Let's assume that $U$ and $V$ are non-negative random variables. Suppose that
\begin{align}
\sup_{t \ge 0 } \frac{| M_U(-t) – M_V(-t)|}{t} \le \epsilon
\end{align}
where $M_U(t)$ and $M_V(t)$ are moment generating functions.
A few facts:
- Technically $M(-t)$ is known as Laplace transform.
- $M(t)$ unique on an open interval. Therefore, this question is well defined.
- $ t \to M(-t)$ is decreasing.
Question: Does this imply that
\begin{align}
\sup_{t \in \mathbb{R} } \frac{| \phi_U(t) – \phi_V(t)| }{|t|}\le f(\epsilon)
\end{align}
where $\phi_U(t)$ and $\phi_V(t)$ are characteristic functions, and $f$ is some function that goes to zero as $\epsilon \to 0$.
I was thinking of using that $\phi(t)=M(it)$, but this doesn't work out.
Best Answer
No. Consider $U_n=\frac{1}{2}\delta_{n-1}+\frac{1}{2}\delta_{n+1}$, $V_n=\delta_{n}$. Then \begin{align*} \frac{\vert M_{U_n}(-t)-M_{V_n}(-t)\vert}{t}=\frac{\vert \frac{e^{-(n+1)t}}{2}+\frac{e^{-(n-1)t}}{2}-e^{-nt}\vert}{t}=\frac{e^{-(n-1)t}}{2}\frac{(1-e^{-t})^2}{t}. \end{align*} Say there are an infinite sequence of $\{t_{n_k}\}_k$ and an $\epsilon>0$ such that \begin{align*} \frac{e^{-(n_k-1)t_{n_k}}}{2}\frac{(1-e^{-t_{n_k}})^2}{t_{n_k}}>\epsilon. \end{align*} Then we must have $t_{n_k}\rightarrow 0$ as $k\rightarrow\infty$. But \begin{align*} \lim_{k\rightarrow\infty}\frac{e^{-(n_k-1)t_{n_k}}}{2}\frac{(1-e^{-t_{n_k}})^2}{t_{n_k}}\leq\frac{1}{2}\lim_{k\rightarrow\infty}\frac{(1-e^{-t_{n_k}})^2}{t_{n_k}}=0 \end{align*} by a simple application of L'Hopital's Rule.
However, \begin{align*} \frac{\vert \phi_{U_n}(t)-\phi_{V_n}(t)\vert}{\vert t \vert}&=\frac{\sqrt{\left(\frac{\cos((n-1)t)}{2}+\frac{\cos((n+1)t)}{2}-\cos(nt)\right)^2+\left(\frac{\sin((n-1)t)}{2}+\frac{\sin((n+1)t)}{2}-\sin(nt)\right)^2}}{\vert t\vert} \\ &=\frac{2}{\pi} \end{align*} when $t=\pi$. So for any $\epsilon,f$ I have a pair $U_n,V_n$ that satisfy your MGF constraint but not your CF constraint.