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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.

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$\operatorname{E}[X] = \int_{-\infty}^\infty x f(x)\, \operatorname {d}x$, where $ f(x)$ is probability density function. –  Mhenni Benghorbal May 14 '13 at 22:46
    
@MhenniBenghorbal Yeah, but what variable should I take the expectation? –  JFK May 15 '13 at 4:20

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