Probability — Question 10

Question 10

A continuous random variable has CDF $F(x)=0$ for $x<0$, $F(x)=x^2/(1+x^2)$ for $x\ge0$. Find the density and $P(1<X<3)$.

Question 10 – Solution

Step 1: Differentiate the CDF on each interval. For $x<0$, $F(x)=0$, so $f(x)=0$. For $x>0$, $$\begin{align*} f(x)&=\frac{d}{dx}\left(\frac{x^2}{1+x^2}\right)\\ &=\frac{2x(1+x^2)-x^2(2x)}{(1+x^2)^2}\\ &=\frac{2x}{(1+x^2)^2}. \end{align*}$$ The CDF is continuous at $x=0$, so there is no point mass there. Hence $$\boxed{ f(x)= \begin{cases} 0,&x<0,\\[2pt] \dfrac{2x}{(1+x^2)^2},&x\ge0. \end{cases}}$$

Step 2: Use the CDF to compute the interval probability. For a continuous random variable, $$P(1<X<3)=F(3)-F(1).$$ Now $$F(3)=\frac{3^2}{1+3^2}=\frac9{10}, \qquad F(1)=\frac{1^2}{1+1^2}=\frac12.$$

Step 3: Subtract. $$P(1<X<3) =\frac9{10}-\frac12 =\frac9{10}-\frac5{10} =\frac25.$$ $$\boxed{P(1<X<3)=\frac25}$$