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I am sure all you integration buffs can do this faster than I can type it. Your help with a quick explanation and solution is appreciated.

$$F _{XY} = \int_0^\infty\int_0^\infty xye^{-\frac{x^2 + y^2}{2}}~dxdy$$

This is an early problem in a problem set for a joint probability distribution. I am pretty sure the $-\frac{x^2 + y^2}{2}$ is the u, but if so, when we plug back in, the dx is in the exponent. I don't suppose that just "comes down" magically.

If there is a shortcut method, I prefer that since I am learning this in preparation for a timed test. (Actuarial P/1).

Anyway, have at it!

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1 Answer

The integral is separable:

$$F_{XY} = \int_0^{\infty} dx \: x e^{-x^2/2} \int_0^{\infty} dy \: y e^{-y^2/2} = \left (\int_0^{\infty} dx \: x e^{-x^2/2} \right )^2$$

Each of the integrals is

$$\int_0^{\infty} dx \: x e^{-x^2/2} = \frac{1}{2} \int_0^{\infty} d(x^2)\: e^{-x^2/2} = \frac{1}{2} \int_0^{\infty} du\: e^{-u/2} = \frac{1}{2} \cdot 2 =1$$

Therefore

$$F_{XY} = 1^2 = 1$$

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