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This is the first time I came across the problem of finding integral of $\propto$. I have a joint distribution $$f_{X,Y}(x,y) \propto \exp\left(13xy - 94x^2 - \frac{1}{2}y^2\right)$$ where $ -\infty< x <\infty, -\infty< y <\infty $

I attempted to find $f_X(x)$ as follows: \begin{align*} f_X(x) &\propto \int_{-\infty}^\infty e^{13xy - 94x^2 - \frac{1}{2}y^2}{\rm d}y\\ &\propto \int_{-\infty}^\infty e^{-\frac{1}{2}(y - 13x)^2 - \frac{19x^2}{2}}{\rm d}y\\ &\propto \frac{1}{e^{\frac{19x^2}{2}}} \int_{-\infty}^\infty e^{-\frac{u^2}{2}}{\rm d}u \end{align*} where $ u = (y - 13x)^2 $

Similarly, I derived $$ f_Y(y) \propto \frac{1}{e^{\frac{19x^2}{376}}} \int_{-\infty}^\infty e^{-u^2}{\rm d}u $$ where $ u = \sqrt{94}x - \frac{13y}{2\sqrt{94}} $

Could you please show me how to proceed to the destination solutions? Thanks in advance.

Nemo
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1 Answers1

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This particular integral is known as Gaussian Integral and we have the (in)famous identity

$$\int_{-\infty}^\infty e^{-x^2}{\rm d}x=\sqrt\pi$$

Proofs may be found here:

mrtaurho
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  • Thanks for the rescue @mrtaurho. I'm so amazed that the solution can be that elegantly simple $\sqrt{\pi}$!!! Just to clarify, regardless whether it is $x$ or $u$ where $u$ is a function of $x$, the answer will always be the same, that is, multiple of $\sqrt{\pi}$ ? – Nemo Mar 19 '20 at 04:44
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    @Nemo Glad to help! Well, as long as $u$ is linear function of $x$, i.e. $u(x)=ax+b$ you can substitute $u=ax+b$ and use the identity from above. If $u$ is a non-linear function of $x$ things might become weird... (consider $u=x+\frac1x$ for example) – mrtaurho Mar 19 '20 at 04:50
  • These things just blow me away with their hidden simplicity under a complex appearance and vice versa. Thanks – Nemo Mar 19 '20 at 04:54
  • @Nemo Yes, definite integrals are sometimes just... weird ;p Anyways, always happy to help! – mrtaurho Mar 19 '20 at 04:55