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.

  • $\begingroup$ thanks you saved me $\endgroup$
    – Steve
    May 16 '20 at 20:13

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