# Double Integral of the Joint PDF of the Gaussian RV

I am trying to show that the double integral over $\mathbb{R}^2$ of the joint PDF of Gaussian Distribution is $1$. I am looking at:

$$\frac{1}{2\pi} \cdot \frac{1}{\sqrt{a^2b^2-c^2}} \iint_{\mathbb{R^2}} \exp\left\{\frac{-(a^2(x-m)^2 - 2c(x-m)(y-n)+b^2(y-n)^2)}{2(a^2b^2 - c^2)}\right\} dx dy$$

where $m, n \in\mathbb{R}$, $a>0, b>0$, and $c \in \mathbb{R}$ s.t. $|c| < ab$.

How I'm attempting to solve this is by trying to get an upper bound of this integral by removing $c$ in the exponent of the integral, and replacing it by $ab$, and then completing the square so that the term in the exponent gives :

$$\exp\left\{\frac{-(a(x-m) + b(y-n))^2}{2a^2b^2-c^2}\right\}$$ and trying to use the fact that the integral of $$\exp\left\{\frac{-(x^2+y^2)}{2}\right\}$$ gives $\sqrt{2\pi}$. But I don't really know exactly how to connect these two statements.

If anyone could help it would be appreciated. I am only looking for ways to do this with integration tricks.

• I tried to improve the formatting. Please verify it, and try to apply it in the future. Example: \left(\sqrt{\frac{abc}{123}}\right) for $\left(\sqrt{\frac{abc}{123}}\right)$. – Em. Nov 6 '16 at 23:34

Consider a $n$-dimensional normal distribution $N(\mu ,\Sigma )$ where $\Sigma$ is a spd matrix. You can assume $\mu =0$ (do a change of variables). Let's compute $$I=\int_{R^n} e^{-1/2 x^T\Sigma ^{-1} x} dx_1 \cdots dx_n$$ where $x$ is a column vector of $x_i$s. Since $\Sigma$ is spd, it admits a Cholesky decomposition $\Sigma ^{-1} =A^T A$ so in the exponent we really have $-1/2 x^T A^TAx=-1/2(Ax)^TAx$. Then we can introduce a new variable (vector) $y=Ax$ and by multivariate chain rule obtain $$I=|A|^{-1}\int_{R^n} e^{-1/2y^Ty}dy_1\cdots dy_n=|A|^{-1}\int_{R^n}e^{-1/2(y_1^2+\cdots y_n^2)}dy_1\cdots dy_n$$ Here $|A|$ denotes the determinant of $A$. But this integral breaks up into $n$ copies of$$\int_Re^{-t^2/2}dt=\sqrt{2\pi}$$ so $$I=|A|^{-1}\sqrt{2\pi}^n$$ Now from $\Sigma^{-1}=A^TA$ follows that $|A|=|\Sigma|^{-1/2}$ and finally $$1=\frac{I|A|}{\sqrt{2\pi}^n}=\frac{I}{\sqrt{2\pi}^n\sqrt{|\Sigma|}}$$ which is precisely the integral of the density of $N(0,\Sigma)$.
• Then yes you complete the square but in a different way. Treat $y$ is a constant and express the exponent as $(px+q)^2+r$ where $p,q,r$ are constants ( = functions of $a,b,c,y$), introduce a new variable $z=px+q$ and integrate over $z$, then you are left with only a 2nd degree polynomial of $y$ in the exponent, which you integrate in much the same way (complete the square). The other way is to take $\Sigma^{-1}=[a^2,-c;-c,b^2]$ and follow the proof above. In any case, convince yourself that you can take $m=n=0$. – Erik Nov 7 '16 at 11:21