Is there a closed form expression for $\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-(1/2) ( x^2+y^2 )} \mathrm{d}x\,\mathrm{d}y$?

Question

I have been trying to evaluate the integral:

$$\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y$$

I know of course that the integral equals $1$ over $[-\infty,\infty] \times [-\infty,\infty]$ but I do not quite know how to handle the present case. Are there any tricks here?

Thank you.

Answer 1

Your integral is the probability: $$\mathbb{P}[X\leq Y]$$ where $X$ and $Y$ are two independent normal variables $N(0,1)$,

hence the value of the integral is just $\frac{1}{2}$, since: $$\mathbb{P}[X\leq Y]=\mathbb{P}[Y\leq X],\qquad \mathbb{P}[X\leq Y]+\mathbb{P}[Y\leq X]=1.$$

Answer 2

Ah, I realise just now I'd misread to start with. Algebraically, we can solve it by noting that the value of the integral is the same under the change of variables $u=-x$, and since summing the two resulting integrals results in the $[-\infty,\infty] \times [-\infty,\infty]$ case, the answer is $1/2$.

i.e.:

$$\int_{- \infty}^\infty \int_{-\infty}^y \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)} \,\mathrm{d}x\,\mathrm{d}y = \int_{- \infty}^\infty \int_{-y}^\infty \frac{1}{2 \pi} e^{-1/2 \left( x^2+y^2 \right)}\,\mathrm{d}x\,\mathrm{d}y.$$

$LHS+RHS=\int_{- \infty}^\infty \int_{-\infty}^\infty \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y=1$, and $LHS=RHS$, so your integral is $1/2$.

EDIT: I guess I was a little bit short. There are several ways of showing $LHS+RHS$ is what I quote, but here is a simple, uninformative one.

$LHS+RHS=\int_{- \infty}^\infty \int_{-\infty}^\infty \frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y-\int_{- \infty}^\infty \int_{-y}^y\frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y$.

Now, Let $I=\int_{- \infty}^\infty \int_{-y}^y\frac{1}{2 \pi} e^{-1/2 (x^2+y^2)}\,\mathrm{d}x\,\mathrm{d}y$. Use substitution $u=-y$. Then

$I=-\int_{\infty}^{-\infty} \int_{u}^{-u}\frac{1}{2 \pi} e^{-1/2 (x^2+u^2)}\,\mathrm{d}x\,\mathrm{d}u=\int_{-\infty}^{\infty} \int_{u}^{-u}\frac{1}{2 \pi} e^{-1/2 (x^2+u^2)}\,\mathrm{d}x\,\mathrm{d}u=-I$, hence as $I=-I$, $I=0$.

I think it could also be done by splitting the integration range up. (e.g. looking at the integrals on $(-\infty,0]$ and $[0,\infty)$)

Answer 3

Jack D'Aurizio's answer is good, but since you said in comments under it that you wanted a different point of view, let's try this: \begin{align} u & = (\cos45^\circ)x-(\sin45^\circ)y = \tfrac{\sqrt{2}}2 x - \tfrac{\sqrt{2}}2 y \\ v & = (\sin45^\circ)x+(\cos45^\circ)y = \tfrac{\sqrt{2}}2 x + \tfrac{\sqrt{2}}2 y \end{align} This is just a $45^\circ$ rotation of the coordinate system, suggested by the fact that your boundary line $y=x$ is just a $45^\circ$ rotation of one of the coordinate axes.

Then simplify $u^2+v^2$ and find that it comes down to $x^2+y^2$.

Solving the system of two equations above for $x$ and $y$, one gets \begin{align} x & = \phantom{-}\tfrac{\sqrt{2}}2 u + \tfrac{\sqrt{2}}2 v \\ y & = -\tfrac{\sqrt{2}}2 u + \tfrac{\sqrt{2}}2 v \end{align} By trivial algebra, the condition that $x\le y$ now becomes $u\le0$.

If you know about Jacobians, you get $$ du\,dv = \left|\frac{\partial(u,v)}{\partial(x,y)}\right|\,dx\,dy = \left|\frac{\partial u}{\partial x}\cdot\frac{\partial v}{\partial y} - \frac{\partial u}{\partial y}\cdot\frac{\partial v}{\partial x}\right|\,dx\,dy = 1\,dx\,dy. $$ Hence your iterated integral becomes $$ \int_{-\infty}^\infty\int_{-\infty}^0 \frac 1{2\pi} e^{-(u^2+v^2)/2}\,du\,dv = \int_{-\infty}^\infty \int_{-\infty}^0 \left\{\frac 1{2\pi} e^{-v^2/2}\right\} e^{-u^2/2}\,du\,dv $$ The part in $\{\text{braces}\}$ does not depend on $u$, so this is $$ \int_{-\infty}^\infty\left( \frac 1{2\pi} e^{-v^2/2} \int_{-\infty}^0 e^{-u^2/2}\,du \right)\,dv. $$ Now the inside integral does not depend on $v$, so it pulls out: $$ \int_{-\infty}^\infty e^{-v^2/2}\,dv \cdot \frac1{2\pi} \int_{-\infty}^0 e^{-u^2/2}\,du $$ and this is of course $$ \int_{-\infty}^\infty \frac1{\sqrt{2\pi}} e^{-v^2/2}\,dv \cdot \int_{-\infty}^0 \frac1{\sqrt{2\pi}} e^{-u^2/2}\,du. $$ The first integral comes to $1$ and the second, by a simple symmetry argument, is $1/2$.

Answer 4

Set $x=r\cos \theta,y=r\sin \theta$, then we have

$$\int_{- \infty}^\infty \int_{-\infty}^y e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y=\int_{0}^\infty \left(\int_{-3\pi/4}^{\pi/4} e^{-r^2/2}\mathrm {d}\theta\,\right)r\,\mathrm{d}r=\pi \int_{0}^\infty e^{-r^2/2}r\,\mathrm{d}r=\pi$$

So the original integral is equal to $(1/2)$.

This method as well as @MichaelHardy's method also works for the integrals like:

$$\int_{- \infty}^\infty \int_{-\infty}^{a y} e^{-(1/2) ( x^2+y^2 )}\mathrm {d}x\,\mathrm{d}y, \text{ }a \in \mathbb{R}$$

The results are the same. This is because the function to be integrated ($e^{-r^2/2}$) is rotational invariant (independent of $\theta$) and $x=a y$ is a straight line going through the origin and divides the plane into 2 halfs of equal size.