Biased/Unbiased estimator

Question

I'm trying to solve a statistic exam and i got lost with this exercise.

1) Consider a sample from a continuos probability distribution with density:

$$ f(x) = \begin{cases} (1+\theta x)/8 & \text{if }-4<x<4, \\ 0 & \text{otherwise.} \end{cases} $$

a) Is $\bar\theta = \frac{3}{16}X$ (this $X$ is the sample median) an unbiased estimator of $\theta$?

Where should i start? Should i integrate the function, and compare the expected value of it with the expected value of the estimator? How do i integrate if i don't know which value should i assign to $\theta$? (should i integrate leaving it as a constant?)

I hope you understand, i translated the exercise, maybe some term are different in english.

Answer 1

We calculate the mean of the estimator $\frac{3}{16}X$. This is $\frac{3}{16}$ times the mean of $X$.

The mean of $X$ is $$\int_{-4}^{4} x\frac{1+\theta x}{8}\,dx.$$ Integrating $\frac{1}{8}(x+\theta x^2)$ is straightforward. The definite integral turns out to be $\frac{16}{3}\theta$. Multiply by $\frac{3}{16}$, and we do get $\theta$. Thus we do have an unbiased estimator of $\theta$.

Unbiasedness is an important quality in an estimator: it is good if the estimator is on average right. (But there are some instances where a slightly biased estimator has compensating properties that make it attractive.)

On your question about $\theta$, yes, it is an unknown distribution parameter. There are some restrictions on $\theta$, since a density function can never be negative. It turns out that we must have $-\frac{1}{4}\le \theta\le \frac{1}{4}$.

There are many distributions in which there is one or more free parameter. The most important is the normal, with parameters $\mu$ and $\sigma^2$, but there is also the binomial, with parameters $p$ and $n$, and so on. Estimating the value of a parameter is an important statistical task.

Answer 2

Often one uses a lower-case $x$ for the argument to the density function or to the c.d.f., and a capital $X$ to refer to the random variable whose density it is. But you've said your capital $X$ is the sample median, and that suggests a sample size more than $1$. So I am uncertain of the meaning of the question.

If we assume you mean simple a sample of one observation, then one can find an expected value by integrating: $$ \mathbb E X = \int_{-4}^4 xf(x)\,dx = \int_{-4}^4 x\frac{1+\theta x}{8}\,dx $$ $$ = \frac18 \left(\int_{-4}^4 \frac x8\,dx + \theta \int_{-4}^4 x^2\,dx \right) $$

The first integral is readliy seen to be $0$ (you're integrating an odd function over an interval symmetric about $0$). The second one comes to $16\theta/3$.

So $\frac{3}{16}\mathbb E X = \theta$. The expression $\frac3{16}X$ is "observable" i.e. does not depend on $\theta$, and its expected value remains equal to $\theta$ if $\theta$ changes.