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
Best Answer
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, $$ \mathbb{P}\left(X_{1}>X_{2}\vee X_{3}\mid X_{1}\wedge X_{2}\wedge X_{3}\geq d\right)=\frac{1}{3}. $$