Question

# The continuous random variables X and Y are statistically independent and have marginal density functions f_{X}(x)=2x, 0 \leq x \leq 1, f_{Y}(y)=\frac{1}{y^{2}}, y\geq 1. Calculate the probability P(X\leq 0.5, Y\leq 2)=?

Random variables
The continuous random variables X and Y are statistically independent and have marginal density functions $$f_{X}(x)=2x, 0 \leq x \leq 1, f_{Y}(y)=\frac{1}{y^{2}}, y\geq 1$$.
Calculate the probability $$P(X\leq 0.5, Y\leq 2)=$$?

2021-06-02
Step 1
If two random variables X and Y, are stochastically independent then the joint probability distribution of X and Y can be expressed as $$f(x,y)=f_{X}(x)f_{X}(y)$$.
Step 2
The probability density function of X is $$f_{X}=2x$$, for $$0 \leq x \leq 1$$.
The probability density function of Y is, $$f_{X}=\frac{1}{y^{2}}$$, for $$1\leq y$$.
Thus,
$$P(X\leq 0.5, Y\leq 2)=(\int_{0}^{0.5}2xdx)(\int_{1}^{2}\frac{1}{y^{2}}dy)$$
$$=(2[\frac{x^{2}}{2}]_{0}^{0.5})([\frac{y^{-2+1}}{-2+1}]_{1}^{2})$$
$$=(0.5^{2}-0)(\frac{1}{1}-\frac{1}{2})$$
=0.125
Thus, the required probability is 0.125.