My textbook, Introduction to Probability by Blitzstein and Hwang, gives the following example:
(Comparing Exponentials of different rates). Let $T_1 \sim \text{Expo}(\lambda_1)$ and $T_2 \sim \text{Expo}(\lambda_2)$ be independent. Find $P(T_1 < T_2)$. For example, $T_1$ could be the lifetime of a refrigerator and $T_2$ could be the lifetime of a stove (if we are willing to assume Exponential distributions for these), and then $P(T_1 < T_2)$ is the probability that the refrigerator fails before the stove. We know from Chapter 5 that $\min(T_1, T_2) \sim \text{Expo}(\lambda_1 + \lambda_2)$, which tells us about when the first appliance failure will occur, but we also may want to know about which appliance will fail first.
Solution:
We just need to integrate the joint PDF of $T_1$ and $T_2$ over the appropriate region, which is all $(t_1, t_2)$ with $t_1 > 0$, $t_2 > 0$, and $t_1 < t_2$. This yields
$$\begin{align} P(T_1 < T_2) &= \int_0^\infty \int_0^{t_2} \lambda_1 e^{-\lambda_1 t_1} \lambda_2 e^{-\lambda t_2} \ dt_1 dt_2 \\ &= \int_0^\infty \left( \int_0^{t_2} \lambda_1 e^{-\lambda_1 t_1} \ dt_1 \right)\lambda_2 e^{-\lambda_2 t_2} \ dt_2 \\ &= \int_0^\infty (1 - e^{-\lambda_1 t_2}) \lambda_2e^{-\lambda t_2} \ dt_2 \\ &= 1 - \int_0^\infty \lambda_2 e^{-(\lambda_1 + \lambda_2)t_2} \ dt_2 \\ &= 1 - \dfrac{\lambda_2}{\lambda_1 + \lambda_2} \\ &= \dfrac{\lambda_1}{\lambda_1 + \lambda_2} \end{align}$$
I have the following questions about this:
Why do we require $T_1$ and $T_2$ to be independent in order to construct the joint PDF?
My multiple integral knowledge is rusty. Why do we go over the region $\int_0^\infty \int_0^{t_2}$? Why is the second integral from $0$ to $t_2$? What happened to $t_1$? Why $\int_0^\infty$ for the first integral?
I would greatly appreciate it if people could please take the time to clarify this.
1)
It is not so much a "requirement" but more a fortunate coincidence that we are only interested in this special case that is characterized by independence.
If $T_1$ and $T_2$ are independent then we are capable to find the joint PDF (as product of the PDF's of $T_1$ and $T_2$).
If there is no independence then we simply cannot calculate $P(T_1<T_2)$ unless the joint PDF is one of our data.
2)
Let $u:\mathbb{R}^{2}\to\mathbb{R}$ be the function prescribed by $\left(x,y\right)\mapsto1$ if $x<y$ and $\left(x,y\right)\mapsto0$ otherwise.
Then:
$$P\left(T_{1}<T_{2}\right)=\mathbb{E}u\left(T_{1},T_{2}\right)=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}u\left(t_{1},t_{2}\right)f_{\left(T_{1},T_{2}\right)}\left(t_{1},t_{2}\right)dt_{1}dt_{2}=$$$$\int_{0}^{\infty}\int_{0}^{\infty}u\left(t_{1},t_{2}\right)f_{\left(T_{1},T_{2}\right)}\left(t_{1},t_{2}\right)dt_{1}dt_{2}$$
The first equality is an application of the general rule $P(A)=\mathbb E\mathbf1_A$ where $\mathbf1_A$ denotes the indicator function of event $A$.
The second equality is nothing more than a definition of expectation.
The third equality rest on the fact that $f_{\left(T_{1},T_{2}\right)}\left(t_{1},t_{2}\right)=0$ if $t_{1}<0$ or $t_{2}<0$.
Moreover we have: $$\int_{0}^{\infty}\int_{0}^{\infty}u\left(t_{1},t_{2}\right)f_{\left(T_{1},T_{2}\right)}\left(t_{1},t_{2}\right)dt_{1}dt_{2}=\int_{0}^{\infty}\int_{0}^{t_{2}}f_{\left(T_{1},T_{2}\right)}\left(t_{1},t_{2}\right)dt_{1}dt_{2}$$ because of the properties of function $u\left(t_{1},t_{2}\right)$: it takes value $1$ if $t_{1}<t_{2}$ and value $0$ otherwise.