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how to do this integral: $$ \mathop{\int\int}_{y+2x>0} x y \frac1{2\pi\sigma_x\sigma_y}e^{ -\frac{(x-\mu_x)^2}{2\sigma_x^2}}\cdot e^{ -\frac{(y-\mu_y)^2}{2\sigma_y^2}} dx dy$$ Both x and y are normally distributed and mutually independent. I need to calculate the above integral.

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For simplicity, I let $\sigma_x=\sigma_y=1$. Then we have \begin{align*} &\mathop{\int\int}_{y+2x>0}\frac{xy}{2\pi}e^{ -\frac{(x-\mu_x)^2}{2}}\cdot e^{-\frac{(y-\mu_y)^2}{2}}dxdy\\ =&\int_{-\infty}^\infty dx\int_{-2x}^\infty\frac{xy}{2\pi}e^{-\frac{(x-\mu_x)^2}{2}}\cdot e^{-\frac{(y-\mu_y)^2}{2}}dy\\ =&\frac{1}{2\pi}\int_{-\infty}^\infty xe^{-\frac{(x-\mu_x)^2}{2}} dx\int_{-2x}^\infty ye^{-\frac{(y-\mu_y)^2}{2}}dy\\ =&\frac{1}{2\pi}\int_{-\infty}^\infty(x-\mu_x)e^{-\frac{(x-\mu_x)^2}{2}} dx\int_{-2x}^\infty ye^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty(y-\mu_y)e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x\mu_y}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty e^{-\frac{(y-\mu_y)^2}{2}}dy\\ =&\frac{1}{2\pi}\int_{-\infty}^\infty(x-\mu_x)e^{-\frac{(x-\mu_x)^2}{2}} dx\int_{-2x}^\infty(y-\mu_y)e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_y}{2\pi}\int_{-\infty}^\infty(x-\mu_x)e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty(y-\mu_y)e^{-\frac{(y-\mu_y)^2}{2}}dy+\frac{\mu_x\mu_y}{2\pi}\int_{-\infty}^\infty e^{-\frac{(x-\mu_x)^2}{2}}dx\int_{-2x}^\infty e^{-\frac{(y-\mu_y)^2}{2}}dy \end{align*}

You can have explicit expression for each term above except for the last one.

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  • $\begingroup$ i really appreciate your help. actually i also got similar expression, but then stuck on the last term. i don't how to calculate that value. $\endgroup$ – user75904 May 5 '13 at 14:00
  • $\begingroup$ @user75904, for the last term I think there is only numerical solution available, table can be looked up from normal distribution function table. $\endgroup$ – Easy May 5 '13 at 14:02
  • $\begingroup$ i don't think it's feasible. see the lower limit of the last integral. it's "-2x", not a constant. please see math.stackexchange.com/questions/368512/… . maybe it might give you some inspiration. but here x is not a standard normal variable. so i don't know how to do $\endgroup$ – user75904 May 5 '13 at 14:08
  • $\begingroup$ @user75904, yeah, you are right. Maybe my wording is confusing, I was trying to say the normal distribution function can be looked up from the table. $\endgroup$ – Easy May 5 '13 at 14:11
  • $\begingroup$ @user75904, yes, I agree with you. $\endgroup$ – Easy May 5 '13 at 14:20

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