# Expectation, Variance and Moment estimator of Beta Distribution

I'm given a beta distributed random variable: $$X \sim \text{Beta}_{(\theta, 1)} =: \mathbb{P}_\theta$$. Where $$\theta \geq q$$ and

$$\mathbb{f}_\theta(x) = \theta \cdot x^{\theta-1} \cdot \mathbb{1}_{[0,1]}(x)$$

I was asked to compute the expectation and variance of $$X$$ and came up with the following solution:

$$\mathbb{E}X = \int_\mathbb{R} x \cdot \mathbb{f}_\theta(x) \, \text{dx} = \theta \int_0^{1} x^{\theta} \, \text{dx} = \frac{\theta}{\theta + 1}$$

$$\mathbb{E}X^2 = \int_\mathbb{R} x^2 \cdot \mathbb{f}_\theta(x) \, \text{dx} = \theta \int_0^{1} x^{\theta+1} \, \text{dx} = \frac{\theta}{\theta + 2}$$

$$\text{Var}X = \mathbb{E}X^2 - (\mathbb{E}X)^2 = \frac{\theta}{(\theta+1)^2(\theta+2)}$$

Are these computations correct or did I end up making a mistake? Now let $$X_1, \dots, X_n \sim \text{Beta}_{(\theta,1)}$$ iid. How can I justify that the moment estimator of $$\theta$$ is given by $$\hat\theta:n = \frac{\bar X_n}{1 - \bar X_n}$$? Is this estimator consistent?

I'd say that the estimator is consistent. The Expectation and Variance of $$X_1$$ is finite. Now after the strong law of large numbers we get

$$\lim_{n \rightarrow \infty} \hat\theta_n = \frac{\mathbb{E}X_1}{1 - \mathbb{E}X_1} = \theta$$

Therefore the estimator should be consistent, but how can I show that the moment estimator of $$\theta$$ is given by $$\hat\theta_n$$? I'm familiar with the method of moments, but don't understand how to apply it in this case.

• Do you know method of moments? – StubbornAtom Apr 9 at 18:56
• @StubbornAtom No, I'm not familiar with it. Honestly I'm not even exactly sure, what is meant with "moment estimator of $/theta$". Is it just an estimator for $\theta$? – user7802048 Apr 9 at 19:00
• @StubbornAtom Ok, now I'm familiar, but can't seem to understand how to apply it in my case. I have the first two moments as computed in my question. – user7802048 Apr 9 at 21:03

The idea behind the Method of Moments estimator is the following.

Let suppose we have $$\underline{X}$$ $$=$$ $$(X_{1},...,X_{n})$$ iid observations, distributed according to $$f(\cdot \mid \boldsymbol{\theta})$$, $$\boldsymbol{\theta}$$ $$\in$$ $$\Theta$$ $$\subseteq$$ $$\mathbb{R}^{d}$$.

Define:

$$\mu_{k} = E(X_{i}^{k}) = \mu_{k}(\theta_{1},...,\theta_{d})$$

the $$k$$-th moment of $$X_{i}$$ and

$$m_{k} = \frac{1}{n}\sum_{i = 1}^{n}X_{i}^{k}$$

the $$k$$-th sample moment. Then we estimate the vector of parameters $$\boldsymbol{\theta}$$ = $$(\theta_{1},...,\theta_{d})$$ solving the following system of equations:

$$\begin{cases} m_{1} = \mu_{1}(\theta_{1},...,\theta_{d})\\ .\\ .\\ m_{d} = \mu_{d}(\theta_{1},...,\theta_{d})\end{cases}$$

leading to $$(\hat{\theta_{1}},...,\hat{\theta_{d}})$$ $$=$$ $$\boldsymbol{\hat{\theta}}$$, our method of moments estimator.

Now, consider a generic beta distribution:

$$f(x_{i} \mid \theta, \alpha) = \frac{\Gamma(\theta + \alpha)}{\Gamma(\theta)\Gamma(\alpha)}x_{i}^{\theta - 1}(1 - x_{i})^{\alpha - 1}I(0 < x_{i} < 1)$$

In this case, $$\boldsymbol{\theta}$$ $$=$$ $$(\theta, \alpha)$$, hence we need to consider the first two moments.

The first moment of $$X_{i}$$ is:

$$E(X_{i}) = \frac{\theta}{\theta + \alpha}$$

while the second moment can be proved to be:

$$E(X_{i}^{2}) = \frac{\theta(\theta + 1)}{(\theta + \alpha)(\theta + \alpha + 1)}$$

Now, at this point, for the first moment in the sample, we have:

$$m_{1} = \frac{1}{n}\sum_{i = 1}^{n}X_{i} = \overline{X_{n}}$$

while for the second moment in the sample we exploit the sample variance:

$$s^{2} = m_{2} - m_{1}^{2} \Rightarrow m_{2} = s^{2} + m_{1}^{2}$$

and, at this point, we apply the system of equations described before to this case, to be then solved for both $$\theta$$ and $$\alpha$$:

\begin{cases} m_{1} = \overline{X_{n}} = E(X_{i}) = \frac{\theta}{\theta + \alpha}\\ m_{2} = s^{2} + m_{1}^{2} = E(X_{i}^{2}) = \frac{\theta(\theta + 1)}{(\theta + \alpha)(\theta + \alpha + 1)}\end{cases}

which would yield the two Method of Moments estimators $$\hat{\theta}$$ and $$\hat{\alpha}$$, and this would be in the general case with both parameters of the beta unknown. In your case, the problem is simplified being $$\alpha$$ $$=$$ $$1$$ known, so that we have to estimate only $$\theta$$ using the first sample moment, that is we have only the first equation to solve with $$\alpha$$ $$=$$ $$1$$:

$$\overline{X_{n}} = \frac{\theta}{\theta + 1} \Rightarrow \hat{\theta} = \frac{\overline{X_{n}}}{1 - \overline{X_{n}}}$$

Hope it clarifies.