# Mean and Variance of Methods of Moment Estimate and Maximum Likelihood Estimate of Uniform Distribution.

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.

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.

#### Solutions Collecting From Web of "Mean and Variance of Methods of Moment Estimate and Maximum Likelihood Estimate of Uniform Distribution."

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$\hat \theta=\int_0^{\theta}x^2 (nx^{n-1}/\theta^n)\,dx – E(\hat\theta)^2$

Your MLE is wrong. You said $X_1,\ldots,X_n$ are i.i.d. That implies $X_1$ or $X_2$, etc., is just as likely to be the maximum observed value as is $X_n$ or any other. The MLE is actually $\max\{X_1,\ldots,X_n\}$.

If you use the conventional notation for the order statistics, with parentheses enclosing the subscripts, so that $X_{(1)}\le X_{(2)} \le \cdots\le X_{(n)}$, then the MLE is $X_{(n)}$.

The density of the uniform distribution on $[0,\theta]$ is $\dfrac 1 \theta$ for $0<x<\theta$, so the joint density is $\dfrac{1}{\theta^n}$ for $0< x_1,\ldots,x_n<\theta$. Look at this as a function of $\theta$: it’s $\dfrac{1}{\theta^n}$ for $\theta>\text{all }x\text{s}$. Thus the likelihood function is
$$L(\theta) = \frac{1}{\theta^n}\text{ for }\theta \ge \max\{x_1,\ldots,x_n\}.$$
This is a decreasing function on the whole interval $[\max\{x_1,\ldots,x_n\},\infty)$. Thus it attains its maximum value at the left endpoint of the interval, which is $\max\{x_1,\ldots,x_n\}$.