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Maximum of a Sample from a Uniform Distribution:

Suppose $X_1, ... , X_n$ is a random sample for a $uniform(0,\theta)$ distribution. Suppose $\theta$ is unknown. An intuitive estimate of $\theta$ is the maximum of a sample. Let $Y_n = max\{X_1, ... , X_n\}$. Exercise 5.1.4 shows that the cdf of $Y_n$ is

$F_{Y_n}(t) = 1$ if $t>\theta$, $F_{Y_n}(t) = \frac{t^n}{\theta^n}$ if $0 < t \leq \theta$, and $F_{Y_n}(t) = 0$ if t is less than or equal to 0.

Hen the pdf of $Y_n$ is $f_{Y_n}(t) = \frac{nt^{n-1}}{\theta^n}$ if $0 < t \leq \theta$ and $f_{Y_n}(t) = 0$ elsewhere.

Based on its pdf, it is easy to show that $E(Y_n) = (n/(n+1))\theta$. Thus, $Y_n$ is a biased estimator $\theta$...Further, based on the cdf of $Y_n$, it is easily seen that $Y_n$ converges to $\theta$ in probability.


How do we know that $Y_n$ converges to $\theta$ in probability? Is it because $E(Y_n) \rightarrow \theta$

Thanks in advnace

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2 Answers

up vote 2 down vote accepted

The definition of convergence in probability is given below :

Let $\{X_n\}$ be a sequence of random variables on a probability space. then we say that $\{X_n\}$ convergence in probability to $\theta$ if for every $\epsilon >0$, $$Pr[|X_n-\theta| \geq \epsilon]\rightarrow0 ~\text{as}~ n\rightarrow \infty$$ and this is equivalent to $$Pr[|X_n-\theta|< \epsilon]\rightarrow 1 ~\text{as}~ n\rightarrow \infty$$

For your problem note that $$\begin{align}P[|Y_{(n)}-\theta|< \epsilon] &=P[\theta-\epsilon<Y_{(n)}<\theta+\epsilon]\\ &= F_{Y_{(n)}}(\theta+\epsilon)- F_{Y_{(n)}}(\theta-\epsilon)\\&= 1-(\frac{\theta-\epsilon}{\theta})^n, 0< \epsilon <\theta \\&=1-0,\epsilon \geq\theta \\& \rightarrow 1 ~\text{as}~ n \to \infty\end{align}$$ Hence $Y_n$ converges to $\theta$ in probability.

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How did you passed from line 2 to line 3 in your answer above? I mean, how did you get $F_Y(\theta+\epsilon) - F_Y(\theta-\epsilon) = 1-(\frac{\theta-\epsilon}{\theta})^n$? –  srodriguex Dec 28 '13 at 14:44
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Assuming $E \left[ Y_n \right] \rightarrow \theta$ so that $E \left[ \left| Y_n - \theta \right| \right] \rightarrow 0$, you could use Markov's inequality to show that for $\varepsilon > 0$ \begin{eqnarray*} \Pr \left[ \left| Y_n - \theta \right| \geq \varepsilon \right] & \leqslant & E \left[ \left| Y_n - \theta \right| \right]/\varepsilon \end{eqnarray*} which goes to zero proving convergence in probability.

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Thanks, but then we need to have epsilon next to $Pr[|Y_n - \theta| \geq \epsilon]$, right? Like this...\begin{eqnarray*} \epsilon \Pr \left[ \left| Y_n - \theta \right| \geq \varepsilon \right] & \leqslant & E \left[ \left| Y_n - \theta \right| \right] \end{eqnarray*} Do we ignore that epsilon, because it is arbitrarily small? –  user58289 Jan 27 '13 at 13:00
I think $Pr[|Y_n-\theta| \geq \epsilon] \leq \frac {E[|Y_n-\theta|]}{\epsilon}$ not $E[|Y_n-\theta|]$, from Markov. –  A.D Jan 27 '13 at 13:04
@Artus and A.D. Sorry about my typo. I added the missing epsilon. –  Learner Jan 27 '13 at 13:09
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