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!

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

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