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first time poster so be nice! Here's the problem:

Let $\phi(t)$ be a characteristic function, then $e^{\lambda(\phi(t)-1)}$ is a characteristic function.

Pretty stuck, any help appreciated!

Sebastian
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    Hint: Poisson distribution. Now it would be much better if you added your thoughts on the problem (and if the answer is "none", something is not right...). – Did Oct 02 '13 at 09:48
  • Hello, welcome to Math.SE. Please, try to make the title of your questions more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. For more information on choosing a good title, see this post. – Lord_Farin Oct 02 '13 at 10:17

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Suppose that $\phi(t)$ is CF of $Y$. Suppose $X=Y_1+...+Y_P$, sum of $P$ i.i.d. $Y_i$'s, where $P$ follows Poisson distribution of parameter $\lambda$. Then we have: $$ \mathbb E(e^{itX})=\sum_{k=0}^\infty \mathbb E(e^{itX};P=k)\frac{e^{-\lambda}\lambda^k}{k!}=\sum_{k=0}^\infty \phi(t)^k \frac{e^{-\lambda}\lambda^k}{k!}=e^{\lambda(\phi(t)-1)}. $$

Arash
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