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Let $\gamma_s(x)=\frac{1}{\sqrt{2\pi s}}\exp(\frac{-x^2}{2s})$ be a gaussian density with variance $s$ and zero mean. Is the integral $$\int_{\{x\geq y\geq 0\}\subset\mathbb{R}^2}\gamma_s(x)\gamma_t(y)dxdy$$ something that can be computed? i am particlarly interested in the case $t=1-s$.

I was asked for context, so the context is trying to do part 3 of this exercise brings up this integral ( $B_t$ is a brownian motion, $S_1$ is its supremum over $0\leq t\leq 1$)

Namely $$\{T\leq s\}=\{sup_{t\leq s}B_t-B_s\leq sup_{1\geq t\geq s}B_t-B_s\}$$ And the two sides of the latter inequality are distributed as (the absolute value of) independent Gaussian variables with variances $s$ and $1-s$. So this gives rise to the integral that interests me. I continue to get close votes, so i'm not sure how much more context you require.

famous mortimer
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  • *Hint:* The integral is the probability that for $X\sim N(0,s)$ and $Y\sim N(0,t),,$ both independent, $$ \mathbb P\big{0\le Y\le X\big}=\mathbb P\Big{0\le Y,0\le X-Y\Big},. $$ Writing $Z=X-Y$ gives a random variable that has variance $s+t$ and $$ \operatorname{Cov}(Z,Y)=-t,. $$ Then think about the CDF of the bivariate normal distribution. – Kurt G. Jul 06 '24 at 12:32
  • @KurtG. thank you. your hint led me to this post https://math.stackexchange.com/questions/255368/px0-y0-for-a-bivariate-normal-distribution-with-correlation-rho. Funnily enough the second answer converts the question back to my question and then solves it with polar coordinates. – famous mortimer Jul 06 '24 at 18:15
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    Yes. Funny sometimes. Neat result, isn't it? – Kurt G. Jul 06 '24 at 18:43

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