Computing the joint moment generating function for two functions of two random variables

Question

Let $X$ and $Y$ be i.i.d random variables in the plane with a pdf

$$f(x) = \frac{1}{\sqrt{2\pi}} \cdot \text{exp}(-x^{2}/2) \hspace{1cm} -\infty < x< \infty.$$

Let $U = X + Y$ and $V = X^{2} + Y^{2}$. Compute the joint mgf of $U$ and $V$.

My first step was to compute the MGF of $U$ as follows:

\begin{align*} \mathbb{E}[e^{sU}] = \mathbb{E}[e^{s(X + Y)}] = \mathbb{E}[e^{sX}e^{sY}] \\ = \mathbb{E}[e^{sX}] \cdot \mathbb{E}[e^{sY}] \\ = M_{X}(s) \cdot M_{Y}(s) \\ = (e^{s^2/2})^{2} \\ = e^{s^{2}}, \end{align*}

where $t \in \mathbb{R}$. Note that we can separate expectations due to independence.

Now I can compute the MGF of $V$ as follows:

\begin{align*} \mathbb{E}[e^{tV}] = \mathbb{E}[e^{t(X^{2} + Y^{2})}] = \mathbb{E}[e^{tX^{2}}] \cdot \mathbb{E}[e^{tY^{2}}] \\ = M_{X^{2}}(t) \cdot M_{Y^{2}}(t), \end{align*}

To proceed, I need the mgf of $X^{2}$ and $Y^{2}$. So, maybe is there an easier way to approach this problem? If not, how would I proceed. Then, at the end, I can just multiply the two mgf I found individually? I'm not sure if this approach is correct though, as it would assume $U$ and $V$ to be independent.

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The joint moment generating function of $U$ and $V$ is given by

$$\mathbb{E}[e^{sU + tV}] $$

Answer 1

You have written down the joint MGF correctly. Now just plug in the definitions of $U$ and $V$, and keep going, using independence where you can to simplify things.

$$E[e^{sU + tV}] = E[e^{sX+tX^2 + sY + tY^2}] = E[e^{sX + tX^2}] E[e^{sY + tY^2}] = E[e^{sX + tX^2}]^2$$

To compute the last term, I believe you can compute the integral directly, i.e. $$E[e^{sX+tX^2}] = \int_{-\infty}^\infty e^{sx+tx^2} f(x) \, dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{sx + tx^2 - x^2/2} \, dx.$$ You might need to do some casework (e.g., when $t=1/2$ this integral diverges, etc.).

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Edit:

When $t \ge 1/2$, I believe the integral $\int_{-\infty}^\infty e^{sx + (t - \frac{1}{2}) x^2} \, dx$ diverges. When $t < 1/2$ we can write $$sx + tx^2 - x^2/2 = -\frac{(x-\mu)^2}{2\sigma^2} + c$$ where $\mu = \frac{s}{1-2t}$ and $\sigma^2 = \frac{1}{1-2t}$ and $c = \frac{s^2}{2(1-2t)}$. So the integral is $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{sx + tx^2 - x^2/2} \, dx = \frac{1}{\sqrt{2\pi}} e^c \int_{-\infty}^\infty e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx = \frac{1}{\sqrt{2\pi}} e^c \sqrt{2 \pi \sigma^2} = \frac{1}{\sqrt{1-2t}} \exp\left(\frac{s^2}{2(1-2t)}\right).$$ It seems like the final answer is thus $$\frac{1}{1-2t} \exp\left(\frac{s^2}{1-2t}\right), \qquad t < 1/2$$ but I may have made some typos.