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