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Let $X_1, \ldots, X_n \sim \mathrm{Uniform}(0,T)$ and $T^\wedge = \max\{X_1, \ldots, X_n\}$, which is the estimator of $T$. What is the bias and se of this estimator? If $n=1$, then the calculating the bias and standard error are straight forward ($\text{bias}=-T/2$, and $\text{se}=\sqrt{T^2/12}$. But I'm not sure how to go for it when we have $n$ samples as in this problem?!

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    $\begingroup$ You're misusing the word "sample". This is not "$n$ samples". This is one sample of size $n$. $\endgroup$ Mar 30, 2013 at 1:42

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We change notation a bit to make typing easier. Let the $X_i$ be uniformly distributed on $[0,a]$, and let $Y=\max(X_1,\dots,X_n)$. We find the distribution of $Y$. For that we will need to assume that the $X_i$ are independent. Without that assumption, or other detailed information about the joint distribution of the $X_i$, we cannot solve the problem.

The probability that $Y\le y$ is the probability that all the $X_i$ are $\le y$. For $0\le y\le a$, this is $\left(\dfrac{y}{a}\right)^n$. That gives density function $\dfrac{ny^{n-1}}{a^n}$ on $(0,a)$, and $0$ elsewhere.

Now we can compute the expectation of $Y$, and the variance of $Y$, by using standard methods. You will find after integrating that $E(Y)=\dfrac{n}{n+1}a$.

For the variance, use $E(Y^2)-(E(Y))^2$. You will find that $E(Y^2)=\dfrac{n}{n+2}a^2$, and therefore $\operatorname{Var}(Y)= \dfrac{n}{(n+1)^2(n+2)}a^2$.

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  • $\begingroup$ with independency assumption or? $\endgroup$ Mar 30, 2013 at 1:15
  • $\begingroup$ You are right, I am usually quite careful about stressing that, slipped. $\endgroup$ Mar 30, 2013 at 1:18

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