# Establishing an upper bound for the tail probability $P(X-\lambda \geq z)$ for any $z>0$, where $X$ is Poisson r.v. w/ parameter $\lambda$

Poisson random variable $$X$$ with the parameter $$\lambda$$ has, respectively, the pmf and the moment generating function of the forms $$P(X = k) = \dfrac{e^{-\lambda}\lambda^k}{k!}, \quad k=0,1,2,\dots \quad\quad\quad E(e^{Xt}) = \exp(\lambda(e^t -1)),$$ respectively. Show that $$X$$ is sub-exponential and establish an upper bound for the tail probability $$P(X - \lambda \geq z)$$ for any $$z>0$$.

In my lecture notes I have that $$X$$ is sub-exponential with parameter $$(\sigma^2,b)$$ if for all $$|\lambda| < 1/b$$, we have that $$\ln(M_{x-\mu}(\lambda)) \leq \frac{\lambda^2\sigma^2}{2} \implies M_{x-\mu}(\lambda) \leq \exp(\frac{\lambda^2\sigma^2}{2})$$. Moreover, if $$X$$ is sub-exponential with paramters $$(\sigma^2,b)$$, then $$P(x-\mu \geq z) \leq \begin{cases} \exp(\frac{-z^2}{2\sigma^2}) &\quad\text{if}\quad 0\leq z \leq \frac{\sigma^2}{b}, \\ \exp(\frac{-z}{2b}) &\quad\text{if}\quad z > \frac{\sigma^2}{b}. \end{cases}$$ However, I am not sure how to translate this to the question at hand while utilizing the pmf and the mgf in my solution. Any help would be appreciated!

• Start by expressing the logarithm of the mgf in terms of $\lambda.$ – whuber Nov 6 '18 at 16:07