Why is the Fisher Information matrix positive semidefinite? Let $\theta \in R^{n}$. The Fisher Information Matrix is defined as:
$$I(\theta)_{i,j} = -E\left[\frac{\partial^{2} \log(f(X|\theta))}{\partial \theta_{i} \partial \theta_{j}}\bigg|\theta\right]$$
How can I prove the Fisher Information Matrix is positive semidefinite? 
 A: WARNING: not a general answer!
If $f(X|\theta)$ corresponds to a full-rank exponential family, then the negative Hessian of the log-likelihood is the covariance matrix of the sufficient statistic.  Covariance matrices are always positive semi-definite.  Since the Fisher information is a convex combination of positive semi-definite matrices, so it must also be positive semi-definite.
A: Check this out: http://en.wikipedia.org/wiki/Fisher_information#Matrix_form
From the definition, we have
$$
  I_{ij} = \mathrm{E}_\theta \left[ \left(\partial_i \log f_{X\mid\Theta}(X\mid\theta)\right) \left(\partial_j \log f_{X\mid\Theta}(X\mid\theta)\right)\right] \, ,
$$
for $i,j=1,\dots,k$, in which $\partial_i=\partial /\partial \theta_i$. Your expression for $I_{ij}$ follows from this one under regularity conditions.
For a nonnull vector $u = (u_1,\dots,u_k)^\top\in\mathbb{R}^n$, it follows from the linearity of the expectation that
$$
  \sum_{i,j=1}^k u_i I_{ij} u_j = \sum_{i,j=1}^k \left( u_i \mathrm{E}_\theta \left[ \left(\partial_i \log f_{X\mid\Theta}(X\mid\theta)\right) \left(\partial_j \log f_{X\mid\Theta}(X\mid\theta)\right)\right] u_j \right) \\
= \mathrm{E}_\theta \left[ \left(\sum_{i=1}^k u_i \partial_i \log f_{X\mid\Theta}(X\mid\theta)\right) \left(\sum_{j=1}^k u_j \partial_j \log f_{X\mid\Theta} (X\mid\theta)\right)\right] \\
= \mathrm{E}_\theta \left[ \left(\sum_{i=1}^k u_i \partial_i \log f_{X\mid\Theta}(X\mid\theta)\right)^2 \right] \geq 0 \, .
$$
If this component wise notation is too ugly, note that the Fisher Information matrix $H=(I_{ij})$ can be written as $H = \mathrm{E}_\theta\left[S S^\top\right]$, in which the scores vector $S$ is defined as
$$
  S = \left( \partial_1 \log f_{X\mid\Theta}(X\mid\theta), \dots, \partial_k \log f_{X\mid\Theta}(X\mid\theta) \right)^\top \, .
$$
Hence, we have the one-liner
$$
  u^\top H u = u^\top \mathrm{E}_\theta[S S^\top] u = \mathrm{E}_\theta[u^\top S S^\top u] = \mathrm{E}_\theta\left[|| S^\top u ||^2\right] \geq 0.
$$
