# Thread: Verifying that Uniform and Pareto distributions are probability densities

1. ## Verifying that Uniform and Pareto distributions are probability densities

Show that the following functions are probability densities, i.e.

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

Uniform distribution:

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

Pareto distribution: For $\displaystyle k>0$

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

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

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

Uniform distribution:

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

Pareto distribution: For $\displaystyle k>0$

$\displaystyle f(x)=\left\{\begin{array}{cc}\frac{k}{x^{k+1}} & \mbox{ for } x \geq 1\\0 & \mbox{ otherwise }\end{array}\right.$
Where exactly are you stuck here?

3. Could someone please verify my working?

Uniform Distirbution:

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

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

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

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

$\displaystyle =1$

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

Why is $\displaystyle f(x) \geq 0$

Now for a Pareto distribution:

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

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

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

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

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

$\displaystyle =-[0-1]$

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

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

$\displaystyle \Longrightarrow \frac{k}{x^{k+1}} \geq 0$

4. Originally Posted by acevipa
Could someone please verify my working?

Uniform Distirbution:

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

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

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

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

$\displaystyle =1$

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

Why is $\displaystyle f(x) \geq 0$ 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:

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

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

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

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

$\displaystyle =-\left[\frac{1}{\infty^k}-\frac{1}{1^k}\right]$ 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?

$\displaystyle =-[0-1]$

$\displaystyle \int_{-\infty}^{infty}\frac{k}{x^{k+1}}dx=1$
It looks OK (except for my second red note).

5. 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?

6. 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?
$\displaystyle = k \lim_{\alpha \to +\infty} \int_{1}^{\alpha}\frac{1}{x^{k+1}}dx$

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

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

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

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

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

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

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

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

$\displaystyle =-[0-1] = 1$.
Thanks a lot. That makes more sense