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Suppose I have a positive integrable random variable $X$ s.t. $$E[e^X]=+\infty$$ Now let's take a series with general term $p_n$, summing to one, and define $$Z=\sum_{n>0}p_ne^{X_n}$$ and $U=\ln Z$ where $X_n$ are i.i.d. copies of $X$.

Now I have two questions :

  1. I think that $$\mathcal{L}_U(\lambda)=\cases{+\infty & if $\lambda>0$ \\ 1 & otherwise}$$ is that true? (Here $\mathcal{L}_U (\lambda)$ is the Laplace transform of $U$ at $\lambda$). My proof here is a fraud I think and before giving it I would like to see other people's ideas.
  2. If 1 is true can I deduce from this fact that $U$ is not integrable, and why? (Here I hope it's true)

Best regards

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1 Answer 1

up vote 2 down vote accepted

Here is a counterexample to 1. Assume that $p_n=0$ for every $n\geqslant k+1$, then, for every $n\leqslant k$, $X_n\leqslant X_1+\cdots+X_k$, hence $Z\leqslant\sum\limits_{n\leqslant k}p_n\mathrm e^{X_1+\cdots+X_k}=\mathrm e^{X_1+\cdots+X_k}$ hence $$\mathbb E(\mathrm e^{\lambda U})=\mathbb E(Z^\lambda)\leqslant\mathbb E(\mathrm e^{\lambda X})^k, $$ where every $X_n$ is distributed like $X$. The hypothesis is that $\mathbb E(\mathrm e^{X})$ is infinite. Assume that $\mathbb E(\mathrm e^{\lambda X})$ is finite for some $\lambda$ in $(0,1)$, then $\mathbb E(\mathrm e^{\lambda U})$ is finite as well. An example of this situation is when $X$ is exponential with parameter $\mu\leqslant1$, then $\mathbb E(\mathrm e^{\lambda X})$ is finite if and only if $\lambda\lt\mu$.

Here is a counterexample to 2. Consider a positive random variable $Z$ such that $\mathbb P(Z\geqslant z)\sim c/z^a$ when $z\to+\infty$, for some $a\gt0$. Then $\mathbb E(\mathrm e^{\lambda Z})$ is infinite for every $\lambda\gt0$ but, if $a\gt1$, then $\mathbb E(Z)$ is finite.

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