Integral:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{t_1x+t_2y}*\frac{1}{2\pi\sqrt{1-\rho^2}}exp\left(-\frac{1}{2(1-\rho^2)}(x^2-2\rho xy+y^2)\right)dxdy$$

Apparently the answer is$$exp\left(\frac{1}{2}(t_1^2+2\rho t_1 t_2+t_2^2)\right)$$

Question

Integral:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{t_1x+t_2y}*\frac{1}{2\pi\sqrt{1-\rho^2}}exp\left(-\frac{1}{2(1-\rho^2)}(x^2-2\rho xy+y^2)\right)dxdy$$

Apparently the answer is$$exp\left(\frac{1}{2}(t_1^2+2\rho t_1 t_2+t_2^2)\right)$$

kirbykirby · Answer

I tried transforming the inner integral into a $Normal$~$(0,\sqrt{1-\rho^2})$ but the outer one is giving me trouble..

kirbykirby · Answer

If you're wondering I'm trying to solve problem on p. 97 here: the double integral part. http://www.am.qub.ac.uk/users/g.gribakin/sor/Chap6.pdf They don't provide any details :S

jhannybean · Answer

I can hardly read it. :|

kirbykirby · Answer

If you check the website^^ it should be clear. I just changed theta1 and theta2 to t1 and t2, and changed u->x and v->y

kirbykirby · Answer

Maybe hints on appropriate substitutions?

zarkon · Answer

try $$w=\frac{x-\rho y}{\sqrt{1-\rho^2}}$$

kirbykirby · Answer

oh that should work well i hope. thankss

kirbykirby · Answer

@Zarkon just letting you know I finally figured it out. What a grueling integral @_@ . the substitution w=x-py worked really well, without the square-root

zarkon · Answer

If you write it in matrix form (the general case for multivariate normal) it is actually easier...but I figured you didn't want to do it that way.

kirbykirby · Answer

Oh it does look easier. If only I knew about it.... I didn't know there was a multivariate generalized form.