Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 4 down vote accepted

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

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

share|cite|improve this answer
Thanks so much! :) – Fred Aug 7 '12 at 22:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.