Question

Let X1, X2, X3 be independent and identically distributed continuous random variables. Compute

(a) P{X1 > X2|X1 > X3};

(b) P{X1 > X2|X1 < X3};

(c) P{X1 > X2|X2 > X3};

(d) P{X1 > X2|X2 < X3}

Short answer

a.$P\{X_1>X_2|X_1<X_3\}=\frac{2}{3}$

b. $P\{X_1>X_2|X_1<X_3\}=\frac{1}{3}$

c. $P\{X_1>X_2|X_2>X_3\}=\frac{1}{3}$

d.$P\{X_1>X_2|X_2<X_3\}=\frac{2}{3}$

Step-by-step solution

Content Introduction

If all Xi are mutually independent and have (or belong to) the same distribution, we say that random variables $X_1,X_2,...,X_n$ are all independent and identically distributed.

Explanation (Part a)

We have that,

$$\begin{gathered} P\{X_1>X_2|X_1<X_3\}=\frac{P(X_1>X_2,X_1>X_3)}{P(X_1>X_2)} \\ =\frac{P(X_3<X_2<X_1)+P(X_2<X_3<X_1)}{P(X_1>X_2)} \\ =\frac{2.{\displaystyle \frac{1}{6}}}{{\displaystyle \frac{1}{2}}} \\ =\frac{2}{3} \end{gathered}$$

Explanation (Part b)

$$\begin{gathered} P\{X_1>X_2|X_1<X_3\}=\frac{P(X_1>X_2,X_1<X_3)}{P(X_1<X_3)} \\ =\frac{P(X_2<X_1<X_3)}{P(X_1<X_3)} \\ =\frac{{\displaystyle \frac{1}{6}}}{{\displaystyle \frac{1}{2}}} \\ =\frac{1}{3} \end{gathered}$$

Explanation (Part c)

$P\{X_1>X_2|X_2>X_3\}=\frac{P(X_1>X_2,X_2>X_3)}{P(X_2>X_3)}$

$$\begin{gathered} =\frac{P(X_3<\hspace{0.17em}{X}_2<X_1)}{P(X_2>X_3)} \\ =\frac{{\displaystyle \frac{1}{6}}}{{\displaystyle \frac{1}{2}}} \\ =\frac{1}{3} \end{gathered}$$

Explanation (Part d)

$P\{X_1>X_2|X_2<X_3\}=\frac{P(X_1>X_2,X_2<X_3)}{P(X_2<X_3)}$

$$\begin{gathered} =\frac{P(X_2<X_3<X_1)+P(X_2<X_1<X_3)}{P(X_2<\hspace{0.17em}{X}_3)} \\ =\frac{2}{3} \end{gathered}$$