Find distribution function of the probability density function : $f(x) = \frac{\alpha}{x_0}(\frac{x_0}{x})^{\alpha+1}$ for $x > x_0 , \alpha>1$ A continuous random variable $x$ has the following probability density function:
$f(x) = \frac{\alpha}{x_0}(\frac{x_0}{x})^{\alpha+1}$ for $x > x_0 , \alpha>1$
The distribution function and the mean of $x$ are given
respectively by:
A. $1 - (\frac{x}{x_0})^{\alpha} , \frac{\alpha - 1}{\alpha}x_0  $
B. $1 - (\frac{x}{x_0})^{-\alpha} , \frac{\alpha - 1}{\alpha}x_0  $
C. $1 - (\frac{x}{x_0})^{-\alpha} , \frac{\alpha}{\alpha - 1}x_0  $
D. $1 - (\frac{x}{x_0})^{\alpha} , \frac{\alpha}{\alpha - 1}x_0  $
I tried finding the Distribution function by integrating $\int_{-\infty}^{\infty}f(x)$ but I don't know whether to treat $x_0$ as a variable or constant also if I should start the integration from $x_0$ or not.
 A: General formula for finding distribution function $F(x)$ when probability density function $f(x)$ is given:
$$F(x)=\int_{-\infty}^{x}f(x)\,dx\tag{1}\label{1}$$
Usually only non zero expression of density is given. In other words, in this case we have this probability density function:
$$f(x) =
\begin{cases}
0,& x\leq x_0 \\
\frac{\alpha}{x_0}\left(\frac{x_0}{x}\right)^{\alpha+1},& x > x_0
\end{cases}\tag{2}\label{2}$$
Now $f(x)$ is defined on the whole $\mathbb{R}$ and thus we can use formula $(1)$. It has two parameters $x_0$ and $\alpha$ (they are constants). These parameters must satisfy inequalities $x_0>0$ and $\alpha>0$ since any probability density $f(x)$ has to satisfy these two properties:
$$\forall x \in \mathbb{R}: f(x) \geq 0$$
$$\int_{-\infty}^{\infty}f(x)\,dx = 1$$
Now lets use $(1)$ and $(2)$. Function $F(x)$ must be defined on the whole $\mathbb{R}$, but by looking at $(2)$ we see that the expression of $f(x)$ is different in these two regions: $x\leq x_0$ and $x>x_0$. That's why these two cases will be dealt separately.
If $x\leq x_0$ then
$$F(x)=\int_{-\infty}^{x}f(x)\,dx = \int_{-\infty}^{x}0\,dx = 0$$
If $x> x_0$ then
$$F(x)=\int_{-\infty}^{x}f(x)\,dx = \int_{-\infty}^{x_0}f(x)\,dx+\int_{x_0}^{x}f(x)\,dx = \int_{-\infty}^{x_0}0\,dx+\int_{x_0}^{x}\frac{\alpha}{x_0}\left(\frac{x_0}{x}\right)^{\alpha+1}\,dx=$$
$$= 0 + x_0^\alpha\frac{-1}{x^\alpha}\bigm|_{x_0}^{x} = 1-\left(\frac{x_0}{x}\right)^\alpha$$
So, we've got this distribution function:
$$F(x) =
\begin{cases}
0,& x\leq x_0 \\
1-\left(\frac{x_0}{x}\right)^\alpha,& x > x_0
\end{cases}\tag{3}\label{3}$$
Mathematical expectation is
$$\mathbb{E}X = \int_{-\infty}^{\infty}x\cdot f(x)\,dx = \int_{-\infty}^{x_0}x\cdot 0\,dx + \int_{x_0}^{\infty}x\cdot \frac{\alpha}{x_0}\left(\frac{x_0}{x}\right)^{\alpha+1}\,dx =$$
$$= 0 + \alpha x_0^\alpha\int_{x_0}^{\infty}\frac{1}{x^\alpha}\,dx = \begin{cases}
\infty,& 0<\alpha\leq 1 \\
\frac{\alpha x_0}{\alpha-1},& \alpha > 1
\end{cases}$$
Bonus tip:
While distribution function has to be defined on the whole $\mathbb{R}$ a probability density function doesn't have to be defined on the whole $\mathbb{R}$, but it still has to be defined almost everywhere on $\mathbb{R}$. It means that the same distribution (with the same parameters of Pareto distribution in this case) can have more than one different probability density function, but the distribution function will be the same. If $f(x)$ and $f_1(x)$ are different density functions of the same distribution then the equality $f_1(x)=f(x)$ must be true for almost all $x$ (it means $f_1(x)\neq f(x)$ only on a set $A$, whose Lebesgue measure, in other words length in this case, is zero).
For example,
$$f_1(x) =
\begin{cases}
0,& x< x_0 \\
1,& x= x_0 \\
\frac{\alpha}{x_0}\left(\frac{x_0}{x}\right)^{\alpha+1},& x > x_0
\end{cases}$$
is also a probability density function of Pareto distribution and using $(1)$ we'll still get the same distribution function $(3)$. The values of functions $f(x)$ and $f_1(x)$ differ only at one point $x=x_0$ (the length of one point is zero).
