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I'm in need of help with this problem. I've been reviewing for an exam and I'm terrible with estimations.

$X_1,X_2,\dots,X_n$ be an i.i.d sample from continuous uniform distribution on the interval $(0,\theta)$ with parameter $\theta>0$. Let us consider $Y=\max(X_1,X_2,\dots,X_n)$ has a c.d.f $F(y) = P(Y \le y) = (\frac y \theta)^n$; $0<y<\theta$.

Is $Y$ an unbiased estimator for $\theta$?

I've made attempts but I think I'm going the opposite directions.

My attempt so far:

$$n\over{(n-r)!(r-1)!}[F(y)^{r-1}][(1-F(y))^{n-r}]f(y)$$

Michael Hardy
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user130431
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2 Answers2

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Since $\Pr(Y<\theta)=1$, one must have $\operatorname{E}(Y)<\theta$, so $Y$ is not an unbiased estimator of $\theta$.

Michael Hardy
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Your question amounts to asking whether $E[Y]=\theta$. But $E[Y]$ can just be calculated, now that you have the CDF:

$$E[Y]=\int_0^\theta x f(x) dx = \int_0^\theta x F'(x) dx = \int_0^\theta \frac{n}{\theta^n} x^n dx.$$

Now calculate this integral. You will necessarily find that it is strictly less than $\theta$; one way to see this without doing the calculation is to notice that $P(Y \leq \theta)=1$ and $P(Y<\theta/2)>0$.

The application of this calculation is that $E[Y]$ is not $\theta$, but is $k_n \theta$ for some number $k_n$. This means that $E[Y/k_n]=\theta$, which gets you an unbiased estimator for $\theta$. This is much like how we correct the "obvious" biased estimator for $\sigma^2$ into the unbiased estimator that is customarily used.

Ian
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