Finding the bias of an estimator for a beta distribution based on method of moments?

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!

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 $$\hat{\theta}$$?
• 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? Aug 18, 2020 at 3:08
• @Yujian MGF sounds like a good idea, here is Wolfram Alpha computing the MGF for one of the $X_k$'s. Aug 18, 2020 at 12:05