# Probability that one random variable is larger than another with known correlation

Let's say I have a normally distributed random variable $X_1$ with known standard deviation $\sigma_1$ and $E[X_1]$ is $0$. Let's say I have another variable with known standard deviation $\sigma_2$ and $E[X_2]$ is $-V$. Let's say the are correlated with $\rho = .9$.

This implies the $\beta$ (defined by $X_2 = \alpha + \beta X_1$) between $X_1$ and $X_2$ is $\beta =\rho \Sigma_2/\Sigma_1$.

This gives a graph that looks like the following....

I want to know the probability that the value $[X_1]$ is less than or equal to $[X_2]$, i.e., $P(X_2 \ge X_1)$.

I tried to conceptualize this as concentric circles on an $X_1-X_2$ plane... but I'm not sure how to combine these variables with a known correlation.

• How are $X_1$ and $X_2$ jointly distributed? If normally, then $Z=X_2-X_1$ is another normal with computable mean and variance, so you know $P(Z\ge 0)$. – A.S. Dec 23 '15 at 2:17

I think I have come up with an answer, and I'd like some feedback.

I used a cholesky decomposition to simulate the correlated random numbers, and my theoretical result appears to match my simulation.

The question $P(X_2 > X_1)$ can be turned into $P(X_2-X_1 > 0)$.

The joint distribution of $X_2-X_1$ then can be summarized as:

$$E[X_1-X_2] = w_1 E(X_1) + w_2 E(X_2)$$

$$\operatorname{Var}[X_2-X_1] = w_1^2 \sigma_1^2 + w_2^2 \sigma_2^2 + 2 w_1 w_2 \sigma_1 \sigma_2 \rho$$

where $\rho$ is the negative of the correlation of the two random variables, because if you think of a portfolio of two assets... we have $X_2 - X_1$, which is going to mean movements in $X_1$ will be the negative of the correlated movements in $X_2$.

For both $w_1$ and $w_2$, I used 1, because each of $X_1$ and $X_2$ have a coefficient of 1 in $F(X_1,X_2) = X_2-X_1$.

Then $P(X_2 - X_1 > 0) = 1-P(X_2 - X_1 < 0)$. And I can use the normal CDF with the given $E[X_2-X_1]$ and $\operatorname{Var}(X_2-X_1)$ to find that value.

Does this sound right? How about the argument for the negative for $\rho$?

• The idea is good but it implicitly assumes $(X_1,X_2)$ have a binormal distribution, whereas your question only states that $X_2$ is "another variable." – whuber Dec 12 '17 at 15:13
• You take $\rho$ negative because $Cor(X1,-X2) = - Cor(X1,X2)$. I think that your expressions would be more easy to follow if you either do not use the terms $w_1$ and $w_2$, or, if you do place them in your equation, then also use 'really' use them by expressing your $X_2 - X_1$ as $w_1 X_1+w_2 X_2$ with $w_2=1$ and $w_1 =-1$ – Sextus Empiricus Dec 12 '17 at 17:06