# Solved – 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.

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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\$?

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