# Does proximity of moment generating functions implies proximity of characteristic functions?

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:

1. Technically $$M(-t)$$ is known as Laplace transform.
2. $$M(t)$$ unique on an open interval. Therefore, this question is well defined.
3. $$t \to M(-t)$$ 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.

• Is your first condition ever true for $U\neq V$? It can't be in the case of $U, V$ both discrete positive. – Pepe Silvia Jun 17 '20 at 2:21
• @PepeSilvia Thanks. I modified the answer to fix this. – Lisa Jun 17 '20 at 12:54
• Is there a specific reason for normalizing by $t$ (as opposed to just looking at maximum distance between the functions)? (Probablly one obvious reason I'm missing) – Clement C. Jun 17 '20 at 18:05
• @ClementC. This type of expression shows up when you try control Levy distance with characteristic functions. I would be fine with a result for just a maximum distance. – Lisa Jun 17 '20 at 18:16

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.
• Thanks. So $U_n$ converges to $V_n$ in moment generating function, but not in the characteristic function? Is this right? – Lisa Jun 22 '20 at 23:41
• I wouldn't use those exact words, that one sequence converges to another. But in principle I see what you're saying, and yes, in the case of $U_n,V_n$, as $n\rightarrow\infty$ your MGF supremum goes to 0 but you CF supremum doesn't. – Pepe Silvia Jun 22 '20 at 23:48
• They're often similar-looking algebraically, take for example the Normal or Exponential distributions. But take a discrete distribution and you'll see that you're taking a weighted average of trig functions (for CF) instead of exponentials (for MGF). There are lots of effects to this. First, $|\phi(t)|\leq 1$ for all $t$. Second, see what happens if $\mathbb{P}[X=x]=a>0$ and you change $x$. Make $x$ large positive and you affect $M_X(t)$ a lot for $t>0$ but very little for $t<0$. For $\phi_X$ you're adding an oscillation of period proportional to $\frac{1}{x}$ in the real and imaginary parts. – Pepe Silvia Jun 23 '20 at 1:17