# Showing a probabilistic inequality using probability generating function

Let Y~Poisson($$\lambda$$) be a random variable. Use $$G(z)$$ to show that: $$\mathbb P(Y\geq2\lambda)\leq e^\lambda 2^{-2\lambda}$$, where $$G$$ is the probability generating function of Y, defined as $$G(z)=\sum z^k\mathbb P (Y=k)$$.

How do you suggest I should approach this inequality? It seems like Markov's inequality at first, but that that does not give a sufficient bound, hence the need to use PGF. Any help is appreciated.

Let $$z >1$$. We have $$G(\lambda)=\sum z^{n} \frac {e^{-\lambda} \lambda^{n}} {n!}=e^{-\lambda (1-z)}$$. Hence$$P\{Y \geq 2\lambda\} =P\{z^{Y} \geq z^{2\lambda}\}\leq \frac 1 {z^{2\lambda}} {EZ^Y}=e^{-\lambda (1-z)} z^{-2\lambda}$$. Take $$z=2$$ in this.