Question.

Suppose that $\displaystyle Y_1, ... Y_n$ is a random sample from a population with density $\displaystyle f_Y(y)=\frac{2y}{\theta^2}$ for $\displaystyle 0<y\leq\theta$, where $\displaystyle \theta>0$.

(a) is $\displaystyle Y_{(n)}$ a sufficient statistic for $\displaystyle \theta$?

(b) Derive the density of the last order statistic $\displaystyle Y_{(n)}$.

(c) Find an unbiased estimator of $\displaystyle \theta$ that is a function of $\displaystyle Y_{(n)}$.

(d) By considering the condition that a function of $\displaystyle Y_{(n)}$ is unbiased for $\displaystyle \theta$, determine whether there is a better unbiased estimator for $\displaystyle \theta$.

I've just started stat inference and my confidence is so low, I feel like threading on seashells... basically I worked through most of proofs but I don't see how I can apply them. This is a sample exam question (to which I don't have an answer). So I'd appreciate your pointers.

Answer.

(a) sufficiency of $\displaystyle Y_{(n)}$

Should I apply factorisation theorem here to factorise the density ofYinto two parts: one is the function of $\displaystyle Y_{(n)}$ and $\displaystyle \theta$, another is a function ofyonly?

Another thought, given the fact that y is limited by $\displaystyle \theta$, could it be that y somehow depends on $\displaystyle \theta$ and therefore the highest order statistic $\displaystyle Y_{(n)}$ cannot be a sufficient statistic simply because of this dependency?

(b) Density for $\displaystyle Y_{(n)}$

find anti-derivative of $\displaystyle f_Y(y)$

$\displaystyle F_Y(y)=\frac{2}{\theta^2}\frac{y^2}{2}, 0<y\leq\theta$ - for a single observation

$\displaystyle F_{Y_{(n)}}(y)=P(Y_{(n)}{\leq}y)=[F_Y(y)]^n=(\frac{y^2}{\theta^2})^n, 0<y\leq\theta$

$\displaystyle f_{Y_{(n)}}(y)=\frac{d}{dy}F_{Y_{(n)}}(y)=n(\frac{ y^2}{\theta^2})^{n-1}\frac{2y}{\theta^2}$

(c) unbiased estimator of $\displaystyle \theta$ that is a function of $\displaystyle Y_{(n)}$

I cannot 'see' it just from looking at the distribution, so I was thinking to use the definition of the unbiased estimator, but the integration seems like a puzzle.

Define a function of $\displaystyle Y_{(n)}: g(Y_{(n)})$. Then

$\displaystyle E(g(Y_{(n)})=\theta$ according to the definition of an unbiased estimator.

Then I would try to find $\displaystyle g(Y_{(n)}$ form the equation

$\displaystyle E(g(Y_{(n)})=\int_0^{\theta}g(Y_{(n)})f_{Y_{(n)}}( y)dy=\theta$

If I do that, I get an integral of a product of $\displaystyle g(Y_{(n)})$ and $\displaystyle y^{2n-1}$, on the left side, and an expression involving theta and n on the right side - is it solvable?

Any advice? Thanks!