Show that if $X\geq 0$ and $Y\geq 0$ satisfy $\mathbb{E}(e^{-tX})=\mathbb{E}(e^{-tY})$ for every $t>0$ then $X=Y$ in distribution.
If $X$ and $Y$ are continuous random variable, then we can $$f(z)=\int_0^\infty e^{-zx}f_Xdx-\int_0^\infty e^{-zy}f_Ydy$$ and an analytic continuation argument to prove $X=Y$ in distribution in that case. If they are discrete random variable, then we have to find an alternate proof. Any hints/ideas? Thanks
Here's a start for the discrete case-- under the assumption that the discrete $X$ and $Y$ both take values in $\{0,1,2,\ldots, n\}$. I'll leave it to you to generalize a bit further.
Suppose that $X$ and $Y$ have the same mgf for all $t$: $$ \sum_{x=0}^{n} e^{tx} f_{X}(x) = \sum_{y=0}^{n} e^{ty} f_{Y}(y). $$
For simplicity, let $s=e^{t}$ and we will define $c_{x}=f_{X}(x)-f_{Y}(x)$ for $x=0,1,\ldots,n$.
Now $$ \sum_{x=0}^{n} e^{tx} f_{X}(x) - \sum_{y=0}^{n} e^{ty} f_{Y}(y) = 0 $$ $$ \Downarrow $$ $$ \sum_{x=0}^{n} s^{x} f_{X}(x) - \sum_{y=0}^{n} s^{y} f_{Y}(y) = 0 $$ $$ \Downarrow $$ $$ \sum_{x=0}^{n} s^{x} f_{X}(x) - \sum_{x=0}^{n} s^{x} f_{Y}(x) = 0 $$ $$ \Downarrow $$ $$ \sum_{x=0}^{n} s^{x} \left[ f_{X}(x)-f_{Y}(x) \right] = 0 $$ $$ \Downarrow $$ $$ \sum_{x=0}^{n} s^{x} c_{x} =0 \qquad \forall \,\, s>0 $$
The above is simply a polynomial in $s$ with coefficients $c_{0}, c_{1}, \ldots, c_{n}$. The only way it can be zero for all values of $s$ is if $c_{0}=c_{1}=\cdots=c_{n}=0$.
So, we have that $$ 0=c_{x} = f_{X}(x)-f_{Y}(x) \qquad \mbox{for} \,\,\, x=0,1,\ldots,n. $$
Therefore $$ f_{X}(x)=f_{Y}(x) \qquad \mbox{for} \,\,\, x=0,1,\ldots,n. $$
In other words the density functions for $X$ and $Y$ are exactly the same.