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Let $X_1, X_2,..., X_n$ be i.i.d $U(0, \theta)$ random variables.

I am attempting to prove that $\theta_1=\frac{n+1}{n} Y_n$ is a consistent estimator for $\theta$, where $Y_n=max(X_1, X_2,...,X_n)$ using the definition of consistency directly.

I am familiar with other techniques but I'm not sure on how to proceed using the definition directly.

I am aware that I have to prove that $P(|\theta_1 - \theta| > \delta)=P(|\frac{n+1}{n}Y_n - \theta| > \delta)< \epsilon$, for all $n>N(\epsilon)$, $\epsilon > 0$.

I'm unsure as to how I can proceed, any help would be greatly appreciated!

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    $\begingroup$ Try to find the density of $Y_n$ first. The rest of the calculations must follow easily, right? $\endgroup$ – Shashi Apr 22 '18 at 22:46
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\begin{align} \mathbb{P}(|\theta_1(n) - \theta| > \delta) &= \mathbb{P}( \theta - \frac{n+1}{n} Y_n > \delta) + \mathbb{P}(\frac{n+1}{n} Y_n - \theta > \delta) \\ &=\mathbb{P}(Y_n < \frac{ n }{ n+1 } ( \theta - \delta)) + \left(1 - \mathbb{P}(\frac{n+1}{n} Y_n \le \theta + \delta)\right)\\ &= F_{Y_n}\left(\frac{n}{ n + 1 } (\theta-\delta)\right) + (1 - F_{Y_n}\left(\frac{n}{ n + 1 } (\theta+\delta)\right) . \end{align} Now, $$ F_{Y_n}(y)=(F_X(y))^n = ( 1/\theta ) ^ n, $$ hence, \begin{align} \lim_{n \to \infty} \mathbb{P}(|\theta_1(n) - \theta| > \delta) &= \lim_{n \to \infty} \left( \frac{n}{n+1}( 1 - \frac{ \delta }{ \theta } ) \right) ^ n + (1 - \lim_{n \to \infty} F_{Y_n}\left( \frac{n}{n+1}( \theta + \delta ) \right) \\ &= ( 1 - \delta/\theta)^\infty + ( 1 - F_{Y_n}( \theta + \delta))\\ &= 0 + (1-1) = 0. \end{align} The last line is abuse of notation, but the idea is that $n/(n+1)$ goes to $1$ and $\delta/\theta >0$, hence you have a fraction that goes to $0$ as the power goes to infinity. Moreover, $\theta + \delta > \theta$ while the support of $Y_n$ is $[0, \theta]$, thus the CDF is $1$.

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  • $\begingroup$ How do one know that $\mathbb{P}(\frac{n+1} {n} Y_n-\theta>\delta)=0$? $\endgroup$ – Shashi Apr 23 '18 at 5:24
  • $\begingroup$ You're right. I've edited the answer to include this possibility as well $\endgroup$ – V. Vancak Apr 23 '18 at 8:59
  • $\begingroup$ Shouldn't $$ F_{Y_n}(y) = (F_X (y))^n = \left(\frac{y}{ \theta}\right)^n? $$ Or have I misunderstood? $\endgroup$ – Theoretical Economist Apr 23 '18 at 9:14
  • $\begingroup$ @TheoreticalEconomist Sure. A typo. Thanks. $\endgroup$ – V. Vancak Apr 23 '18 at 13:42

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