This is an end-of-chapter question from a Korean textbook, and unfortunately it only has solutions to the even-numbered q's, so I'm seeking for some hints or tips to work out this particular joint moment generating function question. or I better state that I cannot find a way to solve this particular double integral)
The question (roughly translated) states that
Given the joint pdf of random variables X and Y,
$$ f(x,y) = \frac{1}{\sqrt(2\pi)}e^{-x}e^{\frac{-(y-x)^2}{2}}, x \ge 0, -\infty \le y \le \infty $$
Find the joint mgf M(s,t), and for what values of s and t does the mgf exist?
The following is how I approached this problem.
$ M(s,t)=\int_{-\infty}^\infty\int_{0}^\infty e^{sx+ty}\frac{1}{\sqrt(2\pi)}e^{-x}e^{\frac{{-(y-x)}^2}{2}}dxdy $
$ =\int_{0}^\infty \int_{-\infty}^{\infty}e^{sx+ty}\frac{1}{\sqrt(2\pi)}e^{-x}e^{\frac{{-(y-x)}^2}{2}}dydx $
$ =\int_{0}^\infty \int_{-\infty}^{\infty} \frac{1}{\sqrt(2\pi)} e^{\frac{2(s-1)x-x^2}{2}}e^{\frac{ty-y^2}{2}}e^{xy} dydx $ $ =\int_{0}^\infty \int_{-\infty}^{\infty} \frac{1}{\sqrt(2\pi)} e^{\frac{-(x-(s-1))^2}{2}}e^{\frac{-(y-t)^2}{2}}e^{xy}e^{\frac{(s-1)^2+t^2}{2}} dydx $
I see $\frac{1}{\sqrt{2\pi}}$ in the intergral and y is to be integrated from $\infty $ to $-\infty$, so I believe that I need to make the expression in the form $\frac{1}{\sqrt(2\pi)}*e^-{\frac{(y-something)^2}{2}}*f(x)$ (pdf of a Normal distribution, and integral will be evaluated to 1 whatever the value something is) and then integrating the expression w.r.t x from 0 to $\infty$ from there.
But I cannot find any useful integration technique to evaluate this double integral. I tried polar coordinate; making substitution $ x = r cos{\theta} $ and $ y = r sin{\theta} $ which will turn the integral to $ =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_0^{\infty} \frac{1}{\sqrt(2\pi)} r*exp(r*((s-1)cos{\theta}+tsin{\theta})-\frac{{r^2}(1-\frac{sin{2\theta}}{2})}{4}) drd{\theta} $
Evaluating this I got $\sqrt{\frac{2\pi}{3}}$ which I feel is incorrect as it's neither of a function of x nor y.
So can anyone suggest me with any hints (possible substitution?) as to approach this question?
The joint density can be expressed as $f(x,y) = e^{-x} N_y(x,1) = f(x) f(y|x)$ where $N_y(\mu,\sigma^2)$ is a normal density on the variable $y$.
That is
$$M(s,t)= \int_{0}^\infty e^{sx} e^{-x}\int_{-\infty}^\infty e^{ty} \frac{1}{\sqrt{2 \pi}} e^{-\frac{(y-x)^2}{2}} dy \, dx$$
The inner integral is the MGF of a normal $N(x,1)$, which is $\exp\left({xt+\frac{t^2}{2}}\right)$
So we have
$$M(s,t)= \int_{0}^\infty e^{(s+t-1)x } e^{\frac{t^2}{2}} dx$$
I leave the rest to you.