Let $X_1, X_2,\ldots, X_n$ be i.i.d. uniform on $[0, \theta ]$.
a. Find the method of moments estimate of $\theta$ and its mean and variance
b. Find the MLE of $\theta$ and its mean and variance.
Thank you for answering, I really appreciate it.
My answers were:
a. $\hat{\theta} = 2 \bar{X}$
b. $\hat{\theta} = X_n$
I'm not just sure about my solution, I don't also know how to start solving for the mean and variance considering the MLE and MME.
Best Answer
OK, so I'll drop a few hints.
First of all: Check that your MLE estimator of $\theta$ is indeed the maximum of the likelihood function. Note that this maximum is not detected by the derivative!
Now, mean and variance of $\hat \theta=2\overline{X}$ can be deduced from those of $\overline{X}.$ The distribution of $\overline{X}$ is difficult to write down, but you don't need the whole pdf, you only need $E(\overline{X})$ and Var $(\overline{X}).$ If you have been given this problem, you probably already know what the mean and the variance of the sample mean is, but just in case, here you are: $E(\overline{X})=E(X),$ Var$(\overline{X})=$Var$(X)/n.$
Concerning the MLE, you probably will have to work out first the pdf of $\hat \theta=\max\{X_1,\cdots,X_n\}$. Fix $x\in [0,\theta].$ From the definition of a maximum, $P[\hat \theta \le x]=P[X_1\le x, \;X_2\le x,\;\cdots, X_n\le x];$ now we use that the $X_i$ are independent copies of the uniform distribution in $[0,\theta]$,hence $P[\hat \theta \le x]=P[X\le x]^n$ where $X$ is a uniform distribution in $[0,\theta].$ Since $P[X\le x]=x/\theta$ (cdf of a uniform variable in $[0,\theta]$), we deduce that the cdf of $\hat\theta$ is $x^n/\theta^n$ in $[0,\theta]$ and thus the pdf is $nx^{n-1}/\theta^n$, also in $[0,\theta]$. Now use this pdf to compute $E(\hat \theta)$ and Var$\hat \theta$ in the usual way. That is, $E(\hat \theta)=\int_0^{\theta}x (nx^{n-1}/\theta^n)\,dx$ and Var$\hat \theta=\int_0^{\theta}x^2 (nx^{n-1}/\theta^n)\,dx - E(\hat\theta)^2$