# Conditional probability of independent & identically distributed random variables

For $$X_i$$ that are independent and identically distributed, compute: $$A=P(x_6>x_2|x_1= \max [x_1,x_2,x_3,x_4,x_5])$$

I have tried using the Law of total probability where

$$P(x_6>x_2)$$ $$= \sum_{i=1}^5 P(x_6>x_2|x_i=\max [x_1, … ,x_5])\cdot P(x_i=\max [x_1,…,x_5])$$

But I’m unsure how to calculate the value of the terms in order to deduce the required value $$A$$.

• Are there further information such that $X_i$'s are continuous random variable? Feb 18, 2022 at 3:26

I assume $$P(x_i = x_j) = 0$$ (e.g., when the $$x_i$$ are continuous random variables).

Each ordering of the six random variables are equiprobable (with probability $$P(x_1 < x_2 < x_3 < x_4 < x_5 < x_6) = 1/6!$$). So just count permutations.

$$P(x_6 > x_2 \mid x_1 = \max\{x_1, \ldots, x_5\}) = \frac{P(x_6 > x_2 ,x_1 = \max\{x_1, \ldots, x_5\})}{P(x_1 = \max\{x_1, \ldots, x_5\})} = \frac{3!(2+3+4+5)/6!}{6 \cdot 4! / 6!} = \frac{7}{12}.$$

e.g., for the denominator, there are $$4!$$ ways to put $$x_1, \ldots, x_5$$ in order such that $$x_1$$ is the largest; then there are $$6$$ places to position $$x_6$$ anywhere amongst the others. For the numerator I had to do some casework based on where $$x_2$$ is positioned among $$x_2, \ldots, x_5$$.

When $$P(x_i = x_j) \ne 0$$ (e.g., when the $$x_i$$ have a discrete component), then we need more information to answer the question.

• Thanks for your answer. In my own working I’ve got close to your answer but I seem to be getting a different multiple of 3! on the numerator. Would it be possible for you to clarify how you worked yours out? Feb 18, 2022 at 3:59
• I seem to be getting a final answer of $5/6$ because my numerator is $2(3!)(4+3+2+1)$ Your answer is correct but I’m not sure where I’ve gone wrong. Feb 18, 2022 at 4:29
• An alternate way to count the numerator's event is to partition on the placement of $x_1$: as the second or first largest. $$\{x_2<x_6, x_1=\max[x_1,..,x_5]\} \\=\\ \{\max[x_2,...,x_5]<x_1<x_6\}\cup\{x_2<x_6, \max[x_2,...,x_6]<x_1\}$$ Feb 18, 2022 at 4:40
• @GrahamKemp I tried to do this but my answer still ended up the same. Because $x_1$ can be in either position 5 or 6, I multiply my answer by 2 to reflect the fact that this doubles the number of outcomes where the constraints hold. Feb 18, 2022 at 5:08
• @FD_bfa Following GrahamKemp's partition into the two cases: the first case is $4!$ (just need to permute $x_2, \ldots, x_5$) and the second case is $5!/2$ (permute $x_2, \ldots, x_6$, but throw away half the cases where $x_2 > x_5$). This yields a total of $84$. Feb 18, 2022 at 5:11