Let $T_1$ has exponential distribution $\sim\text{Exp}(\lambda_1)$ and $T_2$ has exponential distribution $\sim\text{Exp}(\lambda_2)$
Assume $T_1$ = life time of refrigerator $T_2 =$ life time of Stove.
We want to find $P (T_1 < T_2)$ the probability that the refrigerator fails before the stove.
From the book "Introduction to probability" by Joseph K. Blitzstein and Jessica Hwang page 293:
its says: 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 $t_1 < t_2$
$$P(T_1 < T_2) = \int_0^{\infty} \int_0^{t_2} \lambda_1 e^{- \lambda_1 t_1} \lambda_2 e^{- \lambda_2 t_2} \,\mathrm{d}t_1 \,\mathrm{d}t_2 $$
i do not understand why the above equality is valid?
can I think of $t_1$ as random variable in the X axis and $t_2$ as random variable in the Y axis and their joint PDF is a surface lies on 3 dimension space with Z axis indicates probability. if i think of it in that way then if we do integration according to the above equality. assume $t_2 = 3$ then it is $P(T_1 < 3)$ I can find a point over an area that has been integrated ex. point (2.5 , 1) which is the point that $T_1 > T_2$ but why when we calculate $P (T_1 < T_2)$ we integrate the area in the event $T_1 > T_2$ ??? you can see the area of integration in my attached photo highlighted in blue and the point (2.5 , 1) in red colour. The Y axis is $T_2$ and the X-axis is $T_1$ this is the view from top.
please explain to me step by step. Im a newbie.
The picture you created does not represent the region of integration. The point $(2.5,1)$ is not in the region of integration.
The authors state that the region of integration satisfies the inequalities $$t_1 > 0, \quad t_2 > 0, \quad t_1 < t_2.$$ If you let $t_1$ be plotted on the horizontal axis and $t_2$ on the vertical axis, then $t_1 > 0$ is the right half plane, $t_2 > 0$ is the upper half plane, hence their intersection is the first quadrant. Then the inequality $t_1 < t_2$ comprises those points above the diagonal line $t_1 = t_2$. Thus the intersection of this region with the first quadrant excludes the aforementioned point.