How to calculate the Pareto distribution with m=0 I am trying to calculate the Pareto distribution where the random variable has this range: $0 \le x \le \infty$
The book I found the Pareto distribution in has defined it like so:
$Pareto(x|m,k) = k \times m^{k} \times x^{-(k+1)} \times I(x \ge m)$
where $I(true) = 1$ and $I(false) = 0$
The book also has a few example graphs:

My question is this: it appears that the $m$ is the lower limit of the range of the random variable, and I would like to set mine to zero (like the red dotted line in the plots above). The only problem is that when $m$ in the formula for Pareto is set to zero, the whole thing ends up being zero ... so how did they obtain the red dotted line plot (which is clearly not zero everywhere)?
Thanks in advance
 A: You can't set the lower limit to zero. The reason is that the integral $\int_0^a x^{-(k+1)}dx $ diverges at the lower endpoint (for $k\ge 0$). This means the distribution can't be defined with support going all the way down to zero.
That said, there are other ways to regulate the divergence than just taking a hard cutoff value. For instance you could include a convergence factor $e^{-m/x}$ in the density and then take the support $0<x<\infty$. That would be a distribution that is similar to the Pareto (for $x\gg m$) but has support on all the positive reals. However note that the density drops sharply for $x<m$. You aren't really eliminating the cutoff, just smoothing it out a bit.
A: One way around this is to use the Pareto distribution Type II with  the mean set to zero, which is called the Lomax distribution.
The PDF for this is 
$$\begin{array}{cc}
 \Bigg\{ & 
\begin{array}{cc}
 \dfrac{\alpha}{k}  \left(\dfrac{k+x}{k}\right)^{-\alpha -1} & x\geq 0 \\
 0 & \text{Elsewhere} \\
\end{array}
 \\
\end{array}$$
Or, to alternatively use Pareto Types III or IV. A solution for Type IV is shown here. The Pareto distribution Type IV pdf is,
$$
f(k,\alpha,\gamma,\mu;x)=\begin{array}{cc}
 \Bigg\{ & 
\begin{array}{cc}
 \frac{\alpha  k^{-1/\gamma } (x-\mu )^{\frac{1}{\gamma }-1} \left(\left(\frac{k}{x-\mu }\right)^{-1/\gamma }+1\right)^{-\alpha -1}}{\gamma } & x\geq \mu  \\
 0 & \text{Elsewhere} \\
\end{array}
 \\
\end{array}$$
And, then set the location parameter, $\mu$, to zero, which yields
$$f(k,\alpha,\gamma,0;x)=\begin{array}{cc}
 \Bigg\{ & 
\begin{array}{cc}
 \frac{\alpha  k ^{-1/\gamma } x^{\frac{1}{\gamma }-1} \left(\left(\frac{k }{x}\right)^{-1/\gamma }+1\right)^{-\alpha -1}}{\gamma } & x\geq 0 \\
 0 & \text{Elsewhere} \\
\end{array}
 \\
\end{array}
$$
Next, when $\gamma>1$ the limit as $x\rightarrow0^+$ is $\infty$, which is what was requested, with the difference from the ordinary Pareto distribution (Type I) being that the discontinuity at $x=0$ is now integrable whereas the Pareto distribution Type I cannot take an $x=0$ as in that case the discontinuity at $x=0$ would not be integrable, i.e., the cdf would be infinite. For all values of $\gamma$ the Type IV cdf exists for $x\geq0$ and is
$$F(k,\alpha,\gamma,0;x)=\begin{array}{cc}
 \Bigg\{ & 
\begin{array}{cc}
 1-\left(\left(\frac{x}{k }\right)^{1/\gamma }+1\right)^{-\alpha } & x\geq 0 \\
 0 & \text{Elsewhere} \\
\end{array}
 \\
\end{array}
$$
A plot of the Type IV solution quite similar to the OP plot is shown below as the red dashed curve. That its cdf now exists is illustrated as the blue dotted curve.

A: Why don't you just use change of variable? Let $X\sim\;$Pareto($m,k$) and define $Y=X-m$. Hence, the density function of $Y$ is $$f_Y(y|m,k)=f_X(y+m|m,k)=\dfrac{km^k}{(y+m)^{k+1}}I(y\geq 0).$$
