Let $g(x)$ be a continuous function that is analytic on every interval of $\mathbb{R}$ that does not contain $0$.
An example of such a function is: $e^{-|x|}$.
Let $X$ be a random variable and define \begin{align} f(t)= E[g(X-t)] \end{align} That is we are looking at the convolution integral and suppose that $|g(t)|<c<\infty$ for all $t$.
We want to look at analytic properties of $f(t)$ without many assumptions on $X$.
I have two question about this:
- Does $f(t)$ have the same property as $g(t)$? That is $f(t)$ is analytic on every interval of $\mathbb{R}$ except one point.
- If in addition, we assume that $X$ has an absolutely continuous distribution. Is $f(t)$ analytic everwhere?
This question was inspired by a similar question raised here.
My attempts
I was only more or less able to answer 1).
To study this we can fix $t$ and look at \begin{align} f(t)=E[g(X-t)]&=E[g(X-t) 1_{X \neq t}]+ E[g(X-t) 1_{X = t}]\\ &= E[g(X-t) 1_{X \neq t}]+ g(0)P[ X = t]. \quad (*) \end{align}
We see from (*) that $f(t)$ is a.e. differentiable. Moreover, if we assume that $X$ has no atoms then $f(t)$ is differentiable everywhere since $P[X=t]=0$.
Let $T=\{t:P[X=t]>0 \}$. Equation(*) seems to be suggesting that 1) is false if $|T|\ge 2$ and $f(t)$ will not be differentiable for $t$ such that $P[X=t]>0$.
Moreover, , $T=\{t:P[X=t]>0 \}$ can be a dense set so it appears that $f(t)$ may not be analytic on any neightborhood of $\mathbb{R}$.
Unless I am wrong, this appears to be very intersting since I alway thought that convolution makes things "more" smooth. However, in this case, things get "less" smooth.
I would really appreciate if some one can clarify these for me.
For 2). It seems that $f(t)$ is infinitely differentiable for all $t$, since $P[X=t]=0$ for all $t$. So, it seem resonable to ask if $f(t)$ is analytic everywhere?