Log likehood functions - Expected value

Let $X_1,X_2,\ldots,X_n$ be a random sample from a Bernoulli($θ$) distribution with probility function $$P(X=x)= (θ^x)(1-θ)^{1-x},\qquad x=0,1;\ 0 < θ < 1.$$

$dl/dθ = [n \overline{x}/θ] \cdot (n-n\overline{x})/(1-θ)$ <-- Is it this that's wrong? :/ Got help with this too (Perhaps you can tell stats isn't my best subject)

Show that $E[(dl(θ)/dθ)] = 0$

Apologies, exam in a couple of days in a mad scattered panic!

When I first did this I integrated by accident, then I diffentiated and got a very strange answer and wasn't sure how to bring it to zero.

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Order given + no indication of personal thought = no urge to answer. (Note that the expression of dl/dθ is wrong.) –  Did Aug 7 '12 at 21:03

Let $X_1, X_2, \cdots, X_n$ be a random sample from $\mathrm{Bernoulli}(\theta)$ distribution. The log-likehood function for the sample mean $\bar{X}$ is given by

$$l = l(\theta | \bar{x}) = \log \mathcal{L}(\theta | \bar{x}) = \log \mathbb{P}_{\theta}(\bar{X} = \bar{x}) = \log \left[ \binom{n}{n\bar{x}}\theta^{n\bar{x}}(1-\theta)^{n-n\bar{x}} \right],$$

thus we have

$$\frac{dl}{d\theta} = \frac{n\bar{x}}{\theta} - \frac{n-n\bar{x}}{1-\theta}.$$

It is easy to see, from binomial distribution, that $\mathbb{E}(\bar{X}) = \theta$. Thus

\begin{align*} \mathbb{E}\left[\frac{dl}{d\theta}(\bar{X})\right] &= \mathbb{E}\left[\frac{n\bar{X}}{\theta} - \frac{n-n\bar{X}}{1-\theta}\right] = \frac{n\mathbb{E}(\bar{X})}{\theta} - \frac{n-n\mathbb{E}(\bar{X})}{1-\theta} \\ &= \frac{n\theta}{\theta} - \frac{n-n\theta}{1-\theta} = n - n = 0. \end{align*}

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Thanks so much! :) –  Fred Aug 7 '12 at 22:00