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.