# Math Help - Probability Density Function - Unbiased Estimator

1. ## Probability Density Function - Unbiased Estimator

Hi, I don't know how to solve this past exam question, and I can't find any course material that helps whatsoever

Suppose that a random variable X has the probability density function
$f_{X}(x)=\left\{\begin{array}{cc}\frac{e^{{-x/\theta}}}{\theta},& \mbox{} x>0; {} \theta >0;}\\0,& \mbox{} otherwise.\end{array}\right.$

Consider a random sample $X_{1}, X_{2},...,X_{n}$ from this distribution and let $T=k\sum{X_{i}^2}$

Find $k$ so that $T$ is an unbiased estimator of $\theta^2$

Thanks.

p.s. the sum is i=1 to n, i don't know how to latex that

2. $E(T) = E(k \sum(X_i^2)) = k \left( E(X_1^2) + E(X_2^2) + E(X_3^2) +.... +E(X_n^2) \right)$

But Xi are iid so:
$E(T) = nkE(X^2)$

$E(T) = nk \int^\infty_0 x^2 f(x) dx$

integrate then choose the value of k such that $E(T) = \theta^2$

3. thanks, although just one question, why do you multiply the integral part by $x^2$? I thought the estimator was found by just $x$, and $E(X^2)$ was found by $x^2$?

edit- don't worry, just realised i am being blind and it is infact a $E(X^2)$

Thanks a lot!

4. i don't suppose anyone could clarify my answer of k=1/n could they?

5. i should also mention that the pdf you ahve been given is for the exponential distribution.

Depending on the preferences of your professor, you may be allowed to simply look up the fact that $E(X^2) = 2\theta^2$

6. I ended up with $E(T)=nk(\theta ^2)$ meaning i got $E(X^2) = \theta ^2$...damn,

After integrating i got $nk[-x^2e^{-x/\theta}-(\theta xe^{-x/\theta}-(-\theta ^2e^{-x/\theta}))]_{0}^\infty$

so either I can't substitute properly or I integrated wrong? Also, we are only issued the New Cambridge Statistical Tables by Lindley and Scott so we aren't supposed to rely on remembering the distributions.

edit- OMG I think my brain has partially melted, I've been differentiating $x^2$ to just $x$, i feel so stupid right now :P

7. So if $E(T)=nk(2\theta ^2)$ does this mean $k=\frac {1}{2n}$ and that is the answer to the question?

8. yep