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

$A=\dfrac{1}{2}$

$B=\dfrac{1}{3}$

$C=\dfrac{1}{4}$

$D=\dfrac{1}{6}$

My attempt:

$P(X_3<X_2<\max(X_1,X_4))=F_{X_{2}}(\max(X_1,X4))-F_{X_{2}}(X_3)=P(X_{2}<\max(X_1,X_{4}))-P(X_{2}<X_{3}))=1-P(\max(X_1,X_{4})<X_2)-P(X_{2}<X_{3}))=1-P(X_1<X_2)P(X_{4})<X_2)-P(X_{2}<X_{3}))=1-P(X_1<X_2)^2-P(X_{2}<X_{3}))$

I am blank now. No distribution is provided I don't know what I am missing here.

I know the fact that $Z=F_{X}(x) \sim U(0,1)$ but how do I use that here?

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Ignoring equalities (which happen with probability $0$ since this is a continuous distribution) there are $4!=24$ equally likely orders

Of these, half $(12)$ have $X_3>X_2$ and can be excluded, while a quarter $(6)$ have $X_2$ as the largest value and can also be excluded.

Every other case has $X_3<X_2<\max(X_1,X_4)$ and so $$\mathbb P(X_3<X_2<\max(X_1,X_4)) = 1-\frac12-\frac14 = \frac14.$$

The six orders which meet the condition are

  • $X_4 < X_3 < X_2 < X_1$
  • $X_3 < X_4 < X_2 < X_1$
  • $X_3 < X_2 < X_4 < X_1$
  • $X_3 < X_2 < X_1 < X_4$
  • $X_3 < X_1 < X_2 < X_4$
  • $X_1 < X_3 < X_2 < X_4$

so we can also say $$\mathbb P(X_3<X_2<\max(X_1,X_4)) = \frac{6}{24}= \frac14.$$

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  • $\begingroup$ One thing how do I calculate the number of times my case matches? How did you get 6 ? I mean its count but how do I get it mathematically $\endgroup$
    – Daman
    Jan 8, 2021 at 11:32
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    $\begingroup$ @Daman $24-12-6=6.$ $\endgroup$
    – Henry
    Jan 8, 2021 at 11:34

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