Let $X_1, X_2, . . . , X_n$ be a random sample from a uniform distribution on $[0, \theta]$. Suppose results $x_1, x_2, . . . , x_n$ are observed. Since $f(x) = 1/\theta$ for $0 \leq x \leq \theta$, the likelihood function is:
$L(\theta) = 1/\theta^n$ $0 \leq x_1 \leq \theta$, $. . .$ , $0 \leq x_n \leq \theta$ and zero otherwise
As long as $\theta \geq \max(x_i)$, the likelihood is $1/\theta^n$, which is positive, but as soon as $\theta < \max(x_i)$, the likelihood drops to zero.
:
Let $Y = \max(X_i)$. Derive the CDF of $Y$ and then show that the PDF of $Y$ is
$fy(y) = ny^{n-1}/\theta^n$ $0\leq y\leq \theta$ and zero otherwise
My answer: CDF of $Y = (Y/\theta)^n$.
My question is:
how do we prove that the maximum likelihood estimator of θ is biased. and what would be an unbiased esitimater of θ. What im able to calculate is θ/n+1
Best Answer
$$ \begin{align} E(\hat{\theta}) & = E(Y) \\ & = \int_{0}^{\theta}yf_Y(y)dy \\ & = \int_{0}^{\theta}y \frac{ny^{n-1}}{\theta^n}dy \\ & = \int_{0}^{\theta} \frac{ny^{n}}{\theta^n}dy\\ & = \frac{n}{\theta^n}\int_{0}^{\theta} y^n dy\\ & = \frac{n}{\theta^n}\cdot\frac{1}{n+1}y^{n+1}|_0^{\theta} \\ & = \frac{n}{\theta^n}\cdot\frac{1}{n+1}(\theta^{n+1}-0) \\ & = \frac{n}{\theta^n}\cdot\frac{1}{n+1}\theta^{n+1} \\ & = \frac{n}{n+1}\theta \end{align} $$
Therefore, the estimator is biased.
However, if we multiply by $\frac{n+1}{n}$, the estimator would be unbiased.
So, $\frac{n+1}{n}\hat{\theta}$ is an unbiased estimator.
This indicates that $\frac{n+1}{n}\max{(x_i)}$ is unbiased.