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
ANSWERED
asked 2021-06-01
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 \(P(X\leq 0.5, Y\leq 2)=\)?

Answers (1)

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.
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