1
$\begingroup$

If $x_1,x_2,x_3$ and $x_4$ are independent identically distributed random variable with continuous distribution .what is the probability of
$$P=p(x_1 < x_2< \max(x_3,x_4))$$ I know the exhaustive number of cases are $p(x_1<x_2<x_3)+p(x_1<x_2<x_4)$ and every $x$ are iid so both have same probabiltiy so is $2p(x_1<x_2<x_3)$ is right? But i get the $p(x_1<x_2<x_3) = 1/6$ and then my answer is $1/3$ but the answer is $1/6.$ Can we write $P$ as $p(x_1<x_2<x_3<x_4)+p(x_1<x_2<x_4<x_3)$?

$\endgroup$
2
  • 1
    $\begingroup$ I did type-set as math, but what you write is not fully comprehensible. What you mean by iid? Should there be = before 1/6? $\endgroup$ Feb 17, 2020 at 15:39
  • 2
    $\begingroup$ @emacsdrivesmenuts : "iid" is a standard and universally known abbreviation in probability theory. $\endgroup$ Feb 17, 2020 at 15:43

2 Answers 2

3
$\begingroup$

The two events $x_1 < x_2 < x_3$ and $x_1<x_2<x_4$ are not mutually exclusive, and the events $x_1<x_2<x_3<x_4$ and $x_1<x_2<x_4<x_3$ are not exhaustive. That is where your attempts at solution go wrong.

One can look at it like this: \begin{align} & x_1 < x_2 < \max \end{align}

  • Either $x_3$ or $x_4$ could be $\max,$
  • and then the one of those two that is not $\max$ could be greater than $x_2$ or between $x_2$ and $x_1$ or less than $x_1.$

I.e. choose one of two alternatives, then one of three. So there are six possibilities.

So six different orders favor the event $x_1<x_2 < \max\{x_3,x_4\}{:}$ \begin{align} & x_1 < x_2 < x_3 < x_4 \\ & x_1 < x_2 < x_4 < x_3 \\ & x_1 < x_3 < x_2 < x_4 \\ & x_1 < x_4 < x_2 < x_3 \\ & x_3 < x_1 < x_2 < x_4 \\ & x_4 < x_1 < x_2 < x_3 \end{align} Since all $24$ orders are equally probable in this situation, the probability is therefore $6/24 = 1/4 = 0.25.$

$\endgroup$
1
  • $\begingroup$ Can i answer this question by taking 1-p(max(x3,x4)<x2<x1) by breaking the event into two parts p(max(x3,x4)<x2) and p(x2<x1). Also, can anybody provide the answer by taking all mutually exclusive event of p(max(x3,x4)<x2<x1) $\endgroup$ May 6, 2020 at 19:50
0
$\begingroup$

$$p(x_1\lt x_2\lt\max(x_3,x_4))=p(x_1\lt x_2\lt x_3\text{ OR }x_1\lt x_2\lt x_4)$$ $$=p(x_1\lt x_2\lt x_3)+p(x_1\lt x_2\lt x_4)-p(x_1\lt x_2\lt x_3\text{ AND }x_1\lt x_2\lt x_4)$$ $$=p(x_1\lt x_2\lt x_3)+p(x_1\lt x_2\lt x_4)-p(x_1\lt x_2\lt x_3\lt x_4\text{ OR }x_1\lt x_2\lt x_4\lt x_3)$$ $$=\frac1{3!}+\frac1{3!}-\frac1{4!}-\frac1{4!}=\frac14.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .