Unbiased Estimators

So I've been banging my head against the wall trying to figure out where to go with these problems, and I'm looking for a little direction.

Suppose that $Y_1, Y_2, Y_3$ is a random sample where the density of each random variable $Y_i$ is given by $f_Y(y) = 2 \theta ^{-2} y,\ 0 \leq y \leq \theta$ for some parameter $\theta > 0$.

Show that $\hat{\theta}_1 = \frac{3}{2}\cdot\overline{Y}$ is an unbiased estimator of $\theta$.

Show that $\hat{\theta}_2 = \frac{7}{6}\cdot\max\{Y_1,Y_2,Y_3\}$ is an unbiased estimator of $\theta$.

Which of the two unbiased estimators given is preferable for the estimation of $\theta$? Why?

• Thanks for adding teh tag. Sorry about the malformed link. Here's the link to the tag wiki Mar 5 '15 at 22:50
• I know that $B(\hat{\theta}) = E(\hat{\theta}) - \theta$. I'm not sure about computing the expected value of $\bar{Y}$ though. I believe it has something to do with the indefinite integral of our $f_Y(y)$? Mar 5 '15 at 22:51
• Do you know how to compute the expectation of $Y_1$? Mar 5 '15 at 22:52
• Are you familiar with the linearity of expectation? Specifically, are you aware that $E(X+Y)=E(X)+E(Y)$ and $E(aX)=aE(X)$? Mar 5 '15 at 22:54
• Yes, I am familiar with that. How can this help? Mar 5 '15 at 23:02

So that this has an answer:

OP got to here in comments:

$\hat{\theta}_1 = \frac{Y_1+Y_2+Y_3}{2}$

1. Apply expectation to both sides and use the facts $E(X+Y)=E(X)+E(Y)$ and $E(aX)=aE(X)$ to simplify it in terms of expectations of $Y_i$.

2. Compute $E(Y_i)$.

3. Apply the definition of bias of an estimator to compute the bias.

For the second estimator you need the distribution of the maximum (the third order statistic).

There are formulas for the distributions of order statistics (which are easy to discover for yourself if you don't know them). See for example, here:

Distribution of extremal values

However, you can also do this by elementary methods.

$P(\max(Y_1,Y_2,Y_3)\leq y)=P(Y_1\leq y)P(Y_2\leq y)P(Y_3\leq y)$ from which the distribution of the maximum (and hence ts expectation) can be computed.