Let $X$ be a non-negative random variable such that $\mathbb{E}[e^{(kX)}]<\infty$ for some $k>0$.
Then we have to prove that $$\lim_{y\to\infty}\mathbb{P}(X\geq y)\cdot e^{(ky)}=0.$$
What I have tried is that If I use the method used to prove Markov's inequality then I get that $$P(X\geq y)\leq \frac{1}{e^{kt}}\mathbb{E}[e^{kX}].$$
So $$ e^{yk}P(X\geq y)\leq M\,\,\forall y\,\,such\,\,that\,\, y<X. $$ Here $M=\mathbb{E}[e^{(kX)}]$ .
So $$\lim_{y\to\infty}e^{yk}P(X\geq y)\leq M.$$ This gives that the limit is finite.
But how do I proceed from here if I have to show that the limit is $0$?
Best Answer
Here is another way to approach the problem:
The assumption gives $$ \mathbf{E} e^{kX} = \int_0^\infty e^{kx }d\mathbf{P}_X(x) < \infty, $$ where $\mathbf{P}_X$ is the measure defined by $\mathbf{P}_X(E) = \mathbf{P}(X \in E)$ for every measurable set.
Then for every $y > 0$ montonicity of the integral yields $$ \mathbf{P}(X \geq y) e^{ky} = \int_y^\infty e^{ky} d\mathbf{P}_X(x) \leq \int_y^\infty e^{kx} d\mathbf{P}_X(x). $$ Since $\int_0^\infty e^{kx }d\mathbf{P}_X(x)$ is finite the RHS tends to zero for $y \rightarrow \infty$. But then the LHS has to converge to zero as well, as it is nonnegative.