# Find $P(X_1 > \max(X_2, X_3)| X_1, X_2, X_3 \geq d)$

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

• 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 $\frac13$. – Did Feb 8 at 9:14

# 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 $$\begin{multline*} \mathbb{P}(X_{1}\wedge X_{2}\wedge X_{3}\geq d)=\mathbb{P}(X_{1}\geq d)\mathbb{P}(X_{2}\geq d)\mathbb{P}(X_{3}\geq d)\\ =\left(1-\mathbb{P}(X_{1} Assuming the r.v.s admit probability densities, $$\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, $$\mathbb{P}\left(X_{1}>X_{2}\vee X_{3}\mid X_{1}\wedge X_{2}\wedge X_{3}\geq d\right)=\frac{1}{3}.$$

• Do u mean $\mathbb{P}(X_{1}\wedge X_{2}\wedge X_{3}\geq d)$? – d.k.o. Feb 7 at 18:29
• @d.k.o.: +1 yes I did. The dangers of copy and paste! – parsiad Feb 7 at 18:44
• @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. – parsiad Feb 8 at 6:43
• By a simple symmetry argument, the answer $\frac13$ is much more general than you make it... – Did Feb 8 at 9:15
• @jhon_wick: You can look at the edit history of the question. – parsiad Feb 11 at 17:11