# The Pareto and Exponential Distr. Where is the mistake?

1. Apr 10, 2008

### MrGandalf

Hi.

I want to examine the connection between the Pareto Distribution and the Exponential distribution. According to Wikipedia (en.wikipedia.org/wiki/Pareto_distribution) the Pareto distribution's pdf is
$$f(x; k, x_m) = k x_m^k x^{-(k + 1)} = \frac{k x_m^k}{x^{(k + 1)}}$$
$x$ is the variable, $x_m$ is the minimum value and $k$ is a positive parameter.

Wiki tells me that the Pareto distribution is linked to the Exponential distribution (which is $f(x;\lambda) = \lambda e^{-\lambda x}$) by the following:

$$f(x; k, x_m) = Exp(\ln(x/x_m); k)$$

I assume there are no mistakes on the Wiki, so I think I'm doing something wrong. Can anyone see any errors in my reasoning? I start with the Exponential distribution and work my way backwards with the specified parameters. (Sorry if you feel I'm repeating myself at the end, just wanted to get the same form as above).

$$Exp(\ln(x/x_m); k) \;=\; k e^{-k\ln[x/x_m]} \;=\; k \Big(e^{\ln[x/x_m]}\Big)^{-k}$$
$$k\Big(\frac{x}{x_m}\Big)^{-k} \;=\; k\Big(\frac{x_m}{x}\Big)^{k} \;=\; k\frac{x_m^k}{x^k} = k x_m^k x^{-k} \;=\; \frac{k x_m^k}{x^k} \;\not=\; \frac{k x_m^k}{x^{(k + 1)}}$$

Where does the extra $x$ in the denominator come from? I am very confused right now. I can't see anything wrong! I hope someone can enlighten me.

2. Apr 10, 2008

### Pere Callahan

Hm, seems right to me, maybe there is indeed an error on wiki.

3. Apr 17, 2008

### MissTK

I am digging up an old thread here, sorry.

What you are missing is the fact that f is a density and so you can't just substitute a new variable, you have to do a change of variables accounting for the fact that dx != dy. Otherwise your distribution is "stretched".

The factor 1/x that you are missing comes from d [ ln ( X / Xm) ] / dx.