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.

  • 2
    $\begingroup$ Is your first condition ever true for $U\neq V$? It can't be in the case of $U, V$ both discrete positive. $\endgroup$ – Pepe Silvia Jun 17 '20 at 2:21
  • $\begingroup$ @PepeSilvia Thanks. I modified the answer to fix this. $\endgroup$ – Lisa Jun 17 '20 at 12:54
  • $\begingroup$ 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) $\endgroup$ – Clement C. Jun 17 '20 at 18:05
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ I think your conjecture may be true in the case where your random variables are uniformly restricted to a bounded interval rather than simply being non-negative. $\endgroup$ – Pepe Silvia Jun 22 '20 at 17:00
  • $\begingroup$ Thanks. So $U_n$ converges to $V_n$ in moment generating function, but not in the characteristic function? Is this right? $\endgroup$ – Lisa Jun 22 '20 at 23:41
  • $\begingroup$ 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. $\endgroup$ – Pepe Silvia Jun 22 '20 at 23:48
  • $\begingroup$ This is very interesting. I didn't expect this. I always treated the MGF and CF as having very similar properties. What about the other way around? Can we control MGF's with CF's? This is probably for another question. $\endgroup$ – Lisa Jun 23 '20 at 0:44
  • $\begingroup$ 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. $\endgroup$ – Pepe Silvia Jun 23 '20 at 1:17

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