# Finding joint moment generating function of $Y_1$ and $Y_2$

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?

You are missing the step where the hint states

write $$\operatorname{E}[e^{Y_1 t_1 + Y_2 t_2}]$$ in terms of a double integral involving the joint distribution of $$X_1$$ and $$X_2$$.

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.

• thanks you saved me May 16 '20 at 20:13