Working through some homework problems for a Mathematical Statistics course and I'm having a hard time finding good examples in the text to explain some details. This particular problem is:

$Y_1 ... Y_n \sim f_\theta(y) = \frac{6}{\theta^3} y (\theta-y)$

Find a constant a so that $\hat{\theta} = a\overline{Y}$ is unbiased, and show that it is consistent.

I calculated the Method of Moments estimator:

$$ m_1^{\prime} = \frac{6}{\theta^3}\int_0^\infty y^2(\theta - y) dy = \frac{\theta}{2} $$

So, $\hat{\theta}=2\bar{Y} \Rightarrow a=2$

So far so good.

For bias:

$$ B(\hat{\theta}) = E(\hat{\theta}) - \theta $$

$$ B(\hat{\theta}) = E(2\bar{Y}) - \theta $$

$$ B(\hat{\theta}) = 2\mu - \theta $$

And now I'm stuck on what should be obvious. How do I equate $2\mu$ and $ \theta$?

To check consistency, I need bias and variance to be zero, or approach zero as $n \rightarrow \infty$. Bias is above, but I'm stuck calculating variance for $\theta$.

Looking through the text, all the examples and other homework answers make large jumps in logic without any explanation. In cases like this, examples simply state that bias and variance are zero, but there is no work shown for reference.


1 Answer 1


Your density is missing a constraint on the support, namely $0<y<θ$. Indeed, $$\int_0^\theta y(\theta-y)\text{d}y = \dfrac{\theta^3}{2}- \dfrac{\theta^3}{3} = \dfrac{\theta^3}{6}$$ which is the proper normalising constant.

This however does not change the constant $a$ that turns $a\bar{Y}$ into an unbiased estimator: $$\dfrac{6}{\theta^3}\int_0^\theta y^2(\theta-y)\text{d}y = \dfrac{6}{\theta^3}\dfrac{\theta^4}{3}- \dfrac{\theta^4}{4} = \dfrac{6}{\theta^3}\dfrac{\theta^4}{12}=\dfrac{\theta}{2}$$ Which implies that$$\mathbb{E}_\theta[2\bar{Y}_n] = \theta$$is unbiased.

For consistency, you do not need to prove that bias and variance to be zero (bias is indeed zero since the estimator is unbiased) but that $2\bar{Y}_n$ converges almost surely to $theta$. Given that $\bar{Y}_n$ is an average, you should look at the Law of Large Numbers for an argument.

Note that computing the variance of $Y$ is not particularly complicated: $$\dfrac{6}{\theta^3}\int_0^\theta y^3(\theta-y)\text{d}y = \dfrac{6}{\theta^3}\dfrac{\theta^5}{4}- \dfrac{\theta^5}{5} = \dfrac{6}{\theta^3}\dfrac{\theta^5}{20}=\dfrac{3\theta^2}{10}$$You should be able to deduce the variance of $\bar{Y}_n$ on your own from there.

A final comment is that, given that the support of the distribution depends on $\theta$, an estimator based on $Y_{(n)}=\max_{1\le i\le n} Y_i$ should be more efficient that $\bar{Y}_n$.

  • $\begingroup$ Thanks for the response and sorry for dropping the constraint. Regarding consistency, consistency you describe is "weak consistency" in the text and "consistent in MSE" is introduced, which is where I got the bias & variance going to zero. For $E[2\bar{Y}] = \theta being unbiased, is this because the first moment calculated here is \mu and the conclusion is implicit in the way it was derived? I'm just stuck on understanding why it is implied to be unbiased. For variance, the integral you show is the second moment, which is \sigma^2, correct? Same arguments as for the first moment then follow? $\endgroup$
    – KirkD_CO
    Nov 8, 2017 at 14:31
  • $\begingroup$ Meaning you won't answer the questions? $\endgroup$
    – KirkD_CO
    Nov 8, 2017 at 21:03
  • $\begingroup$ No, meaning a negative reply to the questions. You have to read in more details about the very notion of unbiasedness, which you miss at the present time. And my second integral is not the variance, check its definition as well. And the variance of $\bar{Y}_n$ follows directly from the variance of $Y_1$, this is a most standard derivation, check again the basic literature on moment estimators. $\endgroup$
    – Xi'an
    Nov 8, 2017 at 21:07
  • $\begingroup$ Thanks for the clarification. Thanks also for all your help here. I truly appreciate. Now, I need to go study... $\endgroup$
    – KirkD_CO
    Nov 9, 2017 at 0:46

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.