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Find $C$ such that

$Ce^{(-4x^2-xy-4y^2)/2}$

is a joint probability density of a $2$-variable Gaussian.

If someone could give me a jumping off point, or a process as to how to go about this, I'd really appreciate it.

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    How to approach this might depend on what you've seen so far. If you've been given a form of the density function in which the exponent is $-1/2$ times a quadratic polynomial in which the letter $\rho$, for correlation, might have appeared, then the problem would include figuring out what $\rho$ is and what the two variances are. But if you've seen a form of the density that involves matrices, the answer might be expressed differently.2011-11-30

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For any density $f(x,y)$ you should have $\int\limits_{\mathbb R^2}f(x,y)\mathrm dx\mathrm dy = 1.$ From this condition you can find $C$. Recall that $\int\limits_{\mathbb R}\mathrm e^{-\frac{(y-\mu)^2}2}\mathrm d y = \sqrt{2\pi}$ for all $\mu$ - you may need it in your calculations.

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    I've only seen this done with matrices2011-11-30
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Hint: you probably know that a Multivariable Gaussian takes the form:

$ \frac{1}{\sqrt{(2 \pi)^n |\Sigma|}} \exp{ \left(- \frac{({\bf x}-{\bf \mu})^t \Sigma^{-1}({\bf x}-{\bf \mu})}{2} \right)}$

Try to reduce your given pdf to this, so you can find by inspection the matrix $\Sigma^{-1}$ (it should be easy to see that ${\bf \mu}={\bf 0}$), and from that your are almost done (you don't even need to invert the matrix).