Expected Value for variables $x_1<x_2<x_3$

expected valueprobabilityprobability distributions

Q) There are three-sample values $x_1, x_2$ and $x_3$ following P.D.F $f(x_i) = 2x_i, (0 < x_i<1)$ in $(0,1)$

Find the $E(X)$ for $X=x_2$ (Here the $x_1 < x_2 < x_3$ and $x_i$ are independent variables.)

Here is my attempt.

Since the $x_i$ are independent, $f(x_1, x_2, x_3) = 8x_1x_2x_3$

Say the $f(x_2)$ be P.D.F only for one variable $x_2(=X)$

Then $f(x_2) = \int_{x_2} ^1 \int_{0}^{x_2} f(x_1, x_2, x_3) dx_1 dx_3 = 2X^3(1-X^2)$

Hence, $E(X) = \int_0 ^1 Xf(X) dX = {4 \over 35}$

But the answer was $24 \over 35$

What the point do I have a mistake? I can't find what I've missed.

Thanks.

Best Answer

First, let's be careful with our notation. It's totally possible to define $x_1, x_2, x_3$ as independent random variables with the given density function. But if you do that, you can't enforce the condition that $x_1 < x_2 < x_3$, because this contradicts independence. Instead, the right idea is to define the order statistics, which are three other variables:

\begin{align*} x_{(1)} & := \min\{x_1, x_2, x_3\} \\ x_{(3)} & := \max\{x_1, x_2, x_3\} \\ x_{(2)} & := \text{(the other one)} \end{align*}

(It's possible to write down a definition for $x_{(2)}$, of course, but it's not really worth the effort as long as the meaning is clear.) Keeping the notation straight is important; the $x_i$ variables have the given density function and are independent of one another. The $x_{(i)}$ variables, though, are very much dependent on each other, and they do not have the original density function anymore. Intuitively, it should be clear that $x_{(1)}$ is much more likely to be found in, say, the interval $[0, 0.1]$ than $x_{(3)}$ is. These new variables are called the order statistics, and there is a long body of literature on how to deal with these.

The line in your solution that I don't understand is the line where you say

Then $f(x_2) = \int_{x_2} ^1 \int_{0}^{x_2} f(x_1, x_2, x_3) dx_1 dx_3 = 2X^3(1-X^2)$

and as far as I can tell, you may be making an error in conflating the difference between $x_2$ and $x_{(2)}$. You may have noticed that your answer was off by a factor of $6$; this is not an accident, and the factor of $3!$ comes from the ways to rearrange the three variables.

Hopefully I've said enough at this point to identify your mistake and point you on the right track; please let me know if not.

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