Bivariate Gamma Probability Function This might be a very basic question, but my textbook doesn't provide a satisfactory formula for how to calculate the probability for the Gamma function. Any help is appreciated.
$f(t,r) =$ {$\frac{1}{8}te^{\frac{-1}{2}(t+r)}$} for $t,r >0$ and 0 elsewhere
I'm asked to find the probability that $t,r >2$ and after a long integration by parts I was able to get that $P(t,r>2) = \frac{2}{e^2}$
Similarly, I found that $P(t,r>1) = \frac{3}{2e}$ and
$P(t,r>3)= \frac{5}{2e^3}$
There appears to be a clear pattern, is this normal? And is there a formula that allows you to get this answer with the long integration by parts?
 A: You just integrate the pdf, then you wil find the pattern.
\begin{align*}
P(T>a,R>b)&=\int_a^\infty\int_b^\infty \frac{1}{8}te^{\frac{-1}{2}(t+r)}dtdr\\&=\frac{1}{8}\int_a^\infty\int_b^\infty te^{-\frac{1}{2}t}*e^{-\frac{1}{2}r}dtdr\\&=\frac{1}{8}\int_a^\infty te^{-\frac{1}{2}t}\int_b^\infty e^{-\frac{1}{2}r} drdt\\&=\frac{1}{8}\int_a^\infty te^{-\frac{1}{2}t} (-2)e^{-\frac{r}{2}}|_b^{\infty}dt\\&=\frac{e^{-\frac{b}{2}}}{4}\int_a^\infty te^{-\frac{1}{2}t}dt\\&\text{(integration by parts)}\\&=-\frac{e^{-\frac{b}{2}}}{2}\int_a^\infty tde^{-\frac{t}{2}}\\&=-\frac{e^{-\frac{b}{2}}}{2}\left [ te^{-\frac{t}{2}}|_a^{\infty}-\int_a^\infty e^{-\frac{t}{2}}dt \right ]\\&=\frac{a}{2}e^{-\frac{a+b}{2}}+e^{-\frac{a+b}{2}}
\end{align*} 
When $a=b=1$
$$P(T,R>1)=\frac{1}{2}e^{-1}+e^{-1}=\frac{3}{2e}$$
When  $a=b=2$
$$P(T,R>2)=e^{-2}+e^{-2}=\frac{2}{e^2}$$
When  $a=b=3$
$$P(T,R>3)=\frac{3}{2}e^{-3}+e^{-3}=\frac{5}{2e^3}$$
You can see there are nothing speical just 
$$P(T>a,R>b)=\frac{a}{2}e^{-\frac{a+b}{2}}+e^{-\frac{a+b}{2}}=(\frac{a}{2}+1)e^{-\frac{a+b}{2}}$$
