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This integral is [hopefully] between 0 and 1 as it is supposed to represent a probability.

$$\int_{-\infty}^{\infty} \int_{-2-x}^{2-x} \frac{1}{2\pi} e^{\frac{-x^2-y^2}{2}} dy dx$$

I just wanted to check if anyone saw any easy-ish method to evaluate it that I could be missing. The only idea I can come up with is a few messy steps long:

  1. Express the integrand as a series
  2. Evaluate the inner integral to get two series.
  3. Evaluate the outer improper integral by integrating and finding the limit at both positive and negative infinity.

I don’t even know if the 3rd step is doable, so I am hesitant to do the work leading up to it.

2 Answers2

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Change the variables (rotation by $\pi/4$) $$ \begin{cases} u&=&\frac{1}{\sqrt2}(x-y),\\ v&=&\frac{1}{\sqrt2}(x+y) \end{cases} $$ to get $$ \frac{1}{2\pi}\int_{-\infty}^{+\infty}\int_{-\sqrt2}^{\sqrt2}e^{-\frac{u^2+v^2}{2}}\,dvdu=\frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-u^2/2}\,du\int_{-\sqrt2}^{\sqrt2}e^{-v^2/2}\,dv=\frac{1}{\sqrt\pi}\int_{-1}^1e^{-s^2}\,ds=\color{red}{\operatorname{erf}(1)}. $$

A.Γ.
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    How did you come up with that rotation? –  Aug 11 '19 at 14:56
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    @Wessst try drawing the domain of integration on a 2-D graph to see that it is actually a diagonal strip. It should then be obvious why a 45 degree rotation yields a horizontal strip of the given width, which is much nicer to work with. – Bar Alon Aug 11 '19 at 15:09
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    @Wessst The other commenter is right: after you realize how the integration domain looks like, it becomes natural to try to rotate in order to get independent limits of integration. – A.Γ. Aug 11 '19 at 16:26
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The integral is $\ \mathbb{P}\left(-2\le X+Y\le 2\right)\ $, where $\ X,Y\ $ are independent standard normal variates. In these circumstances $\ \frac{X+Y}{\sqrt {2}}\ $ is also a standard normal variate, so the value of the integral is $\ \mathcal{N}_{(0,1)}(\sqrt{2})-\mathcal{N}_{(0,1)}(-\sqrt{2})\approx 0.8427 $.

lonza leggiera
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  • Thank you! That makes complete sense to me, but I don’t understand why I don’t get about that answer using wolfram alpha. I tried making the bounds of the outer integral pretty large (-100 to 100) and it gives ~0.8 as the answer. I would think that using 100 would be very accurate as the standard deviation is 1, but I guess not? –  Aug 11 '19 at 06:58
  • Small correction: it is $\frac{X+Y}{\sqrt 2}\sim N(0,1)$, so the answer changes to $\Phi(\sqrt 2)-\Phi(-\sqrt 2)$. – StubbornAtom Aug 11 '19 at 07:14
  • @StubbornAtom Thanks. I realised this after the OP posted his numerical estimate, and was in the process of correcting the answer as you posted your comment. – lonza leggiera Aug 11 '19 at 07:19