Question

If $X, X_1, X_2$ are independent and identically distributed positive random variables with distribution function $F_X(x)$ then $\int_{0}^{\infty} 2 \cdot x \cdot \overline{F}_X^2(x) dx$ equals

• A. $E(\min(X_1, X_2))$
• B. $E(\max(X_1, X_2))$
• C. $E(\min^2(X_1, X_2))$
• D. $E(\max^2(X_1, X_2))$

Hint:

The survival function of the minimum of $n$ i.i.d. variables is $[\overline{F}(x)]^n$. For the maximum, the CDF is $[F(x)]^n$. Always keep the "Integral of Survival Function" method in mind for non-negative random variable expectations.

Answer 1

The Correct Option is C

We are dealing with a property of expected values for non-negative random variables using the survival function $\overline{F}(x) = 1 - F(x)$.

Step 1: \color{redExpected Value Formula for Non-negative Variables
For any non-negative random variable $Y$, the expected value of $Y^k$ can be expressed as:
$E(Y^k) = \int_{0}^{\infty} k \cdot y^{k-1} P(Y > y) dy$
In this question, $k=2$, so we are looking at something related to $E(Y^2)$.

Step 2: \color{redDefining the Minimized Variable
Let $Y = \min(X_1, X_2)$.
Since $X_1$ and $X_2$ are independent and identically distributed (i.i.d.), the survival function of $Y$ is:
$P(Y > x) = P(\min(X_1, X_2) > x) = P(X_1 > x \text{ and } X_2 > x)$
$P(Y > x) = P(X_1 > x) \cdot P(X_2 > x) = [1 - F_X(x)]^2 = \overline{F}_X^2(x)$

Step 3: \color{redCalculating the Second Moment of Y
Using the formula from Step 1 with $k=2$ for the random variable $Y$:
$E(Y^2) = \int_{0}^{\infty} 2 \cdot x^{2-1} P(Y > x) dx$
$E(Y^2) = \int_{0}^{\infty} 2 \cdot x \cdot \overline{F}_X^2(x) dx$

Step 4: \color{redFinal Identification
The integral given in the question matches the expression for $E(Y^2)$.
Since $Y = \min(X_1, X_2)$, then $E(Y^2) = E(\min^2(X_1, X_2))$.
Thus, the expression equals $E(\min^2(X_1, X_2))$.