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If X is a continuous random variable with range $[x_l,\infty)$ and p.d.f.

$f_x(X) \propto x^{-a}$, for $x\in[x_l,\infty)$

for some values $x_l > 0$ and $a \in \mathbb{R}$.

How do I calculate the range of values for a which $f_x(X)$ is a valid p.d.f.?

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Integrate, and deduce a(xl). – gnometorule Jan 5 '12 at 15:32
The area under the pdf is $1$. So, integrate $f(x) = cx^{-a}$ from $x_1$ to $\infty$ and choose $c$ such that the value of the integral equals $1$. Hint: you probably need to restrict $x_1 > 0, a> 0$ for this to work. – Dilip Sarwate Jan 5 '12 at 15:35
@DilipSarwate I have -c(xl)^(-a) = 1-a but how do I get the range for a which is valid? – jamess Jan 5 '12 at 17:17
First, check your math! Then express $c$ in terms of $a$ and $x_1$, and decide: for fixed $x_1 > 0$, and knowing that $c$ must be positive (think why this must be so!), what is the range of values of $a$ for which your formula for $c$ gives positive values for $c$? Remember in all this that $x_1$ is a fixed positive number. Set it to $1$ if you like so as to not clutter up things. – Dilip Sarwate Jan 5 '12 at 17:29
DUPLICATE: This question is nearly verbatim the same as this one:… – Michael Hardy Jan 5 '12 at 19:30
up vote 1 down vote accepted

A probability density function $f$ must satisfy:

1) $f(x)\ge 0 $ for all $x$,


2) $\int_{-\infty}^\infty f(x)\, dx =1$.

Your density has the form $$f(x)=\begin{cases}c \cdot x^{-a} & x\ge x_l \\ 0 & \text{ otherwise}\end{cases}$$ where $x_l>0$.

We need 1) to hold; $f$ must be non-negative.

When does that happen?

The first thing to note here is that, since $x_l>0$, it follows that $x^{-a}\ge0$; and thus $c$ must be positive in order for 1) to hold.

So far so good. $a$ can be any number (so far as we have surmised) and, for $c>0$, $f$ would define a density as long as condition 2) holds.

Your task now is to figure out when it does.

A hint towards achieving that end would be to consider when the integral appearing in 2) is converges. If the integral does converge, you can then select $c$ so that it converges to 1; and in this case, $f$ would indeed define a density.

If the integral does not converge, then $f$ would not define a density.

Read no further if all you want is a hint...

To determine the range of values of $a$ for which $f$ is a density we need to determine when $$\tag{3}\int_{x_l}^\infty c x^{-a}\,dx$$ converges.

Towards this end, note that the integral in (3) is convergent if and only if $a>1$. This is because the $p$-integral $\int_{x_l}^\infty {1\over x^p}\,dx $ converges if and only if $p>1$ (the lower limit presents no problems, since $x_l>0$).

This answers your question as to what range of values of $a$ (I assume $a$) give a valid density.

If you have $a>1$ and want to find the value of $c$, use 2): set $$ 1=\int_{x_l}^\infty cx^{-a}\,dx =\lim_{b\rightarrow\infty} { -cx^{-a+1}\over -a+1}\biggl|_{x_l}^b={cx_l^{1-a}\over a-1}, $$ then solve for $c$.

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Say you've found $$ \int_{x_\ell}^\infty x^{-a}\;dx. $$

This is a valid pdf if the integral is finite; it is not a valid pdf if the integral is $\infty$.

The family of distributions we're dealing with here are called the Pareto distributions, after the Italian economist Vilfredo Pareto (1848--1923). It arises from Pareto's way of modeling the distribution of incomes. Pareto proposed that $$ \log N = A - a\log x $$ where $N$ is the number of people whose incomes are more than $x$. A bit of trivial algebra shows how the density arises from what Pareto proposed, but Pareto neglected to think about $x_\ell$.

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