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?

  • $\begingroup$ Thanks for adding teh tag. Sorry about the malformed link. Here's the link to the tag wiki $\endgroup$
    – Glen_b
    Mar 5, 2015 at 22:50
  • $\begingroup$ 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 $f_Y(y)$? $\endgroup$ Mar 5, 2015 at 22:51
  • $\begingroup$ Do you know how to compute the expectation of $Y_1$? $\endgroup$
    – Glen_b
    Mar 5, 2015 at 22:52
  • $\begingroup$ Are you familiar with the linearity of expectation? Specifically, are you aware that $E(X+Y)=E(X)+E(Y)$ and $E(aX)=aE(X)$? $\endgroup$
    – Glen_b
    Mar 5, 2015 at 22:54
  • $\begingroup$ Yes, I am familiar with that. How can this help? $\endgroup$ Mar 5, 2015 at 23:02

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.