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So I'm a little stuck with what I feel is a basic question. The calculation is easy, but I've obviously missed a key concept of MLE.


The Question
Consider the family of models for the data X1,...,Xn i.i.d. unif[0,θ], θ ∈ Θ = R+; the true, unknown parameter governing the data generating process is $θ_{0} > 0$.

We are given the estimator for $θ_{0}$: $\tilde{\theta}_n = 2\bar{x}_{n} = 2\frac{1}{n}\sum^{n}_{i=1} X_{i}$

And we are asked to find the variance for $\tilde{\theta}_n$


The Answer
So I have the following solution supplied to me: $$ Var(2\bar{x}_{n}) = \frac{4}{12n}\theta^{2}_{0} = \frac{1}{3n}\theta^{2}_{0} $$
The Attempt #1
Because we have so much information, my first inclination was to forget estimation, and just calculate the distribution's variance in terms of $\theta$ directly: i.e. $$ \begin{align*} \mu_{2} = E[X^{2}] &= \int_{0}^{\theta} X^{2} f(x) dx \\ &= \int_{0}^{\theta} X^{2} \frac{1}{\theta} dx \\ &= \frac{\theta^{2}}{3} = Var(X) \end{align*} $$

and then sub this result into $Var(2\bar{x}_n)$, i.e.: $$ \begin{align*} Var(\tilde{\theta}_{n}) &= Var(2\bar{x}_n) \\ &= 4Var(\bar{x}_n)\\ &= \frac{4}{3}\theta^{2} \end{align*} $$

I'm assuming my reasoning was flawed here, because in calculating the second moment,second moment=!Variance I've calculated $Var(X)$ (or $Var(x_{i})$), and $Var(X) \neq Var(\bar{x}_n)$


The Attempt #2
I could just use the knowledge that for this uniform distribution $Var(X) = \frac{\theta^2}{12}$. i.e.

$$ \begin{align*} Var(\tilde{\theta}_{n}) &= Var(2\bar{x}_n) \\ &= 4Var(\bar{x}_n)\\ &= \frac{4}{12}\theta^{2}\\ &= \frac{1}{3}\theta^{2}\\ \end{align*} $$ This is a little closer, but I still lack the 'n' term in my answer, and am uncomfortable with this approach, for the same reason that my approach #1 failed.


As mentioned, I've completely overlooked/misunderstood a fundamental here. If it's obvious to anyone, a push in the right direction is greatly appreciated.

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    $\begingroup$ Note that this a method of moments question and has nothing to do with maximum likelihood. $\endgroup$ – Dason Apr 13 '15 at 13:37
  • $\begingroup$ You wrote: $$\require{cancel} \xcancel{\vphantom{\int\limits^\int_\int}\mu_2 = E[X^2] = \int_0^\theta X^2 f(x) \, dx}$$ That is wrong. It should say this: $$ \mu_2 = E[X^2] = \int_0^\theta x^2 f(x) \, dx. $$ One should be careful about which is capital $X$ and which is lower-case $x.$ Without attention to this distinction, there are elementary things that one cannot understand. $\endgroup$ – Michael Hardy Oct 25 at 19:31
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$$\newcommand{\Var}{\operatorname{Var}}\newcommand{\E}{\operatorname{E}} \Var(2\bar{X}_n) = 4\Var(\bar{X}_n) = 4 \Var\left(\frac{\sum_{i=1}^n{X_i}}{n}\right) =\ldots\text{?}$$

$$=\frac{4}{n^2} \Var\left(\sum_{i=1}^n{X_i}\right)$$

Then, as the $X_i$ are independent

$$=\frac{4}{n^2} \sum_{i=1}^n{\Var(X_i)}$$

To find $\Var(X_i)$ you can use the identity

$$\Var(X_i) = \E(X_i^2) - [\E(X_i)]^2$$

& will therefore have two expectations to find through simple integrations.

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  • $\begingroup$ That's an excellent push :) $\endgroup$ – Vince Apr 13 '15 at 12:50
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    $\begingroup$ $$ \begin{align*} Var(\tilde{\theta}_{n}) &= Var(2\bar{x}_n) \\ &= 4Var(\bar{x}_n)\\ &= 4 Var\left(\frac{\sum_i^n{X_i}}{n}\right) \\ &= \frac{4}{n^{2}} Var(x_{1} + \ldots + x_{n}) \\ &= \left( \frac{4}{n^{2}} \right ) n \left( \frac{\theta^{2}_{0}}{12} \right ) \mbox{from knowledge of }unif[0,\theta]\\ &= \frac{\theta^{2}_{n}}{3n} \\ \end{align*}$$ $\endgroup$ – Vince Apr 13 '15 at 12:50
  • $\begingroup$ my only remaining disconnect is that when I calculated the second moment in attempt #1...this doesn't seem to be knowledge that I needed, or used? $\endgroup$ – Vince Apr 13 '15 at 12:53
  • $\begingroup$ @Vince Well, you need to calculate $\operatorname{Var}(X_i)$ if you don't already know it. (And note that $\operatorname{E}(X_i^2)\neq\operatorname{Var}(X_i)$.) $\endgroup$ – Scortchi - Reinstate Monica Apr 13 '15 at 13:09
  • $\begingroup$ Understood, I need to use the centralized moment. I'm really happy with this question now. Many thanks for all of your help. $\endgroup$ – Vince Apr 13 '15 at 14:04

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