Let X_{(1)} < X_{(2)}, < \dots, < X_{(5)} be the order statistics based on a sample of size 5 from a continuous distribution with the probability density function
f(x)= \left\{ \begin{matrix}
\frac{1}{x^2} & 1< x<\infty \\
0 &, otherwise
\end{matrix} \right.Then the sum of all possible values of r \in \{ 1,2,3,4,5 \} for which E(X_{(r}) is finite equals ___________________.
The pdf of X having pdf f(x) if given by:
F(x) = \int_{y=1}^x \frac{1}{y^2}dy = \left[ -\frac{1}{y} \right]_{y=1}^x = \frac{x-1}{x}The r-th order statistic (1<r<5) will have the pdf of the form:
f_{X_{(r)}}(x) = \binom 5 r r(F(x))^{r-1} (1-F(x))^{5-r} f(x) \\
=\binom 5 r \left(\frac{x-1}{x}\right)^{r-1} \left(\frac{1}{x}\right)^{5-r} \frac{1}{x^2}= \binom 5 r (x-1)^r x^{-5+r-2}The expectation of X_{(r)} is then given by:
E(X_{(r)}) = \int_{1}^{\infty} x \binom 5 r (x-1)^r x^{-5+r-2}dx \\
= \int_{1}^{\infty} \binom 5 r (x-1)^{r} x^{r-6}dx \\
= \int_{1}^{\infty} \binom 5 r \sum_{j=0}^{r} \binom r j x^{r-j} (-1)^{j} x^{-6+r} \\
= \sum_{j=0}^{r} \binom r j \binom 5 r \int_{1}^{\infty} x^{2r-j-6} (-1)^{j} dx \\Clearly the Expectation exists if:
(2r-j-5) < 0 \\ \implies 2r < j+5 \quad \forall \quad j=1(1)r \\ \implies 2r< r+5 \\ \implies r<5
Thus the required answer is 1+2+3+4 = 10.