Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $X_1;X_2;\cdots;X_n$ are iid Weibull(1;$\beta$).

1) Find the mgf of $\bar{X}$.

2) Use the mgf to and E[$\bar{X}$].

3) Find the VAR[$\bar{X}$] by any method.

4) Suppose the parameter $\beta$ now has a gamma($\alpha$,$\theta$) distribution. Find E[$\bar{X}$] and VAR[$\bar{X}$].

5) Finally, suppose the parameter $\theta$ in the distribution of $\beta$ has a lognormal($\mu, \sigma^2$) distribution. Find E[$\bar{X}$].

For 1-3, I took the moment generating functions and multiplied them to get $[E(X^n)]^n = [B^n\Gamma(1+n)]^n$. My n is given as 1 for each Weibull so I have $[\beta^1]^n=\beta^n$. For the Variance, I took $E(\bar{X^2}) - [E(\bar{X})]^2 = 2\beta^{2n} - \beta^{2n} = \beta^{2n}$.

For 4, I take $E(\bar{X}) = E[E(\bar{X}|\beta)] = E(\beta^n) = \beta^n$. Then $V(\bar{X}) = E[V(\bar{X|\beta})] + V[E(\bar{X}|\beta) = E(\beta^{2n})+ V(\beta^n)$. This doesn't seem right thought. I am also getting similar stuff for 5. Any help is greatly appreciated.

For 5,

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.