Finding $c$ such that $ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} ce^{-(x^2 - xy + 4y^2)/2}\,dx\,dy = 1 $

Question

Let $f(x, y)$ be the given function. $$f(x, y) = ce^{-(x^2 - xy + 4y^2)/2}$$ Determine the value of $c$ such that $f$ is a pdf.

The problem is finding the value of $c$ such that

$$ \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} ce^{-(x^2 - xy + 4y^2)/2}\,dx\,dy = 1 $$

The usual way of integrating a regular gaussian/normal pdf is by using polar coordinates. However, the values on the exponential above make it harder than usual. I assume there must be some transformation of variables that will make the integral simpler, then you'd need to calculate the jacobian obviously, but I can't see the transformation.

Answer 1

It suffices to "mould" your expression onto the multivariate normal distribution formula one can find for example here, with $m=0$ (mean = $0$) and $n=2$ ($n$ = number of variables).

We have, in our case :

$$c \exp(-\dfrac12 (X^T S X))=c \exp(-\dfrac12 (X^T \Sigma^{-1} X))$$

where $X=\binom{x}{y}$ and :

$$S=\begin{pmatrix}1&-\tfrac12\\ -\tfrac12&4\end{pmatrix} \ \implies \ \Sigma=S^{-1}=\tfrac{16}{15}\begin{pmatrix}4&\tfrac12\\ \tfrac12&1\end{pmatrix}$$

($\Sigma$ is the variance matrix extending the concept of unidimensional variance), giving :

$$c=\dfrac{1}{2 \pi \sqrt{\det(\Sigma)}}$$

But in fact, one can completely shortcut the computation of $\Sigma$. Indeed $\Sigma=S^{-1} \ \implies \ \det(\Sigma)=\det(S)^{-1}$, allowing to write plainly :

$$c=\tfrac{1}{2 \pi}\sqrt{\det(S)}=\tfrac{1}{2 \pi}\sqrt{\tfrac{15}{16}}=\tfrac{\sqrt{15}}{8 \pi}$$

Answer 2

1. Try the substitution $ x=2(u+v), y=u-v$

2. Then convert the integral in the $u-v$ plane , with the Jacobian.

3. Use the symmetry of the integrand i.e you should get an even function of $u$ and $v$, to change the limits to $0$ to $ \infty$ for both $u$ and $v$.

4. Separate the two integrals and use Gamma Function to evaluate the integral and get $c$