I got the following problem:
Let $X$ be a continuous random variable with $CDF$ denoted $F_X$ defined as follows:
$F_X(x)=
\begin{cases}
1-x^{-4/3}, & x\in[1,\infty) \\
0, & x\in (-\infty,1)
\end{cases}$
Find the PDF of $X$.
My try:
Since the PDF (denoted $f_X$) is the derivative of the CDF I get that $\forall x\in(1,\infty), f_X(x)=\frac{4}{3}x^{-7/3}$ and that $\forall x\in(-\infty,1), f_X(x)=0$.
Now I don't know what to do. The function $F_X$ is not differentiable at $x=1$ since the derivative from the right and from the left got different values and since the domain of the PDF must be $\mathbb{R}$ .
Is defining $f_X$ to be zero (or any other non-negative value) when $x=1$ is the solution?
Thanks for any help.
Best Answer
As yourself this: If I define $f_X (1) = 1$, will it be true that
$$ \int_{-\infty}^x f_X(t) ~ dt = F_X(x) ? $$
If the answer's "yes", then you've got a PDF for $X$.
Corollary to the result you'll get: the value of the PDF at any particular point doesn't matter. Why?