Partial Derivative of Poisson CDF I am trying to figure out how to take the partial derivative of the Poisson CDF $F(k, \lambda)$ with respect to $k$. I have seen the partial derivative taken with respect to $\lambda$, but am unsure how to do it with respect to $k$. Especially since there is a factorial involved.
Can someone help me out?
Thanks,
Erich
 A: From the comments above:
QUOTE: Sorry, this is so confusing, I'm probably not explaining myself clearly. I'm not in the mathematics field. So, I've seen a paper that shows the partial derivative of $F(k, λ)$ with respect to $λ$, and their conclusion was that as $λ$ increases and $k$ is held constant, $F$ decreases. I want to do the same but for the $k$. END OF QUOTE
Since $k$ takes on only the values $\{0,1,2,3,\ldots\}$ and does not vary continuously, the thing to do here is to use finite differences rather than derivatives.  I.e. instead of asking how fast the function is changing as $k$ varies, ask how much the function changes as $k$ increases by $1$.
So you have
$$
\begin{align}
& {} \qquad \frac{\lambda^{k+1}e^{-\lambda}}{(k+1)!} - \frac{\lambda^k e^{-\lambda}}{k!} = e^{-\lambda} \lambda^k\left(\frac{\lambda}{(k+1)!} - \frac{1}{k!}\right) \\  \\
& = e^{-\lambda} \lambda^k\left(\frac{\lambda}{(k+1)!} - \frac{k+1}{(k+1)!}\right) = \frac{e^{-\lambda} \lambda^k}{(k+1)!}(\lambda-k-1).
\end{align}
$$
This is positive when $k<\lambda -1$ and negative when $k>\lambda-1$, and if $\lambda$ happens to be an integer, it is $0$ when $k=\lambda-1$.
Therefore the function increases on $\{0,1,2,3,\ldots,\lfloor\lambda-1\rfloor\}$ and decreases on $\{\lfloor\lambda-1\rfloor+1,\lfloor\lambda-1\rfloor+2,\lfloor\lambda-1\rfloor+3,\ldots\}$.
You might still need to look at the question of which of two values is bigger.  In that one isolated case of an exact ineger value, the maximum is attained at each of two consecutive integers.
A: This might be as good an occasion as any to recall that the whole family of Poisson distributions can be coupled in such a way that the distribution of each $X_\lambda$ is Poisson$(\lambda)$ and the random function $\lambda\mapsto X_\lambda$ is almost surely nondecreasing. 
Consequences are that $\mathrm P(X_\lambda\geqslant x)$ for every positive $x$ and $\mathrm E(X_\lambda)$ are increasing functions of $\lambda$. More generally, $\mathrm E(u(X_\lambda))$ is a nondecreasing function of $\lambda$, for every nondecreasing function $u$.
