Method of Moments for $p(x) = cx^{\beta} \hspace{1cm} x \in[0,a] $

Question

Problem: Let $X$ be a random variable with pdf $$p(x)= \left \{ \begin{matrix}{} cx^{\beta} \hspace{1cm} x \in[0,a] \\ 0 \hspace{1cm} \text{otherwise} \\ \end{matrix} \right.$$ where $a > 0$, $\beta > -1$ are unknown and $c$ is the normalizing constant. Find the estimator of $a$ and $\beta$ using the method of moments.

Attempt: I found $$\left \{ \begin{matrix}{} \frac{X_1 + X_2+\cdots+X_n}{n}=\frac{ca^{\beta+2}}{\beta +2} \\ \frac{X_1^2 + X_2^2+\cdots+X_n^2}{n}=\frac{ca^{\beta+3}}{\beta +3} \\ \end{matrix} \right.$$ but I can't continue.

Thanks!

Answer 1

First, note that $$c = \frac{\beta+1}{a^{\beta+1}}.$$ Consequently (presuming that your calculations are right), $$E[X] = \frac{ca^{\beta+2}}{\beta+2}=\frac{\beta+1}{\beta+2}a,$$ and $$E[X^2] = \frac{ca^{\beta+3}}{\beta+3}=\frac{\beta+1}{\beta+3}a^2.$$ Also, $$\frac{E[X]^2}{E[X^2]}=\frac{(\beta+1)(\beta+3)}{(\beta+2)^2} = 1-\frac{1}{(\beta+2)^2} \implies \beta = -2 + \sqrt{\frac{E[X^2]}{E[X^2]-E[X]^2}}.$$ Can you find $a$?