Let $X\sim \text{Poi}(\lambda )$, $\lambda>0$. Let $a>2\lambda$.
Show that $P[X\leq a]\leq \text{exp}\left (a-\lambda-a\ln \left (\frac{a}{\lambda}\right )\right )$ using the Markov-inequality with $\phi (x)=e^{\theta x}$ and minimizing over $\theta>0$.
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I have done the following :
For $\epsilon=\alpha>0$ and $\phi (x)=e^{\theta x}$ we have from the Markov-innequality: \begin{equation*}P[X\geq \alpha]\leq \frac{E[\phi (|X|)]}{\phi (\alpha)} =\frac{E[e^{\theta X}]}{e^{\theta \alpha}}\end{equation*}
The expected value of $e^{\theta X}$ is \begin{align*}E[e^{\theta X}]&=\sum_{x\in X(\Omega)} e^{\theta x}\cdot P[X=x] =\sum_{x=0}^{+\infty} e^{\theta x}\cdot \frac{\lambda^x}{x!}\cdot e^{-\lambda} =e^{-\lambda}\cdot \sum_{x=0}^{+\infty} e^{\theta x}\cdot \frac{\lambda^x}{x!} \end{align*}
How can we continue?
Best Answer
Notice that $$ \sum_{x=0}^{+\infty}e^{\theta x}\cdot\frac{\lambda^x}{x!}=\sum_{x=0}^{+\infty}\frac{(\lambda e^\theta)^x}{x!}=\exp\{\lambda e^\theta\} $$ So we have $$ P[X\geq a]\leq \exp\{(e^\theta-1)\lambda-\theta a\} $$ Then minimizing over $\theta>0$