Let $$f(x,y)=\frac{1}{N_0\pi}\exp(-\frac{(x-a)^2+(y-\frac{b}{2})^2}{N_0})$$ I want to compute $$I=\int_{D_2}f(x,y)d\vec{x}$$ where $D_2$ is the half-plane above the line $2x-y=\frac{a}{4}$: $$D_2=\{(x,y):2x-y<\frac{a}{4}\}$$ where $a\ge 0$ and $N_0\gt0$. Actually this problem comes from a probability problem. $X$ and $Y$ are independent Gaussian random variables and the goal is to determine the probability $$I=\text{Pr}\{(X,Y)\in D_2 \}$$ We can expect that there is no closed-form for $I$, so it should be written in terms of $Q(.)$ function $$Q(z) = \frac{1}{\sqrt{2\pi}}\int_{z}^{+\infty}\exp(-\frac{u^2}{2})du$$ My attempt: I think the region $D_2$ can be described as $$D_2 = \{(x,y): -\infty<x<+\infty , \ \ \ 2x-\frac{a}{4}\le y \lt +\infty\}$$ so we have $$I = \int_{-\infty}^{+\infty}\int_{2x-\frac{a}{4}}^{+\infty}f(x,y)dydx$$ Certainly this can't be expressed as a product of two integrals because the first integration bound depends on $x$. I don't know how to proceed from here. Maybe a change of variables is useful. Mathematica gives the following answer: $$I=\frac{1}{N_0\pi}\times \frac{N_0\pi}{2}\text{erfc}\left(\frac{a\sqrt{5}}{4\sqrt{N_0}}\right) = Q\left(\frac{a\sqrt{10}}{4\sqrt{N_0}}\right).$$
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3What distribution does 2x-y have? Use that directly. – Eric Apr 21 '23 at 16:19
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@Eric Thanks. That leads to the same answer. How can we prove that these are equivalent? I mean $$\text{Pr}{(X,Y)\in D_2 } = \text{Pr}{Z \lt \frac{a}{4} }$$ where $Z = 2X - Y$. – S.H.W Apr 21 '23 at 17:05
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1To do it explicitly, let $z=2x-y$ and $w=x+2y$ (I.e. rotate) and substitute. Dilip’s answer is the intuitive result of doing this. – Eric Apr 22 '23 at 14:26
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@Eric Thanks for the great answer. – S.H.W Apr 22 '23 at 22:42
1 Answers
The joint density of $X$ and $Y$ has circular symmetry about the mean point $\left(a,\frac b2\right)$. You are asked for the probability that $Z = 2X-Y$ is larger than $\frac a4$. So, draw a picture of the $x$-$y$ plane after picking your favorite values of $a$ and $b$ and notice that because of the circular symmetry of the joint density about $\left(a,\frac b2\right)$, the whole diagram can be rotated about $\left(a,\frac b2\right)$ until the dividing line between $D_2$ and $D_2^c$ is vertical. All that remains now is to write down the answer in terms of the $Q$ function, the complementary Gaussian CDF, because all that matters is how far that line is from the mean point $\left(a,\frac b2\right)$.
This idea is illustrated in the picture below.
See also this answer of mine which treats the special case $a=b=0, \frac{N_0}{2} = 1.$
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