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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?

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    $\begingroup$ What do you know about $X-Y$ ? $\endgroup$ – Raskolnikov Aug 3 '12 at 9:55
  • $\begingroup$ If $X > Y$ what can you say about $X-Y$? $\endgroup$ – Ilya Aug 3 '12 at 9:58
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    $\begingroup$ 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). $\endgroup$ – Davide Giraudo Aug 3 '12 at 10:05
  • $\begingroup$ Yes, do we have a name for this? $\endgroup$ – Strin Aug 3 '12 at 10:50
  • $\begingroup$ A name for what? $\endgroup$ – Chris Eagle Aug 3 '12 at 11:54
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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: $$ \mathbb{P}(X>Y) = \mathbb{P}(Z>0) = 1- \Phi\left(-\frac{\mu}{ \sigma}\right) = \Phi\left(\frac{\mu}{ \sigma}\right) = \frac{1}{2} \operatorname{erfc}\left(-\frac{\mu}{\sqrt{2}\sigma}\right) $$

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    $\begingroup$ I believe the $\sqrt{2}$ should not appear in the standard CDFs, at the last line. $\endgroup$ – Arthur Jun 3 '18 at 22:16
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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

$$P(Z>0)=\int_0^\infty\frac{1}{\sqrt{2\pi(\sigma_0^2+\sigma_1^2)}}\exp\left(\frac{-(z-x_0+y_0)^2}{2(\sigma_0^2+\sigma_1^2)}\right)\mathrm{d}z$$

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 $$P(Z>0)=\frac{1}{2}\operatorname{erfc}\left(\frac{y_0-x_0}{\sqrt{2(\sigma_0^2+\sigma_1^2)}}\right)$$

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    $\begingroup$ 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. $\endgroup$ – Did Aug 3 '12 at 10:54
  • $\begingroup$ Ok I found the mistake. Updating. $\endgroup$ – Seyhmus Güngören Aug 3 '12 at 11:30
  • $\begingroup$ 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? $\endgroup$ – Dilip Sarwate Aug 3 '12 at 15:46
  • $\begingroup$ Yes it has to be erfc of course. It is obvious in the text. Proof is correct. $\endgroup$ – Seyhmus Güngören Aug 3 '12 at 16:26

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