Question

Consider two continuous random variables X and Y with joint density functionf(x,y)=\begin{cases}x+y\ o \leq x \leq 1, 0 \leq y \leq 1\\0 \ \ \ \ otherwise\end{cases}P(X>0.8, Y>0.8) is?

Random variables

Consider two continuous random variables X and Y with joint density function
$$f(x,y)=\begin{cases}x+y\ o \leq x \leq 1, 0 \leq y \leq 1\\0 \ \ \ \ otherwise\end{cases}$$
$$P(X>0.8, Y>0.8)$$ is?

2021-06-04

Step 1
Given information:
The two continuous random variables X and Y has joint density function defined as:
$$f(x,y)=\begin{cases}x+y\ o \leq x \leq 1, 0 \leq y \leq 1\\0 \ \ \ \ otherwise\end{cases}$$
Then,
$$P(X>x, Y>y)=\int_{y}^{1}\int_{x}^{1}f(x,y)dxdy$$
Step 2
$$P(X>0.8, Y>0.8)=\int_{y=0.8}^{1}\int_{x=0.8}^{1}(x+y)dxdy$$
$$=\int_{y=0.8}^{1}\left\{\int_{x=0.8}^{1}(x+y)dx\right\}dy$$
$$=\int_{y=0.8}^{1}\left[\frac{x^{2}}{2}+xy\right]_{0.8}^{1}dy$$
$$=\int_{y=0.8}^{1}\left\{(\frac{1}{2}+y)-(\frac{0.8^{2}}{2}+0.8y)\right\}dy$$
$$=\int_{y=0.8}^{1}\left\{\frac{1}{2}+y-0.32-0.8\right\}dy$$
$$=\int_{y=0.8}^{1}(0.18+0.2y)dy$$
$$=\left[0.18y+\frac{0.2y^{2}}{2}\right]_{0.8}^{1}$$
$$=\left[(0.18+0.1)-(0.18\times 0.8+0.1\times 0.8^{2})\right]$$
$$=0.072$$
The required probability is 0.072.