# How to obtain this estimator of this random sample?

by Aarón David Arroyo Torres   Last Updated May 15, 2018 20:20 PM

Let $X_1,...,X_n$ be a random sample of $X\sim U(0,\theta), \theta >0,$ and $Y=max\{X_i\}$.

I know that $f_Y(y)=\frac{n}{\theta^n}y^{n-1}I_{(0,\theta)}(y)$ is the density function of $Y$ and also that $E[Y]=\int_{(0,\theta)}y\frac{n}{\theta^n}y^{n-1}dy=\frac{n}{n+1}\theta$.

What I don't understand is why $\widetilde \theta = \frac{n+1}{n}max\{X_i\}$ is an estimator of $\theta$. I guess it's related to the equality $E[Y]=\frac{n}{n+1}\theta$ and the methods of moments but I don't know how to relate the $max\{X_i\}$ with the latter equality.

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#### Answers 1

As you have written $E[max{X_i}] =\frac{n}{n+1}\theta$

Then $\theta = \frac{n+1}{n}E[max{X_i}]$

Markoff Chainz
May 15, 2018 20:17 PM

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