Limits of Integration - Joint probability density

Question

Given : $f(x,y)$ = $e^{-x-y}$ if $x>0 , y>0 $ and $0$ elsewhere

Find $P(X+Y>3)$ .

For limit I proceeded this problem in the same method this was done Joint probability density function and limits of integration

Sol:
$P(X+Y>3)=1-P(X+Y<3)$, now I have tried the integration of $P(X+Y<3)$ , $$\int_0^3 \int_0^{3-x} e^{-x-y} dydx \ $$ But the integration limits are wrong when I checked , the actual limits are $$\int_2^3\ \int_2^3\ e^{-x-y} dydx $$ ,here they have found directly $P(X+Y>3)$

Similarly , for $f(x,y) = 1/y$ for $0<x<y , 0<y<1 $ ,find $P(X+Y>1/2)$ .Integration part is totally different.

Help Please ! How was the 2 lower limit determined ?

And can any one refer a book where I can learn how to solve probability problems like these. Thanks in advance!

Question

Solution given

Answer 1

The problems are more related to evaluation of multiple integrals than probabilities.

So I suggest taking a look at worked out examples of double/triple integrals in any standard multivariable calculus textbook.

Direct evaluation gives

\begin{align} P(X+Y>3)&=\iint_{x+y>3}e^{-(x+y)}\mathbf1_{x,y>0}\,dx\,dy \\&=\int_0^\infty \left(\int_{\max(0,3-y)}^\infty e^{-x}\,dx\right)\,e^{-y}\,dy \\&=\int_0^\infty e^{-\max(0,3-y)}e^{-y}\,dy \\&=e^{-3}\int_0^3\,dy+\int_3^\infty e^{-y}\,dy \\&=\frac{4}{e^3} \end{align}

Answer 2

For $P\{X+Y <3\}$ the correct expression is $\int_0^{3}\int_0^{3-x} e^{-x-y}\, dy\, dx$. note the upper limit $3-x$.

Answer 3

Alternative methods:

1. The given joint pdf is $f_{XY}(x,y)=f_X(x)f_Y(y)$ where $X,Y$ are iid exponential with mean 1. Thus

$$Z=(X+Y)\sim\text{Gamma}(2;1)$$

and

$$2Z\sim \chi_{(4)}^2$$

concluding:

$$P(X+Y>3)=P(2Z>6)\approx 0.199$$

using chi-square tables

---

2. being $Z\sim\text{Gamma}(2;1)$ where 1 is the rate parameter, we have

$$f_Z(z)=z e^{-z}$$

thus

$$\mathbb{P}[Z>3]=\int_3^{\infty}z e^{-z}dz=4e^{-3}\approx 0.199$$

Note that in method 2) $f(z)$ is the convolution of the two marginals, $X,Y$