Pls take a look, this is an exam practice question but no answer is provided.

Question.

Let $\displaystyle Y_1, ... Y_n$ be a random sample from a distribution with density function

$\displaystyle f_Y(y;\theta)=3\theta^{-3}y^2, 0\leq{y}\leq{\theta}$

$\displaystyle f_Y(y;\theta)=0$ otherwise

where $\displaystyle \theta>0$ is an unknown parameter.

(a) Let $\displaystyle X=max_{1\leq{i}\leq{n}}Y_i$. Find the density function of X.

(b) find a simple sufficient statistic for $\displaystyle \theta$.

(c) find the maximum likelihood estimator $\displaystyle \hat{\theta}$ for $\displaystyle \theta$.

(d) compute the mean square error of the maximum likelihood estimator $\displaystyle \hat{\theta}$.

Answer.

(a) I take it X is the maximum Y in the sample ie $\displaystyle Y_{(n)}$ so I find the density of it

$\displaystyle P(Y_{(n)}\leq{y})=P(Y_1\leq{y},... Y_n\leq{y})=[P(Y\leq{y})}^n=[F_Y(y)]^n=(3\theta^{-3}\frac{y^3}{3})^n=\theta^{-3n}y^{3n}$

$\displaystyle f_X(y)=f_{Y_{(n)}}(y)=\frac{d}{dy}F_{Y_{(n)}}(y)=n f_Y(y)[F_y(y)]^n=n*3\theta^{-3}y^2*\theta^{-3n}y^{3n}=3n\theta^{-3-3n}y^{2+3n}$

(b) to find sufficient statistics for $\displaystyle \theta$, I will use Factorisation theorem, ie factorising the joint density function of the sample

$\displaystyle f_Y(y;\theta)=(3\theta^{-3})^n\prod_{i=1}^nY_i^2I_{Y_{(1)}\geq{0}}(y)I_{Y_{ (n)}\leq\theta}(y)=(3\theta^{-3})^nI_{Y_{(n)}\leq\theta}(y)*\prod_{i=1}^nY_i^2I_ {Y_{(1)}\geq{0}}(y)$

the second component depends on the sample only; the first component depends on $\displaystyle \theta$ and on the sample through $\displaystyle Y_{(n)}(y)$ and can be a sufficient statistic for $\displaystyle \theta$.

?? does this makes sense ??

(c) here is where I am stuck and cannot move on to (d)

To find MLE for $\displaystyle \theta$, I will find maximum of log-likelihood of the density of Y

$\displaystyle L_Y(y;\theta)=(3\theta^{-3})^n\prod_{i=1}^nY_i^2$ subject to a constraint $\displaystyle 0\leq{y}\leq{\theta}$

$\displaystyle l_Y(y;\theta)=ln(L_Y)=ln[(3\theta^{-3})^n]+ln[\prod_{i=1}^nY_i^2]=n\ln(3\theta^{-3})+\Sigma_{i=1}^n\ln{Y_i^2}$

$\displaystyle \frac{\partial}{\partial\theta}(l_Y)=n\frac{1}{3\t heta^{-3}}(-9\theta^{-4})=-\frac{3n}{\theta^{-3}}\theta^{-4}=-\frac{3n}{\theta}$

this can be equal to zero only if n=0 and this is not what I am looking for here...

Can you help me to spot where I've gone wrong?

thanks!