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I have three independent random variables $X_1$, $X_2$ and $X_3$. I am trying to derive the probability that $P(X_1 > \max(X_2, X_3)| X_1, X_2, X_3 \geq d)$.

Here, pdf of $X_1$: $f_{x_1} = \frac{1}{\alpha_1} e^{-\frac{x_1}{\alpha_1}}$. For, $X_2$ and $X_3$: $f_{x_2} = \frac{1}{\alpha_2} e^{-\frac{x_2}{\alpha_2}}$ and $f_{x_3} = \frac{1}{\alpha_3} e^{-\frac{x_3}{\alpha_3}}$, respectively.

I start as: $P(X_1 > X_2) P(X_1 > X_3)$ Then writing $P(X_1 > X_2) $ as: $\int_\gamma^{\infty} \int_{\gamma}^{x_2} f_{x_1, x_2} dx_1 dx_2$. Am I doing it correctly

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    $\begingroup$ 1. If one does not assume a common distribution, one can only write down explicit but untractable formulas. 2. If one does assume a common distribution without atoms, by a symmetry argument, the answer is always $\frac13$. $\endgroup$ – Did Feb 8 at 9:14
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Original question

Let $\wedge\equiv\min$ and $\vee\equiv\max$. First, note that $$ \mathbb{P}(X_{1}>X_{2}\vee X_{3}\mid X_{1}\wedge X_{2}\wedge X_{3}\geq d)=\frac{\mathbb{P}(X_{1}>X_{2}\vee X_{3},\,X_{1}\wedge X_{2}\wedge X_{3}\geq d)}{\mathbb{P}(X_{1}\wedge X_{2}\wedge X_{3}\geq d)}. $$ The denominator is \begin{multline*} \mathbb{P}(X_{1}\wedge X_{2}\wedge X_{3}\geq d)=\mathbb{P}(X_{1}\geq d)\mathbb{P}(X_{2}\geq d)\mathbb{P}(X_{3}\geq d)\\ =\left(1-\mathbb{P}(X_{1}<d)\right)\left(1-\mathbb{P}(X_{2}<d)\right)\left(1-\mathbb{P}(X_{3}<d)\right)\\ =\left(1-F_{X_{1}}(d-)\right)\left(1-F_{X_{2}}(d-)\right)\left(1-F_{X_{3}}(d-)\right). \end{multline*} Assuming the r.v.s admit probability densities, \begin{multline*} \mathbb{P}(X_{1}>X_{2}\vee X_{3},\,X_{1}\wedge X_{2}\wedge X_{3}\geq d)\\ =\mathbb{P}(X_{1}>X_{2}\vee X_{3},\,X_{1}\geq d,\,X_{2}\geq d,\,X_{3}\geq d)\\ =\int_{d}^{\infty}\int_{d}^{\infty}\int_{d\vee x_{2}\vee x_{3}}^{\infty}f_{X_{1}}(x_{1})f_{X_{2}}(x_{2})f_{X_{3}}(x_{3})dx_{1}dx_{2}dx_{3}. \end{multline*}

Common distribution with no atoms

If the r.v.s are i.i.d., then by relabelling them we see that $$ \mathbb{P}\left(X_{1}>X_{2}\vee X_{3}\mid A\right)=\mathbb{P}\left(X_{2}>X_{1}\vee X_{3}\mid A\right)=\mathbb{P}\left(X_{3}>X_{1}\vee X_{2}\mid A\right) $$ where $A\equiv\{X_{1}\wedge X_{2}\wedge X_{3}\geq d\}$. The relabelling argument above is also referred to as a "symmetry" argument. If in addition the r.v.s. admit admit no atoms, these three events partition the sample space (i.e., exactly one has to occur). Therefore, each one has probability $1/3$. In particular, $$ \mathbb{P}\left(X_{1}>X_{2}\vee X_{3}\mid X_{1}\wedge X_{2}\wedge X_{3}\geq d\right)=\frac{1}{3}. $$

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    $\begingroup$ Do u mean $\mathbb{P}(X_{1}\wedge X_{2}\wedge X_{3}\geq d)$? $\endgroup$ – d.k.o. Feb 7 at 18:29
  • $\begingroup$ @d.k.o.: +1 yes I did. The dangers of copy and paste! $\endgroup$ – parsiad Feb 7 at 18:44
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    $\begingroup$ @jhon_wick: Updated. There were a lot of places where I could have made a mistake in the algebra, so read carefully. Regardless, the idea is correct. $\endgroup$ – parsiad Feb 8 at 6:43
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    $\begingroup$ By a simple symmetry argument, the answer $\frac13$ is much more general than you make it... $\endgroup$ – Did Feb 8 at 9:15
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    $\begingroup$ @jhon_wick: You can look at the edit history of the question. $\endgroup$ – parsiad Feb 11 at 17:11

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