Let $X_1$ and $X_2$ be independent standard normal random variables. Let $Y_1 = X_1 + X_2$ and $Y_2 = X_1^2 + X_2^2$.
(a) Show that the joint moment generating function of $Y_1$ and $Y_2$ is
$$\frac{\exp[t_1^{\hspace{.1cm} 2}/(1-2 t_2\hspace{.1cm})\hspace{.01cm}]}{1-2 t_2}$$
The answer to this problem is explained in this way " write $E[e^{Y_{\hspace{.1cm}1} \hspace{.1cm} t_1 + Y_{\hspace{.1cm}2}\hspace{.1cm} t_2}\hspace{.2cm}]$ in terms of a double integral involving the joint distribution of X_1 and X_2 . Perform the integration by separating the double integral, completing the square, and expressing in terms of integrals of normals".
I only understand this
$$\int _{-\infty}^{\infty} \int _{-\infty}^{\infty} e^{Y_1 \hspace{.1cm} t_1 + Y_{\hspace{.1cm}2}\hspace{.1cm} t_2}\hspace{.2cm} \frac{e^{-\frac{1}{2} \hspace{.1cm}x_{\tiny \hspace{.1cm}1}^2}}{\sqrt{2\pi}} \hspace{.2cm} \frac{e^{-\frac{1}{2} \hspace{.1cm}x_{ \hspace{.1cm}2}^2}}{\sqrt{2\pi}} \cdot dx_1 \cdot dx_2 $$
how can I do the other steps. to separate the integrals and complete squares, or is there any other way to do this exercise?
Best Answer
You are missing the step where the hint states
You haven't done that. Your setup for the integral is correct but you need to change $y_1$ and $y_2$ into functions of $x_1$ and $x_2$ according to their definitions; i.e.,
$$y_1 t_1 + y_2 t_2 = (x_1 + x_2)t_1 + (x_1^2 + x_2^2)t_2 = (x_1 t_1 + x_1^2 t_2) + (x_2 t_1 + x_2^2 t_2).$$
This now renders the double integral separable as the product of single integrals:
$$\iint_{\mathbb R^2} e^{y_1 t_1 + y_2 t_2} f_{X_1, X_2}(x_1, x_2) \, dx_1 \, dx_2 = \int_{x_1=-\infty}^\infty e^{x_1 t_1 + x_1^2 t_2} \frac{e^{-x_1^2/2}}{\sqrt{2\pi}} \, dx_1 \int_{x_2=-\infty}^\infty e^{x_2 t_1 + x_2^2 t_2} \frac{e^{-x_2^2/2}}{\sqrt{2\pi}} \, dx_2.$$ Both integrals are equal, differing only in the variable of integration, so you only have to evaluate one. To do this, we want to express $$xt_1 + x^2 t_2 - x^2/2 = -a(x - b)^2 + c$$ for suitable constants $a, b, c$ with respect to $x$; that is to say, we wish to complete the square. Upon doing so, we compare the integrand to a normal density with mean $b$ and variance $1/(2a)$, times a constant factor $\sqrt{2a} e^c$, from which it follows that the answer is $2a e^{2c}$. I leave the determination of $a, b, c$ to you as an exercise, as well as following through with the remainder of the calculation.