# Binomial and Poisson issues (Jacod and Potter)

I've been reading through Probability Essentials by Jacod and Potter (2nd edition). I'm on a voyage to do every single exercise in the book. The following problems I am unsure of is as such:

5.11) Let $$X \sim Pois(\lambda)$$, $$\lambda \in \mathbb{Z^+}$$. Show $$E|X-\lambda|=\frac{2\lambda^{\lambda}e^{-\lambda}}{(\lambda-1)!}$$.

5.12) Let $$X\sim Bin(n,p)$$. Show that for $$\lambda,\epsilon>0$$ that $$P(X-np>n\epsilon)\leq E(e^{\lambda(X-np-n\epsilon)})$$

So my work for 5.11 led me down this path

$$E|X-\lambda|=\sum_{x \geq \lambda} (x-\lambda)\frac{e^{-\lambda}\lambda^x}{x!} - \sum_{x < \lambda} (x-\lambda)\frac{e^{-\lambda}\lambda^x}{x!}$$ $$=\sum_{x \geq \lambda} (x-\lambda)\frac{e^{-\lambda}\lambda^x}{x!}-(1-\sum_{x \geq \lambda} (x-\lambda)\frac{e^{-\lambda}\lambda^x}{x!})$$ $$=[2\sum_{x \geq \lambda} (x-\lambda)\frac{e^{-\lambda}\lambda^x}{x!}]-1$$ $$=[2e^{-\lambda}\sum_{x \geq \lambda}\frac{\lambda^x}{(x-1)!}-\frac{\lambda^{x+1}}{x!}]-1$$

My primary issue is that I do not know how to get this sum to converge to $$\lambda^{\lambda}$$. My attempts have led me down a very docile road.

My work for 5.12 is as follows:

Recall the definition of moment generating function: $$M_X(\lambda)=E[e^{\lambda X}]$$ Since we are dealing with a discrete distribution taking on non-negative values, it follows (for n large) that we have $$P(X>0)\leq E[e^{\lambda X}]$$ $$\implies P(X-np-n\epsilon>0)\leq E[e^{\lambda(X-np-n\epsilon}]$$

My issue with the above proof is that really does not address the case where n is small (and so the expected value could potentially be smaller than our probability).

Would anyone be able to guide me in the right direction for either of these problems? I'm assuming that I'm overcomplicating 5.11 and undercomplicating 5.12.

• This question misses a self-study tag, which could have impacted the level of details in the answer below. – Xi'an Mar 20 '16 at 13:10

## 1 Answer

For the first problem I think you had the right idea but went a little bit astray in the second step:

\begin{align} \text{E}(|X - \lambda|) &= \sum_{k=0}^{\lambda - 1} (\lambda - k) e^{-\lambda} \frac{\lambda^k}{k!} + \sum_{k=\lambda + 1}^{\infty} (k - \lambda) e^{-\lambda} \frac{\lambda^k}{k!} \\ &= \sum_{k=0}^{\infty} (\lambda - k) e^{-\lambda} \frac{\lambda^k}{k!} - \sum_{k=\lambda + 1}^{\infty} (\lambda - k) e^{-\lambda} \frac{\lambda^k}{k!} + \sum_{k=\lambda + 1}^{\infty} (k - \lambda) e^{-\lambda} \frac{\lambda^k}{k!} \\ &= 2 \sum_{k=\lambda + 1}^{\infty} (k - \lambda) e^{-\lambda} \frac{\lambda^k}{k!} \\ &= 2 \sum_{k=1}^{\infty} k e^{-\lambda} \frac{\lambda^{\lambda + k}}{(\lambda + k)!} \\ &= 2 e^{-\lambda} \lambda^\lambda \sum_{k=1}^{\infty} k \frac{\lambda^k}{(\lambda + k)!} \\ &= \frac{2 e^{-\lambda} \lambda^\lambda}{(\lambda - 1)!} . \end{align}

The second problem is an application of Markov's inequality:

\begin{align} P(X - np > n\epsilon) &= P(\lambda (X - np - n\epsilon) > 0) \\ &= P(e^{\lambda (X - np - n\epsilon)} > 1) \\ &\leq \text{E}(e^{\lambda (X - np - n\epsilon)}) . \end{align}

• Thank you so much! I see where I went wrong now. We have $\sum_{k\geq 0} (\lambda - k) e^{-\lambda}\frac{\lambda^k}{k!}=E[\lambda-X]$ which is clearly 0. I wasn't quite sure how to obtain $\frac{1}{(\lambda-1)!}$ from $\sum_{k\geq 1} k \frac{k\lambda^k}{(\lambda+k)!}$ so I had it expanded as I did in my work. I will continue to play to see what I can uncover. As for application of Markov Inequality: very nice. I don't know why that didn't occur to me! I just looked at the right hand-side of the inequality and I was like "what's up MGF!". – user146925 Mar 20 '16 at 16:21