[Math] 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.