# Expected value of order statistics for uniform distribution

I have $$X$$ ~ $$U(0,1)$$ interval.

Let n=2, i.e. $$X_1 < X_2$$

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 $$m < 1$$. 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: $$\int_0^1 a^{\frac{m}{1-m}} f_{X_2}(a) da = \int_0^1 a^{\frac{m}{1-m}} 2 a\, da,\\ =\int_0^1 2 a^{\left(\frac{m}{1-m} + 1\right)} da = \int_0^1 2 a^{\frac{1}{1-m}} da = 2 \frac{1-m}{2-m}$$

• This looks perfectly great! I just wanted to ask one last thing. When calculating the pdf, do we always take the original form? Like here X is U(0,1). But we need to calculate ${X_n} ^{m/(1-m)}$. So, even if X is raised to power something, we calculated pdf of X and not X raised to the power, is this correct Apr 29 at 12:11
• Yes. If $X$ is a rv then the expected value of $g(X)$ is given by $\int g(x) f_X(x) dx$ where $f_X$ is the pdf of $X$. In your case, $g(x) = x^{m/(1-m)}$.
– tdm
Apr 29 at 12:18
• And sorry, I forgot to mention that m is between 0 to 1. I edited the question. Your assumption, m<1 is correct. Apr 29 at 12:18
• I can't thank you enough! You cleared all my doubts Apr 29 at 12:22