What is wrong with this derivation of the mean of Poisson distribution?

Question

Let Y~ Poisson($\lambda$). Then,
$$\sum_{k\geq 0}\Pr(Y=k) = \sum_{k\geq 0}\frac{\lambda^k e^{-\lambda}}{k!} = 1. $$ Differentiate both sides w.r.t. $\lambda$,
$$-e^{-\lambda} - \sum_{k\geq 1}\frac{k\lambda^{k-1} e^{-\lambda}}{k!} = 0.$$
Multiply $\lambda$ to both sides,
$$\sum_{k\geq 1}\frac{k\lambda^k e^{-\lambda}}{k!} = -\lambda e^{-\lambda}.$$
Hence Mean of Y = $\sum_{k\geq 0}\frac{k\lambda^k e^{-\lambda}}{k!}$ = $-\lambda e^{-\lambda}$.
Which is obviously not true because Mean of Y must be $\lambda$. Could you kindly point out the error? Thanks!

Answer 1

You have correctly that $\sum_{k\geq 0}\frac{\lambda^k e^{-\lambda}}{k!} = 1 $

You have not differentiated correctly.

You should get $\sum_{k\geq 0} \left( \frac d{d \lambda} \frac{\lambda^k e^{-\lambda}}{k!} \right)= 0$

$\sum_{k\geq 0} \left( \frac{k\lambda^{k-1} e^{-\lambda}}{k!}-\frac{\lambda^k e^{-\lambda}}{k!} \right)= 0$

$\sum_{k\geq 0} \left( \frac{k\lambda^{k-1} e^{-\lambda}}{k!}\right)= \sum_{k\geq 0} \left( \frac{\lambda^k e^{-\lambda}}{k!} \right)$

Multiplying by $\lambda$ gives:

$\sum_{k\geq 0} \left( \frac{k\lambda^k e^{-\lambda}}{k!}\right)=\lambda \times \sum_{k\geq 0} \left( \frac{\lambda^k e^{-\lambda}}{k!} \right)=\lambda \times 1=\lambda$