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If I have the following gaussian integral:

$$ \int_{-\infty}^{a}\int_{-\infty}^{b}\exp\left[-\alpha x^2+\beta x-\theta y^2+\gamma y+2\lambda xy\right] dxdy$$ ,where $X$ and $Y$ are standard normal R.V.

Well I can put it under the form:

$$ \exp\left(\frac{\beta^2}{4\alpha} +\frac{\lambda^2}{4\theta}\right)\int_{-\infty}^{a}\int_{-\infty}^{b}\exp\left[-\left(\sqrt\alpha x-\frac{\beta}{2\sqrt\alpha}\right)^2 -\left(\sqrt\theta y-\frac{\gamma}{2\sqrt\theta}\right)^2+2\lambda xy\right] dxdy$$

But from then... I don't know how to reach a bivariate normal distribution form... and my head starts to hurt.

More precisely, when doing the change of variable. In my second eqn, say I def $\hat{x}:=(\sqrt\alpha x-\frac{\beta}{2\sqrt\alpha})$ and $\hat{y}:=(\sqrt\theta y-\frac{\gamma}{2\sqrt\theta})$, then after shifting limits and adjusting the differentials, I'm left with: $$ C\int_{-\infty}^{a*}\int_{-\infty}^{b*}\exp\left[-\hat{x}^2-\hat{y}^2+2\lambda(\hat{x}+...)(\hat{y}+...)\right] d\hat{x}d\hat{y}$$

And now I have some additional $x$ and $y$ double factors that I don't know what to do with...

It looks like this actually ($n$-dimensional with linear term), and I have no problem finding matrix $A$ and vector $B$ that fit my expression but they just go straight to the solution for integration over $R^2$ and give no link to a bivariate...

I still take any hint. Thanks

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Already in dimension 1 there is no general expression except if one defines a new function as the solution... – Did Feb 23 '13 at 6:31
Some formatting tips: Equations with integrals and displayed fractions look better and are easier to read as displayed equations, which you get by enclosing them in double dollar signs instead of single dollar signs. You can get appropriately sized delimiters (e.g. parentheses) by preceding them with \left and \right, respectively. Names like exp are interpreted as a juxtaposition of variable names and hence italicized; to get the appropriate font and spacing for them use \exp, or \operatorname{name} in cases where there's no predefined command. – joriki Feb 23 '13 at 8:00

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