Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For two Gaussian-distributed variables, $ Pr(X=x) = \frac{1}{\sqrt{2\pi}\sigma_0}e^{-\frac{(x-x_0)^2}{2\sigma_0^2}}$ and $ Pr(Y=y) = \frac{1}{\sqrt{2\pi}\sigma_1}e^{-\frac{(x-x_1)^2}{2\sigma_1^2}}$. What is probability of the case X > Y?

share|improve this question
What do you know about $X-Y$ ? –  Raskolnikov Aug 3 '12 at 9:55
If $X > Y$ what can you say about $X-Y$? –  Ilya Aug 3 '12 at 9:58
Are $X$ and $Y$ independent? And I don't agree when you write $Pr(X=x)=\dots$: the probability that a Gaussian random variable take a particular value is $0$ (but we can write $P(X\in A)=\int_A$ of the function you wrote). –  Davide Giraudo Aug 3 '12 at 10:05
Yes, do we have a name for this? –  Strin Aug 3 '12 at 10:50
A name for what? –  Chris Eagle Aug 3 '12 at 11:54

2 Answers 2

I assume that $X$ and $Y$ are independent. Let $Z=X-Y$ then $Z\sim\cal{N}(x_0-y_0,\sigma_0^2+\sigma_1^2)$. Accordingly


if we use the complementary error function $$\operatorname{erf}c(x)=\frac{2}{\sqrt\pi}\int_x^\infty e^{-t^2}dt$$ with $t=\frac{z-x_0+y_0}{\sqrt{2(\sigma_0^2+\sigma_1^2)}}$, we have $\sqrt{2(\sigma_0^2+\sigma_1^2)}dt=dz$ $$P(Z>0)=\frac{2}{2\sqrt{\pi}\sqrt{2(\sigma_0^2+\sigma_1^2)}}\int_{t=\frac{y_0-x_0}{\sqrt{2(\sigma_0^2+\sigma_1^2)}}}^\infty e^{-t^2}\sqrt{2(\sigma_0^2+\sigma_1^2)}dt$$ and we get finally $$P(Z>0)=\frac{1}{2}\operatorname{erfc}\left(\frac{y_0-x_0}{\sqrt{2(\sigma_0^2+\sigma_1^2)}}\right)$$

share|improve this answer
Which has the odd feature of not being $1/2$ when $x_0=y_0$. I suggest to review this answer, especially the change of variable. –  Did Aug 3 '12 at 10:54
Ok I found the mistake. Updating. –  Seyhmus Güngören Aug 3 '12 at 11:30
Well, there still seems to be something wrong.The odd feature noted by @did still persists: the probability is not $1/2$ when $x_0 = y_0$ and worse yet, the right side is negative when $x_0 > y_0$ and so definitely cannot be a probability. Did you mean to write erfc instead of erf in your final answer? –  Dilip Sarwate Aug 3 '12 at 15:46
Yes it has to be erfc of course. It is obvious in the text. Proof is correct. –  Seyhmus Güngören Aug 3 '12 at 16:26

Suppose $X$ and $Y$ are jointly normal, i.e. no independence is needed. Define $Z = X - Y$. It is well known that $Z$ is Gaussian, and thus is determined by its mean $\mu$ and its variance $\sigma^2$. $$ \mu = \mathbb{E}(Z) = \mathbb{E}(X) - \mathbb{E}(Y) = \mu_1 - \mu_2 $$ $$ \sigma^2 = \mathbb{Var}(Z) = \mathbb{Var}(X) + \mathbb{Var}(Y) - 2 \mathbb{Cov}(X,Y) = \sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2 $$ where $\rho$ is the correlation coefficient. Now: $$ \mathbb{P}(X>Y) = \mathbb{P}(Z>0) = 1- \Phi\left(-\frac{\mu}{\sqrt{2} \sigma}\right) = \Phi\left(\frac{\mu}{\sqrt{2} \sigma}\right) = \frac{1}{2} \operatorname{erfc}\left(-\frac{\mu}{\sqrt{2}\sigma}\right) $$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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