For the set of i.i.d. $X_{1},\ldots,X_{n} \sim \mathrm{Uniform}(0,\theta)$, the method of moments estimator is
$$\hat{\theta} = \frac{2}{n} \sum_{k=1}^{n}X_k.$$
now to show that $\hat{\theta}$ is a consistent estimator of $\theta$, I want to show
$$\hat{\theta} \rightarrow \theta$$
in probability.
That is to say,
$$\forall \epsilon >0, \lim_{n\rightarrow \infty} \mathbb{P}(|\hat{\theta} - \theta|>\epsilon) = 0$$
I'm having trouble showing this limit, as I substitute the method of moments estimator into that braacket, but have no idea how to go from here.
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Twenty-six colours
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1If you want a direct proof - I assume that $\theta$ is known to be finite and positive. Note that $\mathbb{P}(|\widehat{\theta}n - \theta| > \epsilon) = \mathbb{P} \left( | \sum{k\le n} (X_k - \theta/2) | > n\epsilon/2\right) =: P_n$. Now the $X_k - \theta/2$ are iid bounded, centered random variables, use Hoeffding's inequality to upper bound this probability. You get that $P_n \le 2e^{-6n\epsilon^2/\theta^2},$ and taking the limit does the trick. – stochasticboy321 Jun 09 '17 at 03:10
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HINT
You have $$ \theta = 2E(X)\\\hat\theta = 2 \bar X.$$ Use the weak law of large numbers.
spaceisdarkgreen
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Ah I see. Is it the fact that (by the Weak/Strong Law of Large Numbers), the sample mean's mean will converge to $\frac{\theta}{2}$ so $\hat{\theta}$ converges to $\theta$? – Twenty-six colours Jun 09 '17 at 02:40
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If so, was there a way to show the above (in the OP) probability convergence definition? – Twenty-six colours Jun 09 '17 at 02:41
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1WLLN says that (if $X_i$ are i.i.d. and integrable), $\frac{X_1+X_2+\ldots+X_n}{n} \rightarrow_P E(X).$ The definition of convergence in probability (i.e. $Y_n\rightarrow_P Y$) is that for all $\epsilon >0,$ $P(|Y_n-Y| > \epsilon) \to 0.$ – spaceisdarkgreen Jun 09 '17 at 03:06
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