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.

  • 2
    $\begingroup$ The PDF is defined only up to a set of measure zero. So one point is no problem. Indeed, a null set where it is undefined is also no problem. (Or course the CDF must be absolutely continuous, or there is no PDF at all.) $\endgroup$
    – GEdgar
    Dec 19, 2014 at 15:40

1 Answer 1


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?


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