Finding a Joint Moment Generating Function

Question

Please consider the following problem and my solution to it:
Problem:
Let $(X,Y)$ be a continues bivariate r.v. with joint pdf \begin{eqnarray*} f_{XY}(x,y) &=& \begin{cases} e^{-(x+y)} & x > 0 , y > 0 \\ 0 & \text{otherwise} \\ \end{cases} \\ \end{eqnarray*} Find the joint moment generating function of $X$ and $Y$.
Answer:
\begin{eqnarray*} M_{XY} &=& E(e^{t_1 X} + e^{t_2 Y}) \\ M_{XY} &=& \int_{0}^{\infty} \int_{0}^{\infty} (e^{t_1 x} + e^{t_2 y})(e^{-(x+y)}) \, dy \, dx \\ M_{XY} &=& \int_{0}^{\infty} \int_{0}^{\infty} e^{t_1x - x - y} + e^{t_2y - x - y} \, dy \, dx \\ M_{XY} &=& \int_{0}^{\infty} -e^{t_1x - x - y} - \frac{ e^{t_2y -x - y} }{t_2-1} \Big|_{y = 0}^{y = \infty} \, dx \\ M_{XY} &=& \int_{0}^{\infty} ( -0 + 0) - ( -e^{t_1x - x} - \frac{e^ { -x } }{t_2-1} ) \, dx \\ M_{XY} &=& \int_{0}^{\infty} e^{(t_1-1)x} + \frac{ e^{-x}}{t_2-1} \, dx \\ M_{XY} &=& \frac{e^{(t_1-1)x}}{{t_1-1}} - \frac{ e^{ -x } } {t_2-1} \Big|_{x = 0}^{x = \infty} \\ M_{XY} &=& ( 0 + 0 ) - ( \frac{1}{t_1 - 1} - \frac{1}{t_2 - 1}) \\ M_{XY} &=& \frac{1}{t_2 - 1} - \frac{1}{t_1 - 1} \\ M_{XY} &=& \frac{ t_1 - 1 - ( t_2 - 1 ) }{ (t_1 - 1)(t_2 - 1) } \\ M_{XY} &=& \frac{ t_1 - t_2 }{ (t_1 - 1)(t_2 - 1) } \\ \end{eqnarray*} However, the book gets: \begin{eqnarray*} M_{XY} &=& \frac{1}{ (1 - t_1)(1 - t_2) } \\ \end{eqnarray*} Please note that my answer could be rewritten as: \begin{eqnarray*} M_{XY} &=& \frac{t_1-t_2}{ (1 - t_1)(1 - t_2) } \\ \end{eqnarray*} What did I do wrong?
Thanks,
Bob

Answer 1

What is wrong is your expression for the MGF. It should be $$ M_{XY}(t_1,t_2) = \mathbb{E}[e^{\big\langle \begin{pmatrix} t_1\\t_2\end{pmatrix},\begin{pmatrix} X\\Y\end{pmatrix} \big\rangle}] = \mathbb{E}[e^{t_1X+t_2Y}] \tag{$\dagger$} $$ not $\mathbb{E}[e^{t_1X}+e^{t_2Y}]$ (which, by linearity, would always be equal to $\mathbb{E}[e^{t_1X}]+\mathbb{E}[e^{t_2Y}]$—this should strike you as strange).

Using the expression from $(\dagger)$, you 'll get, for $t_1,t_2<1$, $$\begin{align} M_{XY}(t_1,t_2) &= \int_{0}^\infty \int_{0}^\infty dxdy e^{t_1x+t_2y}e^{-(x+y)} = \int_{0}^\infty \int_{0}^\infty dxdy e^{(t_1-1)x}e^{(t_2-1)y}\\ &= \int_{0}^\infty dx e^{(t_1-1)x} \int_{0}^\infty dye^{(t_2-1)y} = \frac{1}{1-t_1} \cdot \frac{1}{1-t_2} \end{align}$$ as expected.

Answer 2

You seek $$\begin{align}\mathsf M_{X,Y}(s,t) &= \mathsf E(\mathrm e^{sX+tY}) \\ &= \mathsf E(\mathrm e^{sX}~\mathrm e^{tY})\\ &=\mathsf E(\mathrm e^{sX})~\mathsf E(\mathrm e^{tY}) & \star\\ & =\mathsf M_X(s)~\mathsf M_Y(t)\end{align}$$

$\star~$ Since $f_{X,Y}(x,y)=\mathrm e^{-x}\mathrm e^{-y} \mathbf 1_{x\geq 0}\mathbf 1_{y\geq 0}$ indicates that the random variables are independent, and infact something.   Show this to be so, and thus use the MGF for that distribution to find the joint MGF.