If $X$ is an exponential random variable with mean 1 and let $f(x)$ be a function such that the $k^{th}$ derivative of the function is bounded. Then can I say that the expectation $\mathbb E[f(X)]$ is bounded?
My intuition is that this is true. Since the $k^{th}$ derivative is bounded I think we can bound the function by some polynomial and since the moments of the exponential are bounded, that will imply $\mathbb E[f(X)]$ is bounded. Is this correct?
Yes, it's correct. Applying the fundamental theorem of calculus, we get that $f$ is bounded by a polynomial of degree $k$, say $p_k(x)$. Since $E(X^k) < \infty$ for all $k \in \mathbb{Z}^+$ for an exponentially distributed random variable $X$, it follows that $E(f(X)) \leq E(p_k(X)) < \infty$.