1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Since the random variables are independent can we just multiply P(X1>=X2) times P(X1>=X3)? Or am I missing something. $\endgroup$ – Glen Mar 14 '18 at 15:38
  • $\begingroup$ @Glen, I added some more details of my thinking. I am not sure where things went wrong. $\endgroup$ – Guoyang Qin Mar 14 '18 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. $\endgroup$ – Dougal Mar 14 '18 at 15:50
  • 1
    $\begingroup$ @MartijnWeterings The question formerly said "standard normal." Did you intentionally remove that, @AbrahamChin? $\endgroup$ – Dougal Mar 14 '18 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). $\endgroup$ – Sextus Empiricus Mar 14 '18 at 17:04
0
$\begingroup$

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 $p_{13}$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Guoyang Qin Mar 15 '18 at 5:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.