Expectation of Order Statistic
Let $X$ be a random variable with probability density function (pdf) given by $f\left( x\mid\alpha, \beta\right) = \frac{\alpha\beta\exp \left[ \alpha\left( 1-1/x^{\beta}\right) \right] }{x^{1+\beta}}, \alpha >0, \beta>0, 0<x<1.$
Clearly, its cdf is given by $F\left( x\mid\alpha, \beta\right) =\exp \left[ \alpha\left( 1-1/x^{\beta}\right) \right] . $
Let $X_{(1)}\leq X_{(2)}\leq \cdots \leq X_{(n)}$ denotes the order statistic of a random sample $X_{1}, X_{2}, \cdots, X_{n }$ from this continuous population with cdf $F(x) $ and pdf $f(x) . $
I want to find the $ k-\mathrm{th} $ moment of $ X_{(j)}, $ i.e $ E\left( X_{(j)}^{k}\right) . $ Help will be appreciated.
Initially, we need to find the pdf of $ X_{(j)}. $ I tried and got the following as the pdf:- $f_{X_{j}}(x)=\frac{n!}{(j-1)!(n-j)!}\frac{\alpha \beta}{x^{1+\beta}}\exp \left[ -\alpha j\left( \frac{1}{x^{\beta}}-1\right) \right]\left\lbrace 1-\exp \left[ -\alpha \left( \frac{1}{x^{\beta}}-1\right) \right]\right\rbrace ^{n-j} $
Is this pdf of $ X_{(j)} $ which I have obtained correct?
Now to evaluate the $ k-\mathrm{th} $ moment of $ X_{(j)}$, I need to the expectation of $X_{(j)}^{k}; $ i.e. $ E\left( X_{(j)}^{k}\right)$ . Now,
$$ E\left( X_{(j)}^{k}\right)= \int_{0}^{1} x^{k}f_{X_{j}}(x)dx $$
When I integrate the above, it get very complicated and what is worse, I found that if $ k $ is even, the expectation turns out to be zero. I believe that this is incorrect because how can the expectation of a non-negative random variable turn out to be zero? Or is it that I am missing something?
I shall appreciate if I am helped.