Second moment of the Poisson random variable

[Dustinsfl]

With a Poission random variable, we know that \(E[X] = var(X) = \lambda\). By definition of the variance, we can the second moment to be
\[
var(x) = E[X^2] - E^2[X]\Rightarrow E[X^2] = var(X) + E^2[X] = \lambda(1 + \lambda).
\]
The characteristic equation for the Poisson distribution is \(\phi_X(\omega) = \exp(\lambda e^{i\omega - 1})\) and we can find the second moment by
\[
i^{-2}\frac{d^2\phi}{d\omega^2}\bigg|_{\omega = 0} = -e^{\lambda e^{-1} - 2}\lambda(e + \lambda)
\]
Why am I not getting the same answer?

 

[Siron]

I think the problem is that your characterisic function is wrong, it has to be:
$\phi_X(\omega) = \exp(\lambda (e^{i \omega}-1))$.