Question

# Suppose that X and Y are continuous random variables with joint pdf f(x,y)=e^{-(x+y)} 0<x<\infty\ and\ 0<y<\infty and zero otherwise.Find P(X+Y>3)

Random variables

Suppose that X and Y are continuous random variables with joint pdf $$f(x,y)=e^{-(x+y)} 0$$ and zero otherwise.
Find $$P(X+Y>3)$$

2021-05-24

Step 1
Introduction:
The joint density function of two random variables X and Y is given below:
$$f(x,y)=\left\{\begin{matrix}e^{-(x+y)}.\ \ \ 0\leq x \leq \infty, 0\leq y < \infty \\0.\ elsewhere \end{matrix}\right\}$$
The marginal density function of X is,
$$f(x)=\int_{0}^{\infty}e^{-(x+y)}dy$$
$$=e^{-x}\int_{0}^{\infty}e^{-y}dy$$
$$=-e^{-x}[-e^{-y}]_{0}^{\infty}$$
$$=-e^{-x}[e^{-\infty}-e^{-0}]$$
$$=-e^{-x}[0,1]$$
$$=e^{-x}$$
Step 2
The probability of $$P(X + Y > 3)$$ is obtained as 0.1991 from the calculation given below:
$$P(X+Y>3)=1-P(X+Y\leq 3)$$
$$=1-\int_{0}^{3}\int_{0}^{3-x}e^{-(x+y)}dxdy$$
$$=1-\int_{0}^{3}\int_{0}^{3-x}e^{-x}e^{-y}dxdy$$
$$=1-\int_{0}^{3}e^{-x}dx\int_{0}^{3-x}e^{-y}dy$$
$$=1-\int_{0}^{3}e^{-x}dx[-e^{-y}]_{0}^{3x}$$
$$=1-(-\int_{0}^{3}e^{-x}dx[e^{-(3-x)}-e^{-0}])$$
$$=1-(-\int_{0}^{3}e^{-x}dx[e^{-3+x}-1])$$
$$=1-\int_{0}^{3}e^{-x}dx[1-e^{-3+x}]$$
$$=1-\int_{0}^{3}e^{-x}-e^{-3}e^{x}e^{-x}dx$$
$$=1-([-e^{-x}]_{0}^{3}-e^{-3}[x]_{0}^{3})$$
$$=1-(-[e^{-3}-e^{-0}]-e^{-3}[3-0])$$
$$=1-([1-e^{-3}]-3e^{-3})$$
$$=1-1+e^{-3}+3e^{-3}$$
$$=4e^{-3}$$
=0.1991
Thus, the probability of $$P(X + Y > 3)$$ is 0.1991.