Consider Rayleigh distribution
$$f(x; \theta) = \dfrac{1}{\theta} e^{\dfrac{-x^2}{(2\theta)}}, \ \ \ x > 0 \ \ \text{and} \ \ \theta > 0$$
- Show that $E(X^2) = 2\theta$
- On the basis of the proceeding item, construct an unbiased estimator $\hat{\theta}_1$ for $\theta$ based on i.i.d sample $X_1,...,X_n$ for this model
My solutions:
- I have used substitution by parts and u-substitution:
$E(X^2) = \int_0^x x^2 \frac1\theta e^{\frac{-x^2}{(2\theta)}} dx = [e^{\frac{-x^2}{2\theta}}(x^2-2\theta)]_0^x = e^{\frac{-x^2}{2\theta}}(x^2-2\theta) + 2\theta $
- our professor provided us with an answer that $\hat{\theta}_1 = \frac12 \frac1n \sum_{i=1}^n \chi_i^2 $
Questions:
- What am i doing wrong when calculating the $E(X^2)$
- I am not even sure how to approach this problem. I guess i should be using either MSE or the property that $E(\theta)= \theta$ for unbiased estimators.
My mathematical background is rather weak, stating the obvious is very welcome.
There are a couple of things going on here. First of all, the Rayleigh distribution is given by $$f(x,\theta) = \frac{\color{red}{x}}{\theta} \cdot e^{-\frac{-x^2}{2\theta}},$$ so there is a factor of $x$ missing in your stated definition. Furthermore, the expect expected value of $X^2$ can be calculated by $$E(X^2) = \int_{0}^{\color{red}{\infty}} y^2 \cdot f(y,\theta) dy = 2\theta.$$ Something went wrong with your integration bounds there. I think it is sometimes good practice to give your integration variable a different name than ones you used before in the same problem to avoid confusion (I used $y$ here instead of $x$).
For the second part, notice now that $\frac{1}{2} X^2$ is an unbiased estimator for the parameter $\theta$, since its expected value is exactly $\theta$. Using linearity of the expected value we find that $$\hat{\theta} := \frac{1}{n} \sum_{i=1}^n \frac{1}{2}X_i^2$$ also has an expected value of $\theta$, which means that it is an unbiased estimator for $\theta$.