Suppose I know that $(\mathbb{E} X^k)^{1/k} \le \alpha_k$ and $X \ge 0$ a.s. and I want to get a an upper bound on $\mathbb{E}e^{tX}$. There are a number of ways to lower bound $\mathbb{E}e^{tX}$(i.e. MGF), but nothing about about generic upper bounds.
What I've tried:
- I tried Jensen's inequality (idea of tangent of a function), but the derivative of exponent is again exponent, so it didn't seem to workout (I'll add details later on).
- Taylor expansion for exponent, but I need in a non-asymptotic setup, so I don't know how to proceed.
No, there is no such bound possible. Indeed, I will give an example of a random variable such that all moments are finite, but the exponential moment is infinite.
Let $X$ be the random variable satisfying $\mathbb P(X\geq x)=e^{-\sqrt{x}}$ for all $x\geq 0$. Then $$ \mathbb EX^k=\int_0^{\infty}kx^{k-1}\mathbb P(X\geq x)\ dx=k\int_0^{\infty}x^{k-1}e^{-\sqrt{x}}\ dx<\infty,\qquad \forall k\geq 1, $$ whereas $$ \mathbb Ee^{tX}=\int_0^{\infty}te^{tx}\mathbb P(X\geq x)\ dx=t\int_0^{\infty}e^{tx-\sqrt{x}}\ dx=\infty,\qquad \forall t>0, $$ since the integrand blows up to $\infty$ as $x\to\infty$.
This is part of a general philosophy that you can't bound higher moments in terms of lower moments, you can't bound exponential moment in terms of polynomial moments, basically a faster growing function can't be bounded by a slower growing function.