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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?

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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.2012-04-15
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    But how do we know $g$ is well defined? The sequence might not converge.2012-04-15
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    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.2012-04-15
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    Hmm, I see, we use monotone convergence to prove integrability. Thanks!2012-04-15

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