Compute joint moment of standard normal $E(e^{uX+vX^2}) = \frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ux+vx^2-\frac12x^2}$

Question

Given X,Y i.i.d with standard normal distributions, we want to find the joint mgf of $U=X+Y$ and $V=X^2+Y^2$ $$E(e^{uU+vV}) = E(e^{uX+uY+vX^2+vY^2}) = (E(e^{uX+vX^2}))^2$$ by independant and identicalness of the random variables X,Y

$$E(e^{uX+vX^2}) = \frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ux+vx^2-\frac12x^2}$$. Clearly this integral is finite only if $v \lt 1/2$. But how to solve this integral? I tried completeing the square and using polar coordinates, but the integral seems too complex to solve cleanly.

Any ideas?

Answer 1

Let $a=\frac12 -v$ to reexpress the integral as,

$$\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ux+vx^2-x^2/2}dx$$ $$=\frac1{\sqrt{2\pi}}e^\frac{u^2}{4a}\int_{-\infty}^{\infty}e^{-a(x-\frac u{2a})^2 }dx =\frac1{\sqrt{2\pi}}e^\frac{u^2}{4a}\int_{-\infty}^{\infty}e^{-at^2 }dt$$ $$= \frac1{\sqrt{2\pi}}e^\frac{u^2}{4a} \sqrt{\frac\pi a}=\frac1{\sqrt{2a}}e^\frac{u^2}{4a}=\frac1{\sqrt{1-2v}}e^\frac{u^2}{2(1-2v)}$$

where the Gaussian integral result $\int_{-\infty}^{\infty}e^{-at^2 }dt = \sqrt{\frac\pi a}$ is used.