Cumulative distribution function - trouble finding constant

Question

f(x) = 0.25x for 0 < x < 1

   1/(4x^n) for x > 1

the question is find n.

I'm having some trouble arriving at an answer. The correct answer is 9/7 whereas I'm getting -0.75. Should I be evaluating the second argument from 1 to infinity?

Answer 1

Here is my interpretation of your question:

Find $n$ such that the piecewise function $$ f(x)=\begin{cases}\frac{x}{4},&0<x\leq 1\\\frac{1}{4x^n},&x>1\end{cases} $$ is a density function (i.e., satisfies $\int_0^{\infty}f(x)\ dx=1$).

To solve it, we just calculate the integral by splitting it up into two pieces as follows. $$ \int_0^\infty f(x)\ dx=\int_0^1 f(x) dx+\int_1^\infty f(x) dx=\left[\frac{x^2}{8}\right]_0^1+\left[\frac{1/(1-n)}{4x^{n-1}}\right]_1^\infty. $$ The first term evaluates to $\frac{1}{8}$, and the second evaluates to $\frac{1}{4(n-1)}$. Setting the sum equal to $1$ gives $$ \frac{1}{4(n-1)}=\frac{7}{8}\quad\implies\quad\frac{2}{7}=n-1. $$