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, $P(X_1\ge X_2 \cap X_1\ge X_3)=p_{12}p_{13}$.


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

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

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?

  • $\begingroup$ Since the random variables are independent can we just multiply P(X1>=X2) times P(X1>=X3)? Or am I missing something. $\endgroup$
    – Glen
    Commented Mar 14, 2018 at 15:38
  • $\begingroup$ @Glen, I added some more details of my thinking. I am not sure where things went wrong. $\endgroup$ Commented Mar 14, 2018 at 15:45
  • $\begingroup$ $X_1 \ge X_2$ is not independent of $X_1 \ge X_3$; if one event is true, then it's more likely that $X_1$ happens to be a large draw from its distribution, so it's more likely that $X_1 \ge X_3$. I don't know if there's a nice way to compute this; you might have to do something like define $Y = \max(X_2, X_3)$, find its distribution, then find $\Pr(X_1 \ge Y)$ (which is the same event). Or maybe there's a nice way to compute $\Pr(X_1 \ge X_2 \mid X_1 \ge X_3)$, but I don't know one offhand. $\endgroup$
    – Danica
    Commented Mar 14, 2018 at 15:50
  • 1
    $\begingroup$ @MartijnWeterings The question formerly said "standard normal." Did you intentionally remove that, @AbrahamChin? $\endgroup$
    – Danica
    Commented Mar 14, 2018 at 17:01
  • 1
    $\begingroup$ But if X_1, X_2 and X_3 are standard normals then this knowledge is sufficient to calculate the P(X_1 > max(X_2,X3)). We do not need to make a relation with P(X1 > X2) and P(X1 > X3). $\endgroup$ Commented Mar 14, 2018 at 17:04

1 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 $p_{13}$.

  • $\begingroup$ This answer and the link you offered bring me a different way of thinking of the derivation of the probability. In this way, $p_{12}$ and $p_{13}$ are indeed not necessary. Thanks for your timely help, which addresses this problem which confused me for days. $\endgroup$ Commented Mar 15, 2018 at 5:30

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