# Validity of a Probability Density Function [duplicate]

Possible Duplicate:
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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## marked as duplicate by David Mitra, Asaf Karagila, Nate Eldredge, Srivatsan, t.b.Jan 5 '12 at 21:41

Thanks, Chris for replying to that :) Do you happen to have any hints for me, please? Nikhil, I saw that one, but I couldn't comment on it as I wanted to talk to Dilip. So I thought I could get to talk to him if he commented here. – Sorin Cioban Jan 5 '12 at 17:53
For convergence of the integral we need $a>1$. That's the only general condition on $a$. If $f_X(x)=kx^{-a}$, we find that $\int_{x_l}^\infty kx^{-a}\,dx=\frac{k}{(a-1)x_l^{a-1}}$. For any $a$, $x_l$ there is a unique $k$ such that the integral is (as needed) equal to $1$. – André Nicolas Jan 5 '12 at 18:00
Thanks, but how do we find the range of values for a? Also, do you have an email/gtalk account we could use to talk more about this, please? Oh wait, so you're saying that the range is basically just (1, infinity)? – Sorin Cioban Jan 5 '12 at 18:03
@SorinCioban I replied to your previous query about how to find the range for $a$. Why didn't you read that instead of posting the same question again? I am recommending closure of this question. – Dilip Sarwate Jan 5 '12 at 18:08
Hi Dilip. I'm not the same person as the one that posted in the other thread. I couldn't comment on that one to talk to you so I thought I could get to talk to you by posting the question again. Could we talk privately about the question in any way? Thanks :) – Sorin Cioban Jan 5 '12 at 18:14

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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You managed to write a lot without mentioning that this is called the Pareto distribution, after Vilfredo Pareto. – Michael Hardy Jan 5 '12 at 19:14
I thought the OP's issue was a calculus one. So I wanted to do those details only. Didn't know his first name was Vilfredo. – André Nicolas Jan 5 '12 at 19:18
I don't think it's generally a good idea to restrict answers to what the OP explicitly asks about. – Michael Hardy Jan 5 '12 at 19:29
@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. – André Nicolas Jan 5 '12 at 19:35

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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