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 as to approach this question?

p.s. Sorry that my English isn't good. I only spent my junior year of high school in Australia, and rest in Korea.

  • 1
    $\begingroup$ Your English seems fine. You're making your life hard with this calculation though. You probably will want to write all the terms in x and y as a single quadratic. $\endgroup$ – Glen_b Feb 24 '14 at 2:54
  • $\begingroup$ @Glen_b, so I should write something like $e^{y-x}^2$ ? I took a different approach and got $=\int_{0}^\infty \int_{-\infty}^{\infty} \frac{1}{\sqrt(2\pi)} e^{\frac{-(y-(x-s)^2}{2}}e^{(s-2t-1)x}e^{s^2} dydx$. I'm not sure if I can use that $\int_{-\infty}^{\infty} 1/sqrt(2\pi)*e^{-(y-(x-s))^2}$ is 1 as integrand is Y~N((x-t),1). $\endgroup$ – user40736 Feb 24 '14 at 4:18
  • $\begingroup$ Re first question in comment: no. Re second sentence: you have $x$ in both the first two exponents, so you can't just integrate w.r.t $x$. If the quadratic in both $x$ and $y$ isn't working, try just writing either a quadratic in $y$ or a quadratic in $x$ (no matter if the other variable is in the quadratic, treat it as constant), taking everything else out the front of the inner integral (changing order of integration first, if needed) and thereby integrating one variable out completely, and going from there. $\endgroup$ – Glen_b Feb 24 '14 at 4:30
  • $\begingroup$ The integration w.r.t. $y$ is straightforward -- collect all the terms involving $y$: $\exp(-\frac{1}{2}\{y^2-2(t+x)y\})$, and complete the square (by adding and subtracting a constant term), then multiply and divide by the required constant, and you'll be left with a term in $t$ and $x$ as the value of the $y$ integral, which is then a term in the $x$-integral. $\endgroup$ – Glen_b Feb 24 '14 at 9:18
  • $\begingroup$ It may be noted from the joint density of $(X,Y)$ that $Y\mid X\sim\mathcal{N}(X,1)$ and $X\sim\text{Exp}(1)$. The joint mgf then can be easily derived , as is shown here. $\endgroup$ – StubbornAtom Apr 27 '18 at 13:04

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