Question

Let X1, X2, . . . , Xn represent a random sample from a Rayleigh distribution with pdf

f(x; θ) = (x/θ) * e^x^2/(2θ) for x > 0

(a) It can be shown that E(X2 P ) = 2θ. Use this fact to construct an unbiased estimator of θ based on n i=1 X2 i .

(b) Estimate θ from the following n = 10 observations on vibratory stress of a turbine blade under specified conditions:

16.88 10.23 4.59 6.66 13.68

14.23 19.87 9.40 6.51 10.95

Answer 1

(a) It can be shown that $E\left(X^{2}\right)=2 \theta$.

$$\begin{aligned} E\left(X^{2}\right) &=2 \theta \Rightarrow E\left(\frac{X^{2}}{2}\right)=\theta \\ E(\hat{\theta}) &=E\left(\frac{\sum X_{i}^{2}}{2 n}\right) \quad\left(\because \hat{\theta}=\frac{\sum X_{i}^{2}}{2 n}\right) \\ &=\frac{\sum E\left(X_{i}^{2}\right)}{2 n} \\ &=\frac{\sum 2 \theta}{2 n} \\ &=\frac{2 n \theta}{2 n} \\ &=\theta \end{aligned}$$

Thus, $E(\hat{\theta})=\theta$. That is, $\hat{\theta}$ is an unbiased estimator of $\theta$.

(b) Estimate $\theta$ from the following $\mathrm{n}=10$ observations on vibratory stress of a turbine blade under specified conditions.

$$\begin{aligned} \hat{\theta} &=\frac{\sum X_{i}^{2}}{2 n} \\ &=\frac{\left(\begin{array}{l} (16.88)^{2}+(10.23)^{2}+(4.59)^{2}+(6.66)^{2}+(13.68)^{2}+ \\ (14.23)^{2}+(19.87)^{2}+(9.40)^{2}+(6.51)^{2}+(10.95)^{2} \end{array}\right)}{2(10)} \\ &=\frac{1490.106}{20} \\ &=74.50529 \end{aligned}$$

Thus, $\hat{\theta}=74.50529$.