# Does the second moment estimator of the uniform distribution parameter have the same properties as that of the first moment?

For independent and identically distributed samples $[y_1,...,y_m]$ where $y$ is uniformly distributed between $[0,\theta]$ with $0 \lt \theta \lt \infty$, finding the method of moments estimator for $\theta$ is very straightforward using the first moment, $E[Y]$, and its natural estimator.

I am interested in knowing whether it possible to use the natural estimator for the second moment, $\frac{1}{m}\sum_{i=1}^{m}y_i^2$, to arrive a the same estimate of $\theta$ and whether or not its properties are the same as the estimator using the first moment.

Since $Var(Y) = \frac{\theta^2}{12} = E[Y^2] - E^2[Y] = E[Y^2] - \frac{\theta^2}{4}$ we have that $E[Y^2] = \frac{\theta^2}{3}$.

Then using the method of moments theory and the known support for $\theta$, the invertible function operating on $E[Y^2]$ to produce $\theta$ is $\sqrt{3E[Y^2]}$. Because we do not know the exact value of $E[Y^2]$, we use its natural estimator given above to produce the estimate. $$\hat{\theta} = \sqrt{\frac{3}{m}\sum_{i=1}^{m}y_i^2}$$

I would also like to show whether or not this is biased, but I am not sure if the way I would like to go about it is correct. I have that... $$E[\hat{\theta}] = \sqrt{\frac{3}{m}} E\left[\sqrt{\sum_{i=1}^{m}y_i^2}\right]$$

$$\sqrt{\frac{3}{m}}E\left[\sqrt{\sum_{i=1}^{m}y_i^2}\right] = \sqrt{\frac{3}{m}}\sqrt{\frac{m\theta^2}{3}} = \theta$$

Particularly, I am unsure of whether the following equality holds which I see as being the only way that the estimator could be unbiased. $$E\left[\sqrt{\sum_{i=1}^{m}y_i^2}\right] = \sqrt{\frac{m\theta^2}{3}}$$

• An easier task would be to address the issue of whether the MM estimator of $\theta^2$ is biased. The answer to that will give you clear information about whether or not your $\hat \theta$ is biased.
– whuber
Dec 1, 2015 at 2:14
• Well where $\hat{\mu}_2$ is the MM estimator for $E[Y^2] = \frac{\theta^2}{3}$, $E[\hat{\mu}_2] = E\left[\frac{1}{m} \sum_{i=1}^{m}Y_i^2\right]= \frac{1}{m} \sum_{i=1}^{m}E[Y_i^2] = \frac{1}{m} \sum_{i=1}^{m}E[Y^2] = E[Y^2]$ so the estimator for $\theta^2$ is unbiased. Yet its still not clear to me how this helps in handling the square root in the expectation for $\hat{\theta}$. Dec 1, 2015 at 4:03
• If $\hat\theta$ were unbiased, then you would have complete information about its expectation and the expectation of its square. Use this information to compute the variance of $\hat\theta$, then draw a conclusion.
– whuber
Dec 1, 2015 at 15:46

There's a particular result known as Jensen's inequality which relates $E(g(X))$ to $g(E(X))$ (Mathworld, Wikipedia)
Alternatively, can you show that $E[X^2]-E(X)^2\geq 0$?
Can you then see a way to show that if $E[Y^2]=\theta^2\!/3$, then the estimator you consider must be biased?