Consistent estimator - bias and variance calculations

Question

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.

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 $\bar{Y}_n$.