# Upper bounds for probability

expected valueinequalityprobabilityprobability theory

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 $$\theta>0$$.



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 $$\theta>0$$