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CDF is the probability that a random variable takes on a value less than or equal to a fixed $x = a$.

Assuming we have a a random variable $X$ that has a PDF, both CDF and PDF have the same information as the following PDF gives us the exact information as CDF.

$$ \int_{-\infty}^{a} f(x) \,dx $$

Do we still need the CDF in this case or are there still calculations that can only be done with or information that can merely be derived from CDF?

Infinity
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    Related (not intended to answer the question): https://stats.stackexchange.com/questions/214485/are-cdfs-more-fundamental-than-pdfs – Peter O. Feb 11 '22 at 09:21
  • Consider $X$ with density $1$ on $[0,1]$ and $0$ elsewhere, and $Y$ with density $1$ on irrationals in $(0,1)$ and $0$ elsewhere. They have the same CDF and support, but their densities differ infinitely often. Are they the same distribution? – Henry Feb 11 '22 at 09:58
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    In some cases the CDF is the natural tool for analysis and the PDF won't do. See https://stats.stackexchange.com/questions/86429 for instance. But since there is a one-to-one correspondence between the CDF and an equivalence class of PDFs, they are mathematically equivalent objects, whence the question about "calculations that can only be done" is an empty one. All we might be discussing (if it's worth any discussion at all) would be practical convenience, numerical accuracy for computing, or conceptual simplicity. – whuber Feb 11 '22 at 16:11

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I think you are making a false distinction here.

$$ \int_{-\infty}^{a} f(x) \,dx $$

is the CDF. When you take the integral, you are calculating the CDF. Having closed-form CDF just means that someone already did this for you.

Tim
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There are situations where the CDF comes in handy. E.g. for computing quantiles or copulas or transforming distributions into each other. Some proofs work better/only with CDFs.

But you are right, in case there is an integrable PDF, you also have a CDF.

Of course, the interesting case is when you have a CDF for which there is no pdf.

frank
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