0
$\begingroup$

Suppose I have three random variables $X_1$, $X_2$, and $X_3$ that are i.i.d $N(0,1)$ distributed. The chance that $X_3$ is larger than $X_2$, given that $X_3$ is also larger than $X_1$ should equal $\frac{2}{3}$ in theory. However, when trying to calculate this conditional probability, I always end up with an probability equal to 1, which obviously is not correct. Perhaps you could point out where my mistake is?

$\Pr[X_3 > X_2 \mid X_3 > X_1] = \Pr[X_2 < X_1]\Pr[X_3 > X_2 \mid X_3 > X_1, X_2 < X_1]+\Pr[X_2 \geq X_1]\Pr[X_3 > X_2 \mid X_3 > X_1, X_2 \geq X_1]$.

I then rewrite the last term as

$\Pr[X_2 \geq X_1]\Pr[X_3 > X_2 \mid X_3 > X_1, X_2 \geq X_1] = \Pr[X_2 \geq X_1]\frac{\Pr[X_3 > X_2]}{\Pr[X_3 > X_1]}$.

This gives me

$\Pr[X_3 > X_2 \mid X_3 > X_1] = 0.5 * 1 + 0.5\frac{0.5}{0.5} = 1 \neq \frac{2}{3}$.

$\endgroup$
0
$\begingroup$

The easiest thing is to look at it combinatorically. If you rank the numbers $X_i$ by 1,2 and 3 according to their magnitude (3 is the largest), then you have 6 ranked combinations of ($X_1,X_2,X_3$) in total: (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2) and (3,2,1).

The ones with $X_3>X_1$: (1,2,3),(1,3,2) and (2,1,3).

Out of these two combinations also have $X_3>X_2$: (1,2,3) and (2,1,3).

Hence your probability $P(X_3>X_2\mid X_3>X_1)$: 2/3

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.