What is $P(X_1\ge X_2 \cap X_1\ge X_3)$ given $P(X_1\ge X_2)$ and $P(X_1\ge X_3)$?

Question

Suppose random variables $X_1$, $X_2$, and $X_3$ are independent and normally distributed, and $P(X_1\ge X_2)=p_{12}$ and $P(X_1\ge X_3)=p_{13}$. So what is $P(X_1\ge X_2 \cap X_1\ge X_3)$?

Using multiplication rule for conditional probability, I have

$P(X_1\ge X_2 \cap X_1\ge X_3)=P(X_1\ge X_3)P(X_1\ge X_2 | X_1\ge X_3)=p_{13}P(X_1\ge X_2 | X_1\ge X_3)$.

The question goes to what the probability of $P(X_1\ge X_2 | X_1\ge X_3)$ is. My initial idea is that the event $X_1\ge X_2$ is independent from the event $X_1\ge X_3$, so that $P(X_1\ge X_2 | X_1\ge X_3)=P(X_1\ge X_2)=p_{12}$. Thus, $P(X_1\ge X_2 \cap X_1\ge X_3)=p_{12}p_{13}$.

Similarly,

$P(X_2\ge X_1 \cap X_2\ge X_3)=p_{21}p_{23}$

$P(X_3\ge X_1 \cap X_3\ge X_2)=p_{31}p_{32}$

And, it is obvious that the sum of the three exclusive and exhaustive probabilities should be equal to $1$. But $p_{12}p_{13}+p_{21}p_{23}+p_{31}p_{32}$ is not necessarily $1$ (i.e. $p_{12}=0.5$, $p_{13}=0.7$, and $p_{23}=0.4$, due to the sum's being $1$, $p_{21}=0.5$, $p_{31}=0.3$, and $p_{32}=0.6$, thus $p_{12}p_{13}+p_{21}p_{23}+p_{31}p_{32}=0.5\times 0.7+0.5\times 0.4+0.3\times 0.6=0.73\neq 1$.

It seems the premise that "the event $X_1\ge X_2$ is independent from the event $X_1\ge X_3$" doesn't hold. So how could I get the probability if considering them having dependent relationship?

Answer 1

Suppose $X_i \sim \mathcal N(\mu_i, \sigma_i^2)$ for each $i = 1, 2, 3$.

Let $Y = X_2 - X_1$ and $Z = X_3 - X_1$. Then the event $X_1 \ge X_2 \cap X_1 \ge X_3$ is equivalent to the event $Y \le 0 \cap Z \le 0$.

Note that $$ \begin{bmatrix}Y \\ Z\end{bmatrix} = \begin{bmatrix}X_2 - X_1 \\ X_3 - X_1\end{bmatrix} = \begin{bmatrix}X_2\\ X_3\end{bmatrix} - \begin{bmatrix}1 \\ 1\end{bmatrix} X_1 ,$$ and since we have $$ \begin{bmatrix}X_2\\ X_3\end{bmatrix} \sim \mathcal N\left( \begin{bmatrix}\mu_2\\\mu_3\end{bmatrix}, \begin{bmatrix}\sigma_2^2 & 0\\0 & \sigma_3^2\end{bmatrix} \right) $$ and $$ \begin{bmatrix}1\\ 1\end{bmatrix} X_1 \sim \mathcal N\left( \begin{bmatrix}\mu_1\\\mu_1\end{bmatrix}, \begin{bmatrix}\sigma_1^2 & \sigma_1^2\\ \sigma_1^2 & \sigma_1^2\end{bmatrix} \right) $$ we then know that $$ \begin{bmatrix}Y \\ Z\end{bmatrix} \sim \mathcal N\left( \begin{bmatrix}\mu_2 - \mu_1 \\ \mu_3 - \mu_1\end{bmatrix}, \begin{bmatrix} \sigma_2^2 + \sigma_1^2 & \sigma_1^2\\ \sigma_1^2 & \sigma_3^2 + \sigma_1^2 \end{bmatrix} \right) .$$ The probability of the event you want is then just the CDF of the above bivariate normal, evaluated at $(0, 0)$.

I don't think it has any nice relationship to $p_{12}$ or $p_{13}$.