Answer:
A. p.d.f(x,y)= 1/2
B. P(X>Y)= 3/4
Step-by-step explanation:
A. We are told that p.d.f(x,y) is constant on the rectangle 0<x<2, 0<y<1. To find that constant the property of probabilities that is useful here is the following:
[tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}p.d.f(x,y)\ dx\ dy=1 [/tex]
We know that f(x,y)=c (c is a constant) in the rectangle and outside of it f(x,y)=0, so the integral above could be rewritten easily as:
[tex]\int_{0}^{2}\int_{0}^{1}c\ dy\ dx=1[/tex]
Solving the integral it´s possible to find the value c:
[tex]\int_{0}^{2}\int_{0}^{1}c\ dy\ dx=\int_{0}^{2}c\ dx[/tex]
[tex]\int_{0}^{2}c\ dx= 2c [/tex]
[tex]2c=1\ \rightarrow c=1/2[/tex]
This is the value of f(x,y) in the rectangle
B. Now that we know our p.d.f(x,y), we are asked to find P(X>Y) in the rectangle. There´s a graph of the area (probability) that we are looking for, it´ll make easy the understanding of the move that we are going to make.
We are asked for the probability that X>Y, i.e the part of the rectangle below the graph of X=Y. Because x is between 0 and 2, whenever x>1, it will immediately be greater than y. If x<1, we just need to assure that 0<y<x. With this data we can find the probability this way:
[tex]P(X>Y)=\int_{0}^{1}\int_{0}^{x}1/2\ dy\ dx + \int_{1}^{2}\int_{0}^{1}1/2\ dy\ dx[/tex]
The first integral represents the area (that´s equal to the probability) of the triangle at the left of the blue line, and the other integral it´s related to the square that´s right of the blue line.
And now we are ready to solve the problem:
[tex]P(X>Y)=\int_{0}^{1}\frac{x}{2}\ dx + \int_{1}^{2}1/2\ dx [/tex]
[tex]P(X>Y)=\frac{x^{2}}{4}|_{0}^{1} + 1/2 [/tex]
[tex]P(X>Y)=1/4 + 1/2[/tex]
We conclude that P(X>Y)=3/4
