Negative Expectation of a constant is positive or negative Say, I have a normal distribution $ X  \sim N(\mu, \sigma)$, where $\mu$ unknown and $\sigma^2$ known
Then, $-E[2/\sigma]=2/\sigma$ or $-2/\sigma$
This is in reference to the Fisher Information where you have $-E(\cdot)$ of something.
Edit:
I was taking expectation of Fisher Information.
$$-E[\frac{1}{2\sigma^4}-\frac{(X-\mu)^2}{\sigma^6}|\mu, \sigma^2]=\frac{1}{2\sigma^4}$$
where did the negative sign go?
 A: We know $X-\mu$ has a $N(0,\sigma^2)$ distribution. So 
$$E( (X-\mu)^2 ) = {\rm var}(X-\mu) + [E(X-\mu)]^2 = \sigma^{2} + 0 = \sigma^2$$
therefore 
$$E\left(\frac{(X-\mu)^2}{\sigma^6}\right) = \frac{1}{\sigma^6} \cdot E( (X-\mu)^2 )  = \frac{\sigma^2}{\sigma^{6}} = \frac{1}{\sigma^4}$$
which implies
$$-E \left [\frac{1}{2\sigma^4}-\frac{(X-\mu)^2}{\sigma^6}\right] = -\frac{1}{2\sigma^4} + \frac{1}{\sigma^4} = \frac{1}{2\sigma^4}$$
A: If $\sigma$ is the standard deviation then it is positive (or at least non-negative). 
If $\sigma$ is a constant then $$-E\left[ \frac 2 \sigma \right] = -\frac 2 \sigma$$
In the expression for the Fisher Information you can often find
$$
\mathcal{I}(\theta) = - \operatorname{E} \left[\left. \frac{\partial^2}{\partial\theta^2} \log f(X;\theta)\right|\theta \right]\,
$$
the sign is present because the expectation is negative.  There is another expression $$\mathcal{I}(\theta)=\operatorname{E} \left[\left. \left(\frac{\partial}{\partial\theta} \log f(X;\theta)\right)^2\right|\theta \right]$$ withought the negative sign.
