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.?
Best Answer
A probability density function $f$ must satisfy:
1) $f(x)\ge 0 $ for all $x$,
and
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 $c$.