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Probability Density Function Validity

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 $a \in \mathbb{R}$.

After integrating $f(x)$, how can I find the range of values for $a$ that would make $f(x)$ a valid pdf?

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    A [complete solution](http://math.stackexchange.com/a/96712/15941) to the _other_ identical question has just been posted by David Mitra2012-01-05

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Let $a$ be a real number, and let $f_X(x)=kx^{-a}$, where $k>0$. Since we are dealing with positive quantities, the only thing that we require in order for $f_X(x)$ to be a density function is $\int_{x_l}^\infty kx^{-a}\,dx=1.\qquad\qquad(\ast)$ The above integration, like many others that arise in probability, is over an infinite interval, so may fail to exist.

We will show that the above integral converges if $a>1$ and diverges otherwise.

Let $I(M)=\int_{x_l}^M kx^{-a}\,dx.$ By definition, if $\lim_{M\to\infty}I(M)$ exists, the integral in $(\ast)$ converges and has value equal to that limit.

Suppose first that $a>1$. Integrating, we find that $I(M)=\int_{x_l}^M kx^{-a}\,dx=\left.\frac{-k}{(a-1)x^{a-1}}\right|_{x_l}^M$ Thus $I(M)=\frac{k}{(a-1)x_{x_l}^{a-1}}-\frac{k}{(a-1)M^{a-1}}. \qquad\qquad(\ast\ast)$ Since $a-1>0$, we can see that $\frac{k}{(a-1)M^{a-1}}\to 0$ as $M\to\infty$. It follows that if $a>1$, then $\int_{x_l}^\infty kx^{-a}\,dx=\frac{k}{(a-1){x_l}^{a-1}}.$

For any $a>1$, and any positive $x_l$, we can find a unique constant of proportionality $k$ such that $\frac{k}{(a-1){x_l}^{a-1}}=1.$ Just take $k=(a-1)x_l^{a-1}$. So everything is fine if $a>1$.

We complete the analysis by showing that if $a \le 1$, then our integral does not converge. There are two somewhat different cases, $a=1$ and $a<1$. Suppose first that $a=1$. Then $I(M)=\int_{x_l}^M kx^{-1}\,dx=\left.k\ln x\right|_{x_l}^M=k\ln M-k\ln x_l.$ As $M\to\infty$, $\ln M\to\infty$, so $I(M)$ does not have a finite limit, and therefore $\int_{x_l}^\infty kx^{-1}\,dx$ does not exist.

Finally, we deal with $a<1$. In this case, $I(M)=\frac{kM^{1-a}}{1-a}-\frac{kx_l^{1-a}}{1-a}.$ As $M\to\infty$, $I(M)\to\infty$, so the integral from $x_l$ to $\infty$ diverges (does not have a finite value).

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    @Michael Hardy: Agreed! I usually add a "Comment" at the end of most answers, pushing a bit beyond the actual question. Didn't this time.2012-01-05
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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 $x_\ell$.