Suppose that $X_1, X_2, ..., X_n$ is a sample from a beta distribution with the PDF as $f(x) = (\theta + 1)x^\theta, 0 < x < 1, \theta > -1$. With the method of moments, I can derive that

$$\bar{X} = \mathbb{E}[X] = \frac{\theta + 1}{\theta + 2},$$

so an estimator for $\theta$ is

$$\hat{\theta} = \frac{1}{1 - \bar{X}} - 2.$$

But I got stuck when I tried to compute the expectation of $\hat{\theta}$. I don't know how to derive the distribution of $\bar{X}$. Any help would be appreciated!


1 Answer 1



What you are really saying is that estimating $\mathbb{E}[X]$ using the average of the $X_i$'s, you get the formula above. In other words, $$ \hat{\theta} = \frac{1}{1- \frac1n \sum_{k=1}^n X_k} - 2 = \frac{n}{n- \sum_{k=1}^n X_k} - 2, $$ which is a random variable, depending on the particular values of the $X_k$'s. But you know the distribution of the $X_K$'s, can you now compute the expected value of $\hat{\theta}$?

  • $\begingroup$ My idea was to find the distribution for $\sum_{n=1}^k X_k$ with moment generating function, then the distribution for $\frac{n}{n-\sum_{n=1}^k X_k}$. But the MGF seemed to be very complicated. Could you give me more hints? $\endgroup$
    – Yujian
    Aug 18, 2020 at 3:08
  • $\begingroup$ @Yujian MGF sounds like a good idea, here is Wolfram Alpha computing the MGF for one of the $X_k$'s. $\endgroup$
    – gt6989b
    Aug 18, 2020 at 12:05

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