bbb{E}[\varphi(e^{-tX})] = \Bbb{E}[\varphi(e^{-tY})]. \tag{1} $$

This claim indeed implies that $X \stackrel{d}{=} Y$. To see this, for any sufficiently small $\epsilon > 0$ define $\varphi_{\epsilon}$ as a continuous function on $[0, 1]$ whose graph is obtained by joining $(0, 0)$, $(e^{-1}-\epsilon, 0)$, $(e^{-1}, 1)$ and $(1, 1)$ as below:
$\hspace{9em}$
Then for any $t > 0$, $\text{(1)}$ gives
$$ \Bbb{E}[\varphi_{\epsilon}(e^{-X/t})] = \Bbb{E}[\varphi_{\epsilon}(e^{-Y/t})]. $$
Now take $\epsilon \downarrow 0$. Then $\varphi_{\epsilon}(x) \downarrow \mathbf{1}_{[e^{-1}, 1]}(x)$ and by the dominated convergence theorem we have
$$ \Bbb{P}(X \leq t)
= \Bbb{E}[\mathbf{1}_{[e^{-1}, 1]}(e^{-X/t})]
= \Bbb{E}[\mathbf{1}_{[e^{-1}, 1]}(e^{-Y/t})]
= \Bbb{P}(Y \leq t). $$
Since $X$ and $Y$ are non-negative, this proves that $X$ and $Y$ have the same CDF and hence the same distribution.
Step 2. So it remains to prove the claim. Let $\varphi$ be as in the assertion of the claim. Then $x^{-1}\varphi$ is also continuous, so by Stone-Weierstrass theorem there exists a sequence of polynomial $p_n(x)$ that converges uniformly to $x^{-1}\varphi(x)$ on $[0, 1]$. Then $xp_n(x)$ is a polynomial without constant term that converges uniformly to $\varphi(x)$ on $[0, 1]$. Consequently,
$$ \Bbb{E}[\varphi(e^{-tX})]
= \lim_{n\to\infty} \Bbb{E}[e^{-tX}p_n(e^{-tX})]
= \lim_{n\to\infty} \Bbb{E}[e^{-tY}p_n(e^{-tY})]
= \Bbb{E}[\varphi(e^{-tY})] $$
and the claim follows.
A: Consider the function
$$
\{z\in \Bbb {C} \mid Re (z) <0\}\to \Bbb {C}, z \mapsto \Bbb {E}(e^{z X}).
$$
and the analogoulsy defined function for $Y $.
Show that these are holomorphic. By the identity theorem and your assumption, they coincide. By continuity, we also get coincidence for $Re (z)=0$. 
Thus, the Fourier transforms of $X,Y $ coincide and so does there distribution.
