Suppose that $X$ is a compound Poisson random variable of rate $\lambda$ where the size of the jumps are independent random variables with density function $f$.
ie. $X$ is a sum of $N$ i.i.d. random variables with density $f$, which are independent from $N$, and $N$ is distributed as $Pois(\lambda)$. Here we have $\phi_X (u) = e^{\lambda(\phi_f(u)-1)}$.
I would like to show that conditional on the event that there is at least one jump (to remove the spike at 0), $X$ has a density. It seems natural that convolution should smooth out the distribution and preserve the existence of a density. Intuitively, by conditioning on $N$, we'd imagine it'd have a density $g$, where
\begin{equation}
g(x) = \frac{1}{P(N>0)} e^{-\lambda} \sum\limits_{n=1}^\infty \lambda^n f^{*^n}(x)/n!
\end{equation}
where $f^{*^n}$ is the n-fold convolution of the density with itself. If, for example, $f\in L^2(\mathbb{R})$, we know that $\|f*f\|_\infty \le \|f\|_2^2$, so the series above converges uniformly and I think we can argue that it is indeed the true density of $X$ .
If $f$ is just an arbitrary non-negative element of $L^1(\mathbb{R})$ I'm not sure how to proceed though. We know from conditioning that
\begin{equation}
\mu_{X|N>0} (X \le x) = \frac{1}{P(N>0)} e^{-\lambda} \sum\limits_{n=1}^\infty \lambda^n \mu_f^{*^n}(X \le x)/n!
\end{equation}
so we just need to show it's differentiable. Any ideas?
Best Answer
Another approach would be to work a little more abstractly. By the Radon-Nikodym theorem, a random variable $X$ has a density if and only if $P(X \in A)=0$ for every Borel set $A$ with Lebesgue measure zero.
Suppose $m(A) = 0$. If we let $Y_n$ denote the size of the $n$th jump, by conditioning we have $$P(X \in A \mid N > 0) = \sum_{n=1}^\infty P(Y_1 + \dots + Y_n \in A) P(N=n).$$ But for each $n$, $Y_1 + \dots + Y_n$ has a density, so $P(Y_1 + \dots + Y_n \in A)=0$.