# Unbiased Estimator for Uniform Distribution

$X_1$ , a sample size 1 is drawn from a uniform distribution over $[0,\theta]$. Find an unbiased estimator for the variance of the population. Find a function for $X_1$, $\tau(X_1)$ such that $E(\tau(X_1))=\theta^2/12$.

I know the expected value of this uniform distribution can be found $$\int_{0}^{\theta} x(\frac{1}{\theta}) dx$$. Would I just be looking for some x such that $$\int_{0}^{\theta} x(\frac{1}{\theta}) dx=E(\tau(X_1))=\theta^2/12$$

• Is this from a course or a textbook? If so, please add the [self-study] tag & read its wiki. – gung - Reinstate Monica Mar 5 '15 at 21:52
• Dan, your last sentence is wrong. You would want $\tau$ in that integral (via the law of the unconscious statistician). Let's be ultra naive -- we have a statistic $X_1$ whose expectation is a multiple of $\theta$. What would be a first, wild guess at one whose expectation might be proportional to $\theta^{\,2}$? (We don't expect to guess right ... but we might guess close enough to figure out something better.) – Glen_b -Reinstate Monica Mar 5 '15 at 22:04
• Once you have made a (fairly obvious) guess, work out its actual expectation, and go from there. – Glen_b -Reinstate Monica Mar 5 '15 at 22:11

You're not looking for some $x$ that satisfies that equation at the end. Indeed, $x$ is a dummy variable in the equation (you could write $u$ for $x$ everywhere in the integral without altering the value), so it won't even appear in the equation after you evaluate the integral.
The question simply seeking a function of $X_1$ whose expectation is the variance of the distribution, which is $\theta^2/12$.
By inspection, $E(X_1)=\theta/2$, so no linear function of $X_1$ would do. The obvious "guess" at a functional form is a multiple of $X_1^2$. So evaluate $E(X_1^2)$ and see if you get some multiple of $\theta^2$, and if you do, then see what you'd multiply it by to get the correct variance.