Expected value of order statistics for uniform distribution I have $X$ ~ $U(0,1)$ interval.
Let n=2, i.e. $X_1 &lt; X_2$
I have to calculate the expected value of ${X_2}^{m/(1-m)}$.
Where, $0≤m≤1$
I want to confirm if I have calculated it correctly?
$$\int_0^1 2x^{2m/(1-m)} \,dx = \frac{2(1-m)}{(1+m)}$$
Is the calculation correct?
Thanks!
I guess $m &lt; 1$. Otherwise $x^{1/(m-1)} \to \infty$ as $x \to 0$.
There's a possibility that I made a mistake, so let me know if you spot one.
Let's first compute the cumulative distribution function of $X_2$:
$$
F_{X_2}(a) = \Pr(X_2 \le a) = \Pr(X_1 \le a \text{ and } X_2 \le a) = \Pr(X_1 \le a)\Pr(X_2 \le a) = a^2.
$$
This assumes that the two uniformly random variables are independent.
Then the density is given by:
$$
f_{X_2}(a) = \frac{\partial F_{X_2}(a)}{\partial a} = 2 a.
$$
The expected value of $X_2^{\frac{m}{1-m}}$ is then:
$$
\int_0^1 a^{\frac{m}{1-m}} f_{X_2}(a) da = \int_0^1 a^{\frac{m}{1-m}} 2 a\, da,\\
=\int_0^1 2 a^{\left(\frac{m}{1-m} + 1\right)} da = \int_0^1 2 a^{\frac{1}{1-m}} da = 2 \frac{1-m}{2-m}
$$