Proving that a compound Poisson distribution has a density

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

$$g(x) = \frac{1}{P(N>0)} e^{-\lambda} \sum\limits_{n=1}^\infty \lambda^n f^{*^n}(x)/n!$$

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

$$\mu_{X|N>0} (X \le x) = \frac{1}{P(N>0)} e^{-\lambda} \sum\limits_{n=1}^\infty \lambda^n \mu_f^{*^n}(X \le x)/n!$$

so we just need to show it's differentiable. Any ideas?

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If $f_1$ and $f_2$ are nonnegative, then $\|f_1\ast f_2\|_1=\|f_1\|_1\cdot\|f_2\|_1$. Hence your $g$ is integrable (with integral $1$) and in particular $g$ is finite almost everywhere. – Did Apr 15 '12 at 13:06
But how do we know $g$ is well defined? The sequence might not converge. – Dominic Apr 15 '12 at 13:09
One combines two arguments: (i) the increments of the series defining $g(x)$ are nonnegative hence either this series converges or it goes to infinity; (ii) $g$ is integrable, hence $g(x)$ is finite except possibly on a set of measure zero. – Did Apr 15 '12 at 13:14
Hmm, I see, we use monotone convergence to prove integrability. Thanks! – Dominic Apr 15 '12 at 13:18

1 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 $P(Y_1 + \dots + Y_n \in A)=0$.

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