Question

Let $X_1, X_2, X_3, X_4$ be i.i.d. random variables having a continuous distribution. Then $P(X_3 < X_2 < \max(X_1, X_4))$ equals

• A. $\dfrac{1}{2}$
• B. $\dfrac{1}{3}$
• C. $\dfrac{1}{4}$
• D. $\dfrac{1}{6}$

Hint:

For problems involving i.i.d. continuous random variables, use symmetry arguments since all orderings are equally probable.

Answer 1

The Correct Option is C

Since the random variables are i.i.d. with continuous distribution, by symmetry, all orderings of the four values are equally likely.

There are $4! = 24$ possible orderings of $X_1, X_2, X_3, X_4$.

Analyzing the event: $X_3 < X_2 < \max(X_1, X_4)$

This means:

• $X_3 < X_2$
• $X_2 < \max(X_1, X_4)$

The second condition means $X_2$ is less than at least one of $X_1$ or $X_4$ (or both).

Equivalently: $X_2 < X_1$ OR $X_2 < X_4$ (or both).

Method: Count favorable orderings

We need orderings where:

1. $X_3 < X_2$
2. Either $X_2 < X_1$ OR $X_2 < X_4$ (or both)

Let's denote the positions in the ordering from smallest to largest as positions 1, 2, 3, 4.

Systematic counting:

For the event to occur, we need:

• $X_3$ in a lower position than $X_2$
• At least one of $X_1$ or $X_4$ in a higher position than $X_2$

Favorable cases:

• If $\max(X_1, X_4) = X_1$: then $X_3 < X_2 < X_1$ with $X_4$ anywhere
• If $\max(X_1, X_4) = X_4$: then $X_3 < X_2 < X_4$ with $X_1$ anywhere

Count orderings where $X_3 < X_2 < X_1$:

• Choose 3 positions out of 4 for these three: $\binom{4}{3} = 4$ ways
• Each choice has 1 ordering (since order is fixed)
• $X_4$ fills the remaining position
• Total: 4 orderings

Similarly, orderings where $X_3 < X_2 < X_4$: 4 orderings

But we need to avoid double counting when $X_3 < X_2 < X_1$ AND $X_3 < X_2 < X_4$, which means $X_3 < X_2 < \min(X_1, X_4)$.

Using inclusion-exclusion or direct counting:

The total number of orderings where $X_3 < X_2 < \max(X_1, X_4)$ is 6.

Probability: $$P = \frac{6}{24} = \frac{1}{4}$$

Answer: (C) $\frac{1}{4}$