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I found a (to me) strange notation concerning marginal probabilities I don’t understand. Unfortunately I will include picture of the notation.

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Does it mean x=2.5, shouldn’t it be 0 then? Do they mean the cumulative distribution fuchtion?

Thanks

Arnaud Mortier
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Lillys
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1 Answers1

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They mean the density, which is equal to the derivative of the CDF wherever this one is differentiable.

The notation for a CDF would be $F_X(x)$ (with a capital $F$).

Both are linked by the fundamental $$F_X(x)=\int_{-\infty}^x f_X(t)dt$$

Note that such a density does not necessarily exist. When it does, $X$ is called absolutely continuous.

Note also that $f_X$ has little to do with the probability mass function (as you seem to believe when you say that it has to be $0$), which is denoted by $p_X$.

Arnaud Mortier
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  • @Lillys $f_X$ is a function, so it can be evaluated at a point. No, it's not necessarily $0$. It is not a PMF. For instance an exponential RV has CDF $1-e^{-\lambda x}$ and PDF $\lambda e^{-\lambda x}$. Replace $x$ by $2.5$ to get the value of the PDF at $2.5$. – Arnaud Mortier Jul 14 '18 at 23:51
  • than a value of 2.5 would be the probability that x takes on a value of 2.5? I understand that a function can be evaluated at that point. But what does that mean in terms of probability. – Lillys Jul 14 '18 at 23:55
  • The pmf exists for all distributions, the only thing is that for continuous RVs it does not carry any information so you never talk about it. But there are mixed-type RVs as well, not continuous, but not discrete, where the PMF does carry some incomplete information. – Arnaud Mortier Jul 14 '18 at 23:57
  • Thanks for your answer ! – Lillys Jul 15 '18 at 00:00
  • @Lillys You're welcome. I see that you edited your comment. Note that $f_X(x)$ is a density of probability, which is different from a probability. – Arnaud Mortier Jul 15 '18 at 00:02