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.

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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