The relationship between pdf and cdf of a continuous random variable

If the pdf(probability density function) of Y is continuous, it can be obtained by differentiating the cdf(cumulative distribution function). --"Statistical Inference"

My question is: when the pdf of Y is not continuous, can't we obtain the pdf by differentiating the cdf?

• If the pdf exists at all--no matter what properties it might have--then by definition it is the derivative of the cdf (provided one understands that the pdf represents a density and not a function).
– whuber
May 6 '20 at 11:03

Yes! The density of a continuous distribution is the derivative of the CDF.$$^{\dagger}$$

Example: the uniform distribution, say on $$(0,1)$$, which has PDF $$f(x)= \left\{ \begin{array}{ll} 1 & x\in (0,1)\\ 0 & x\notin (0,1) \\ \end{array} \right.$$.

Then the CDF is $$F(x)= \left\{ \begin{array}{ll} 0& x\le0\\ x & x\in (0,1)\\ 1 & x\ge0) \\ \end{array} \right.$$.

You can see that $$\dfrac{dF(x)}{dx} = f(x)$$, as you'd expect.

We don't usually talk about the PDF as being continuous, however. Continuous vs discrete concerns the CDF. Fair warning: the details of this quickly get you into heavy real analysis, including measure theory.

$$^{\dagger}$$In some sense, you always can get the density through a derivative. Measure theory unifies discrete, continuous, and even funkier distributions and gives their densities through something called the Radon-Nikodym derivative.

• Re "you can see that:" unless you take more care defining what you mean by the derivative, your assertion is incorrect when $x=0$ or $x=1,$ because (according to elementary definitions of the derivative) $F$ is not differentiable at either of those points.
– whuber
May 6 '20 at 11:04