Let X1, X2, ... be a sequence of independent and identically distributed continuous random variables. Find

a) $P\left\{X_6>X_1\mid X_1=\max \left(X_1,\dots ,X_5\right)\right\}$

b)$P\left\{X_6>X_2\mid X_1=\max \left(X_1,\dots ,X_5\right)\right\}$

The value of$P\left\{X_6>X_1\mid X_1=\max \left(X_1,\dots ,X_5\right)\right\}$

Let $X_1,X_2.....$be continuous random variables with independent and identical distributions.

Conditional probability is defined as follows:

$\begin{array}{r}P\left\{X_6>X_1\mid X_1=max\left(X_1,\dots ,X_5\right)\right\}=\frac{P\left\{X_6>X_1,X_1=max\left(X_1,\dots ,X_5\right)\right\}}{P\left\{X_1=max\left(X_1,\dots ,X_5\right)\right\}}\\ =\frac{P\left\{\left(X_6>X_1\right)\cap \left[X_1=max\left(X_1,\dots ,X_5\right)\right]\right\}}{P\left\{X_1=max\left(X_1,\dots ,X_5\right)\right\}}\\ \approx \frac{P\left\{\left(X_6=max\left(X_1,\dots ,X_6\right)\right)\cap \left[X_1=max\left(X_1,\dots ,X_5\right)\right]\right\}}{P\left\{X_1=max\left(X_1,\dots ,X_5\right)\right\}}\\ \approx \frac{P\left(X_6=max\left(X_1,\dots ,X_6\right)\right)\times P\left(X_1=max\left(X_1,\dots ,X_5\right)\right)}{P\left\{X_1=max\left(X_1,\dots ,X_5\right)\right\}}\\ =\frac{\left(\frac{1}{6}\right)\left(\frac{1}{5}\right)}{\left(\frac{1}{5}\right)}\\ =5\cdot \left(\frac{1}{6}\times \frac{1}{5}\right)\\ =\frac{1}{6}\end{array}$

Hence$P\left\{X_6>X_1\mid X_1=\max \left(X_1,\dots ,X_5\right)\right\}\text{is}\frac{1}{6}$

The value of$P\left\{X_6>X_2\mid X_1=\max \left(X_1,\dots ,X_5\right)\right\}$

Given: $P\left\{X_6>X_2\mid X_1=\max \left(X_1,\dots ,X_5\right),X_6>X_1\right\}$

Calculation: Let $X_1,X_2.....$be continuous random variables with independent and identical distributions.

Where $X_6>X_1$

$P\left\{X_6>X_2\mid X_1=\max \left(X_1,\dots ,X_5\right),X_6>X_1\right\}=1$

By symmetry,

$P\left\{X_6>X_2\mid X_1=\max \left(X_1,\dots ,X_5\right),X_6<X_1\right\}=\frac{1}{2}$

From part (a),

$P\left\{X_6>X_1\mid X_1=\max \left(X_1,\dots ,X_5\right)\right\}=\frac{1}{6}$

Therefore

$\begin{array}{r}P\left\{X_6>X_2\mid X_1=max\left(X_1,\dots ,X_5\right)\right\}=\frac{1}{6}+\left(\frac{1}{2}\times \frac{5}{6}\right)\\ =\frac{7}{12}\end{array}$

Hence$P\left\{X_6>X_2\mid X_1=\max \left(X_1,\dots ,X_5\right)\right\}=\frac{7}{12}$