# Verifying that Uniform and Pareto distributions are probability densities

• Jun 8th 2010, 06:38 PM
acevipa
Verifying that Uniform and Pareto distributions are probability densities
Show that the following functions are probability densities, i.e.

$f(x)\geq 0$ and $\int_{-\infty}^{\infty}f(x) dx =1$

Uniform distribution:

$f(x)=\left\{\begin{array}{cc}\frac{1}{b-a} & \mbox{ for } a\leq x \leq b\\0 & \mbox{ otherwise }\end{array}\right.$

Pareto distribution: For $k>0$

$f(x)=\left\{\begin{array}{cc}\frac{k}{x^{k+1}} & \mbox{ for } x \geq 1\\0 & \mbox{ otherwise }\end{array}\right.$
• Jun 8th 2010, 07:13 PM
mr fantastic
Quote:

Originally Posted by acevipa
Show that the following functions are probability densities, i.e.

$f(x)\geq 0$ and $\int_{-\infty}^{\infty}f(x) dx =1$

Uniform distribution:

$f(x)=\left\{\begin{array}{cc}\frac{1}{b-a} & \mbox{ for } a\leq x \leq b\\0 & \mbox{ otherwise }\end{array}\right.$

Pareto distribution: For $k>0$

$f(x)=\left\{\begin{array}{cc}\frac{k}{x^{k+1}} & \mbox{ for } x \geq 1\\0 & \mbox{ otherwise }\end{array}\right.$

Where exactly are you stuck here?
• Jun 9th 2010, 12:33 AM
acevipa
Could someone please verify my working?

Uniform Distirbution:

$\int_{-\infty}^{\infty}\frac{1}{b-a}dx = \int_{a}^{b}\frac{1}{b-a}dx$

$=\left[\frac{x}{b-a}\right]_a^b$

$=\frac{b}{b-a}-\frac{a}{b-a}$

$=\frac{b-a}{b-a}$

$=1$

$\int_{-\infty}^{\infty}\frac{1}{b-a}dx=1$

Why is $f(x) \geq 0$

Now for a Pareto distribution:

$\int_{-\infty}^{\infty}\frac{k}{x^{k+1}}dx = \int_{1}^{\infty}\frac{k}{x^{k+1}}dx$

$=k\int_{1}^{\infty}\frac{1}{x^{k+1}}dx$

$=k \left[\frac{1}{-kx^k}\right]_1^{\infty}$

$=-\left[\frac{1}{x^k}\right]_1^{\infty}$

$=-\left[\frac{1}{\infty^k}-\frac{1}{1^k}\right]$ Not too sure if you sub $\infty$ in since it's a definite integral. What's the accepted way of evaluating this indefinite integral

$=-[0-1]$

$\int_{-\infty}^{\infty} \frac{k}{x^{k+1}}dx=1$

Since $k>0$ and $x\geq 1 \Longrightarrow x^{k+1}\geq 0$

$\Longrightarrow \frac{k}{x^{k+1}} \geq 0$
• Jun 9th 2010, 12:46 AM
mr fantastic
Quote:

Originally Posted by acevipa
Could someone please verify my working?

Uniform Distirbution:

$\int_{-\infty}^{\infty}\frac{1}{b-a}dx = \int_{a}^{b}\frac{1}{b-a}dx$

$=\left[\frac{x}{b-a}\right]_a^b$

$=\frac{b}{b-a}-\frac{a}{b-a}$

$=\frac{b-a}{b-a}$

$=1$

$\int_{-\infty}^{\infty}\frac{1}{b-a}dx=1$

Why is $f(x) \geq 0$ Mr F says: Although Russell and Whitehead would disagree (if they were still alive), it should be obvious that f(x) = 1/(b - a) > 0 if b > a ....

Now for a Pareto distribution:

$\int_{-\infty}^{\infty}\frac{k}{x^{k+1}}dx = \int_{1}^{\infty}\frac{k}{x^{k+1}}dx$

$=k\int_{1}^{\infty}\frac{1}{x^{k+1}}dx$

$=k \left[\frac{1}{-kx^k}\right]_1^{\infty}$

$=-\left[\frac{1}{x^k}\right]_1^{\infty}$

$=-\left[\frac{1}{\infty^k}-\frac{1}{1^k}\right]$ Mr F says: This is poor notation. The integral is an improper integral and needs to be carefully calculated using limits. Do you know how to do that?

$=-[0-1]$

$\int_{-\infty}^{infty}\frac{k}{x^{k+1}}dx=1$

It looks OK (except for my second red note).
• Jun 9th 2010, 12:51 AM
acevipa
Quote:

Originally Posted by mr fantastic
It looks OK (except for my second red note).

Ok, yeah your first red note seems kind of obvious now.

I thought my notation wasn't that good. How would you write it as?
• Jun 9th 2010, 12:56 AM
mr fantastic
Quote:

Originally Posted by acevipa
Ok, yeah your first red note seems kind of obvious now.

I thought my notation wasn't that good. How would you write it as?

$= k \lim_{\alpha \to +\infty} \int_{1}^{\alpha}\frac{1}{x^{k+1}}dx$

$=k \lim_{\alpha \to +\infty} \left[\frac{1}{-kx^k}\right]_1^{\alpha}$

$=- \lim_{\alpha \to +\infty} \left[\frac{1}{x^k}\right]_1^{\alpha}$

$=- \lim_{\alpha \to + \infty}\left[\frac{1}{\alpha^k}-\frac{1}{1^k}\right]$

$=-[0-1] = 1$.
• Jun 9th 2010, 01:01 AM
acevipa
Quote:

Originally Posted by mr fantastic
$= k \lim_{\alpha \to +\infty} \int_{1}^{\alpha}\frac{1}{x^{k+1}}dx$

$=k \lim_{\alpha \to +\infty} \left[\frac{1}{-kx^k}\right]_1^{\alpha}$

$=- \lim_{\alpha \to +\infty} \left[\frac{1}{x^k}\right]_1^{\alpha}$

$=- \lim_{\alpha \to + \infty}\left[\frac{1}{\alpha^k}-\frac{1}{1^k}\right]$

$=-[0-1] = 1$.

Thanks a lot. That makes more sense