Suppose my likelihood function is (multiplication of $n$ Poisson function with $\lambda=\mu x$): $$L(\mu;x)=(e^{-\mu x})^a(1-e^{-\mu a})^{n-a}$$
I have calculated $$\log L=-a\mu x+(n-a)\log (1-e^{-\mu x})$$ $$\frac{d\log}{d\mu}=-ax+\frac{n-a}{1-e^{-\mu x}}xe^{-\mu x}$$ $$\frac{d^2\log L}{d\mu ^2}=-(n-a)x^2(e^{\mu x}-1)^{-2}e^{\mu x}.$$ So $$Var(\hat\mu)=\frac{1}{-\mathbb E(\frac{d^2\log L}{d\mu ^2})}.$$
But what is the expectation here equals to? I am stucked here.
