I am having some trouble with understanding the following:
I have a probability density function $f_X=\frac{1}{(1+x)^2}$ when $x>0$ and $0$ otherwise.
I want to find the CDF by integrating, so the integral of that would be $-1\over1+x$, but I am getting confused with the limits. I know I am meant to integrate from $-\infty$ up to $x$.
Now since everything below $x=0$ is $0$ I can just integrate from $0$ to $x$.
if instead $f_X=\frac{1}{(1+x)^2}$ when $x\geq0$ then I would get 1-1/(1+x) but since at $x=0$, $f_X=0$. I am getting confused as that would give me just $-\frac{1}{1+x}$ which is negative.. I might be wrong.
Could someone help me with this?
Best Answer
The cumulative distribution is 0 for $x<0$. Your expression is only valid for positive $x$.
The cumulative distribution for non-negative $x$ will be $$F(x)=\frac{-1}{1+x}+C$$ but since $F(0)=0$, then $$0 = \frac{-1}{1+0}+C$$ So $C=1$, and the cumulative distribution is given by $1-\frac{1}{1+x}$.