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Let $X_1, X_2, X_3$ be exponential random variables with rates $\lambda_1, \lambda_2, \lambda_3$. We have the following

$$\Pr(X_1 < X_2 \mid \max(X_1, X_2, X_3) = X_3) = \frac{\lambda_1 + \lambda_3}{\lambda_1 + \lambda_2 + 2 \lambda_3}$$

Which I found by considering $f(1, 2, 3) = \Pr(X_1 < X_2 < X_3)$ and noticing that the above is equal to $\frac{f(1, 2, 3)}{f(1, 2, 3) + f(2, 1, 3)}$. However, what is a more intuitive explanation of this result? It looks like the probability that $Y_1 < Y_2$, if $Y_1$ is exponential with rate $\lambda_1 + \lambda_3$ and $Y_2$ is exponential with rate $\lambda_2 + \lambda_3$.

MT_
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  • Did you perhaps leave out some condition from this question? – Glen_b Mar 21 '17 at 02:12
  • @Glen_b Don't think so, but it's possible -- why? – MT_ Mar 21 '17 at 02:28
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    (Assuming the variables are independent as is often intended when only marginals are mentioned) Uh, $X_1 < X_2 \land \max(X_1, X_2,X_3)=X_3$ is equivalent to $X_1<X_2<X_3$, so your result does not hold - for example if $\lambda_1=\lambda_2=\lambda_3$ the probability should be $1/6$, not $1/2$. Are you perhaps talking about the conditional probability $Pr(X_1 < X_2 \mid \max(X_1,X_2,X_3)=X_3)$? – Juho Kokkala Mar 21 '17 at 06:18
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    @JuhoKokkala Oh you're right. I put a comma instead of a conditional -- my mistake! Please unclose the question :) – MT_ Mar 22 '17 at 01:44

1 Answers1

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We have that

$$\Pr[X_1 < X_2 \mid \max(X_1, X_2, X_3) = X_3] =\frac{\Pr[X_1 < X_2 , \max(X_1, X_2, X_3) = X_3]}{\Pr[\max(X_1, X_2, X_3) = X_3]}$$

$$=\frac{\Pr[X_1 < X_2 < X_3]}{\Pr[X_1<X_3, X_2<X_3) ]}$$

$$=\frac{\Pr[X_1 < X_2 < X_3]}{\Pr[X_1<X_2<X_3]+\Pr[X_2<X_1<X_3]} \quad(1)$$

This looks "intuitive": given $X_3$ is the largest, we are left with two other options (that appear in the denominator) and we have the one that interests us in the numerator. As for calculation,

Let

$$f_i(x_i) = \lambda_ie^{-\lambda_ix_i},\;\;\; i=1,2,3$$

Then, assuming that the three variables are independent,

$$\Pr[X_1 < X_2 < X_3] = \int_0^{\infty}f_3(x_3)\int_0^{x_3}f_2(t)\int_0^{t}f_1(s)ds\,dt\,dx_3$$

etc, and noting that the distribution function can be written $F_i = 1- \frac 1{\lambda_i}f_i$. A quick Monte Carlo verifies that the calculation of the probability by the OP is correct.

Moreover, the OP correctly noted that the obtained probability is equal to $\Pr(Y_{1|3} < Y_{2|3})$, if $Y_{1|3}$ is Exponential with rate $\lambda_1 + \lambda_3$ and $Y_{2|3}$ is Exponential with rate $\lambda_2 + \lambda_3$.

Well the following is a well known result for i.i.d Exponentials

$$\min\{X_1,X_3\} = Y_{1|3}, \min\{X_2,X_3\}=Y_{2|3}$$

So we are looking at

$$\Pr [\min\{X_1,X_3\} < \min\{X_2,X_3\}] = \Pr[X_1 < X_2 \mid \max(X_1, X_2, X_3) = X_3]$$

...which makes sense, if we start pondering what options does the left-hand-side allows for.

Alecos Papadopoulos
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  • OK, yes, I guess the solution that I mention in the question body is perhaps "intuitive" by itself, once you do the calculations. I guess I was more wondering if one could bridge the connection between this probability and the $Y_1 < Y_2$ probability (or a similar, simpler one) mentioned in my post. – MT_ Mar 22 '17 at 14:21
  • This answer seems unrelated to the question - OP has already computed the probability, using (1) (though it could perhaps be explained more clearly in the question). – Juho Kokkala Mar 22 '17 at 19:09
  • The derivation of the formula helps a reader understand better. Then I argue that this is a very intuitive formula already. The last part which provides the integrals permits anyone interested (the OP included) to verify the actual value of this probability – Alecos Papadopoulos Mar 22 '17 at 19:19
  • Re the first part: ok, though (1) appears in OP, too, and the derivation is imo trivial (but I see this is subjective. Re the second part: the probability that $X_1$ is smallest is $\lambda_1/(\lambda_1+\lambda_2+\lambda_3)$, and the probability that $X_ 2 < X_3$ conditional on "both are greater than $X_1$" is $\lambda_2/(\lambda_2+\lambda_3)$, just multiply these. – Juho Kokkala Mar 22 '17 at 19:24
  • @JuhoKokkala Are you saying that the required probability is $$\frac {\lambda_1\lambda_2}{(\lambda_2+\lambda_3)\cdot (\lambda_1+\lambda_2+\lambda_3)}$$? – Alecos Papadopoulos Mar 22 '17 at 19:28
  • No. I am saying $f(1,2,3)=P(X_1<X_2<X_3) = \frac{\lambda_1,\lambda_2}{(\lambda_2+\lambda_3),(\lambda_1+\lambda_2+\lambda_3)}$ (note the denominator is different) where $f(1,2,3)$ is OP's notation, and plugging this and $f(2,1,3)$ into $(1)$ leads to the expression in OP (unless I made a mistake in the manipulations that happened to lead to the same result). But none of this answers the question which (I thought!) asks why the final result has this form (especially why it coincides with $P(Y_1<Y_2)$ – Juho Kokkala Mar 22 '17 at 19:33
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    @JuhoKokkala Ok, now I understand. Well, the "ratio" formula obtained by the OP (and replicated in my answer as (1)) is pretty intuitive in my opinion. Why it so happens that we would obtain the same result with two Exponentials that "combine" the characteristics of the first three, as the OP (correctly) writes, hmmm, I will give it some more thought. – Alecos Papadopoulos Mar 22 '17 at 19:39
  • For what it's worth, @JuhoKokkala is right about the method I used to calculate the probability. – MT_ Mar 23 '17 at 05:39