Maximum Likelihood - Information Matrix Identity Derivation

Question

I try to derive the information matrix equality for the Poisson distribution with the log-Likelihood:

$$\mathcal{L}(\lambda; x_1, x_2, \ldots, x_n) = \sum_{i=1}^{n} \left[-\lambda + x_i \log(\lambda) - \log(x_i!)\right]$$

The derivation by the Hessian matrix is clear to me and yield $I(\lambda) = \frac{n}{\lambda}$. So the goal is to show that

$$E\left[\sum_{i=1}^{n} \left(\frac{\partial \mathcal{L}_i(\lambda)}{\partial \lambda} \frac{\partial \mathcal{L}_i(\lambda)}{\partial \lambda^T}\right)\right] = E\{s(\lambda) \cdot s(\lambda)^T\} \overset{!}{=} \frac{n}{\lambda}$$

Now while the derivation with the above method on the left side of the equation yields the same outcome I am struggling showing that $ E\{s(\lambda) \cdot s(\lambda)^T\} = E\{s(\lambda)^2\}$ must also be the same. Since:

$s(\lambda)^2 = \left(\left(\sum_{i=1}^{n} x_i\right) / \lambda - n\right)^2 = \left(\sum_{i=1}^{n} x_i\right)^2 / \lambda^2 - 2n \left(\sum_{i=1}^{n} x_i\right) / \lambda + n^2$

And now this is where I am stuck, since $\left(\sum_{i=1}^{n} x_i\right)^2 \neq \left(\sum_{i=1}^{n} x_i^2\right) $, right? I think the rest would be like the left variant and I could easily derive it. So it's likely just this tiny bit missing.

Answer 1

This is lot easier to show if you use the fact that the expectation of the score is always zero, so $$ \mathbb{E}(s(\lambda;\mathbf{X})^2)=\mathrm{Var}(s(\lambda;\mathbf{X})) $$

You then have $$ \mathrm{Var}\left(\frac{\sum_{i=1}^{n} X_i}{\lambda} - n\right)=\frac1{\lambda^2}n\lambda=\frac{n}{\lambda} $$ as required.