# Expected value of a third order statistic?

1. Mar 11, 2009

### JamesF

Hi all. I'm struggling with this HW question. I've searched through the textbook and on the web and have been unable to find a solution

1. The problem statement, all variables and given/known data
I've got 4 i.i.d. random variables, X1, X2, X3, X4. Uniformly distributed on [0,1]
so the pdf = 1
and cdf F(x_i) = x_i

Let Y3 = the third highest value of the 4 variables.

What is E(Y3)? I know the answer is 2/5. But I can't figure out how to show this.

2. Relevant equations

I know the formula for E(Y2) (second highest) and E(Y1) (highest), but I cannot find anything about the expectation of the third highest.

$$E(Y_1) = \int_{0}^{1} y \cdot 4y^3 \, dy = 4/5$$
$$E(Y_2) = \int_{0}^{1} y \cdot (12y^2 - 12y^3) \, dy = 3/5$$

based on the pattern, one would assume that E(Y3) = 2/5 (plus I know that's the answer), but I can't derive the formula.

3. The attempt at a solution
I've tried a bunch of different approaches, but they've all been dead ends. Maybe I'm misunderstanding the problem. Or maybe just misunderstanding the concept of order statistics entirely.

2. Mar 11, 2009

### JamesF

alright I figured out my problem. I had it backwards. Our teacher uses (Y1) to denote the highest value of the 4 random variables, but the stuff I found on the web used Y1 to denote the LOWEST value.

so $$f_{Y_3} = \frac{4!}{1! \cdot 2!} y \cdot (1-y)^2 \cdot 1$$

integrate to find the expected value, and we get

$$E(Y_3) = \int_{0}^{1} y \cdot (12y(y-1)^2) dy = \frac{2}{5}$$

tahdah!