Can we show that the CDF of the Poisson distribution with mean $\lambda$ is decreasing in $\lambda$?

Question

Can we show that the CDF of the Poisson distribution with mean $\lambda$ is decreasing in $\lambda$?

So far this is what I have:

\begin{align*} \dfrac{\partial}{\partial \lambda} \bigg(e^{- \lambda} \sum_i^t \dfrac{\lambda^i}{i !} \bigg) < 0 &\iff -e^{-\lambda} \sum_i \dfrac{\lambda^i}{i !} + e^{-\lambda} \sum_i i \dfrac{\lambda^{i-1}}{i !} < 0 \\ &\iff \sum_i i \dfrac{\lambda^{i-1}}{i !} < \sum_i \dfrac{\lambda^i}{i !} \\ &\iff \sum_i \dfrac{\lambda^{i-1}}{(i-1) !} < \sum_i \dfrac{\lambda^i}{i !} \\ &\iff \sum_i \dfrac{\lambda^{i-1}}{(i-1) !} < \lambda \sum_i \dfrac{\lambda^{i-1}}{i !} \\ \end{align*} This is where I get stuck so if anybody has any help to offer with respect to proceeding from here please let me know. Thanks!

Answer 1

Start by doing it for $t\in[2,3).$ So you save $$ \frac{d}{d\lambda}\left(e^{-\lambda}(1+\lambda + \frac{1}{2}\lambda^2)\right) = -e^{-\lambda}(1+\lambda+\frac{1}{2}\lambda^2)+e^{-\lambda}(1+\lambda) = -\frac{1}{2}\lambda^2e^{-\lambda}.$$ Yep, negative.

If you don't already see how it's going to go for arbitrary $t$, try it for $t\in [3,4).$ It might help to remember that the $(1+\lambda + \frac{1}{2}\lambda^2+\ldots)$ factor is the Taylor series for $e^{\lambda}.$

Answer 2

You were almost there. At the very last step, do NOT factor out $\lambda$. Rather, notice that the two sums are identical, except that the sum in the RHS (right-hand side) has ONE MORE TERM than the one on the LHS. All the other terms are exactly the same. It would also help to keep careful track of the limits of summation; the LHS starts from $i=1$. Substitute $j$ for $i-1$ in the LHS to understand what is happening.

Answer 3

In fact, it should be $$\dfrac{\partial}{\partial \lambda}\left(e^{-\lambda}\sum_{i=0}^k\frac{\lambda^i}{i!}\right) = -e^{-\lambda}\sum_{i=0}^k\frac{\lambda^i}{i!} + e^{-\lambda}\sum_{i=1}^ki\frac{\lambda^{i-1}}{i!} = -e^{-\lambda}\sum_{i=0}^k\frac{\lambda^i}{i!} + e^{-\lambda}\sum_{i=0}^{k-1}\frac{\lambda^{i}}{i!} = -e^{-\lambda}\frac{\lambda^k}{k!} < 0.$$

Answer 4

This question can also be answered probabilistically using a coupling. Suppose that $\lambda_1 < \lambda_2$. Let $X$ have Poisson distribution with parameter $\lambda_1$ and let $D$ be independent of $X$ and have Poisson distribution with parameter $\lambda_2-\lambda_1$.

A well know fact about Poisson random variables gives that $Y=X+D$ is Poisson distributed with parameter $\lambda_2$.Since $D \ge 0$ we have $\{Y \le c\} \subseteq \{X \le c\}$ for all $c \in \mathbb{R}$.

Now let $F(\cdot,\lambda)$ denote the CDF of the Poisson distribution with parameter. We thus have for all $c \in \mathbb{R}$, $$F(c,\lambda_2) = \mathbb{P}(Y \le c)\le \mathbb{P}(X \le c) = F(c,\lambda_1).$$ This shows that the Poisson CDF is decreasing in $\lambda$.