Finding limits of CDF with indicator function 
*

*Suppose the continuous random variable, X, has a probability density function (pdf)
defined as $f_x(X) = cx^{-2} \mathbb{1}[1,2](x)$ ; c is a constant, c ∈ $\mathbb{R}$+


(a) Determine c so that the function, $f_x$, is a pdf.
(b) Determine the cumulative distribution function (cdf), $F_X(x)$, and verify
i. $F_x$(−∞) = 0,
ii. $F_X$(+∞) = 1, and
iii. $F_X(x)$ is a non-decreasing monotonic function of x.
So I've started this by assuming that because I'm trying to prove we have a pdf, and because the indicator function attached to my function is on the interval [1,2], I need to solve for c by
$\int_{1}^{2} cx^{-2} dx = 1 \rightarrow c= 2$ and $F_x(x) = -2x^{-1}$
I also am not sure, but I believe it would still only be defined on the interval [1,2].  This is making part b confusing. Do I then take the limit as $x \rightarrow -\infty$ or $x \rightarrow 1$? Same thing for ii, do I take the limit as $x \rightarrow \infty$ or $x \rightarrow 2$? In either case for part ii, I can't get it to equal 1, which makes me think that I did part a wrong.
Can anyone shed some light on where I'm going wrong?
 A: We have $f_x(X) = cx^{-2} \mathbb{1}[1,2](x)$, so 
$$\int_{-\infty}^\infty cx^{-2} \mathbb{1}[1,2](x)dx=\int_1^2 cx^{-2}dx=-\frac{c}{x}\Bigm|_1^2=\frac{c}{2}=1$$ 
so $c=2$ as you found. 
Next, $F_X(x)=\int_{-\infty}^x f_X(t)dt$.
Consider $x<1$ first
$$
F_X(x)=\int_{-\infty}^x 2t^{-2} \mathbb{1}[1,2](t)dt=\int_{-\infty}^x 0\,dt = 0.
$$
For $1\leq x \leq 2$
$$
F_X(x)=\int_{-\infty}^x 2t^{-2} \mathbb{1}[1,2](t)dt=\int_1^x 2t^{-2}\,dt =\color{red}{ -\frac1t\Bigm|_1^x=1-\frac{1}{x}}.
$$
For $x>2$
$$
F_X(x)=\int_{-\infty}^x 2t^{-2} \mathbb{1}[1,2](t)dt=\int_1^2 2t^{-2}\,dt = 1.
$$
So the CDF is 
$$
F_X(x)=\begin{cases}0, & x<1\cr 1-\frac1x, & 1\leq x\leq 2 \cr 1, & x>2 \end{cases}
$$
and you can check all properties: 
$$
\lim_{x\to-\infty} F_X(x) = \lim_{x\to-\infty} 0 = 0
$$ 
and so on. 
A: $F_X(x)=\int_{-\infty} ^xf_X(u)du$, $F_X(x)=0$ for $x\lt 1$ and $F_X(x)=1$ for $x\gt 2$.
Moreover, since the integrand is non-negative, $F_X(x)$ is non-decreasing.
A: A probability density function must integrate to one, so we have
$$
1 = \int_1^2 f(x)\ \mathsf dx = \int_1^2 cx^{-2}\ \mathsf dx = \frac c2\implies c=2.
$$
The distribution function $F(x) = \mathbb P(X\leqslant x)$ is zero for $x\leqslant 1$ and one for $x\geqslant 2$. From this it is clear that $\lim_{x\to-\infty}F(x)=0$ and $\lim_{x\to\infty}F(x)=1$. For $1<x<2$ we compute $F(x)$ by integrating the density:
$$
F(x) = \int_1^x f(y)\ \mathsf dy = \int_1^x 2y^{-2}\ \mathsf dy = 2 \left(1-\frac{1}{x}\right).
$$
Note that $F$ is monotonically increasing on $[1,2]$ because its derivative, the density, is strictly positive.
