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If I have a continuous distribution, $N$.

I say $f$ is a sample from $N$, $f \sim N$.

Now I want to determine the probability of $f$ having a value of $x$:

$$ P(x)=\lim_{t\rightarrow0} {\frac{\text{Percent Chance of: } x \le f \le x+t}{t}} $$

Is my understanding of probability correct? If so, is there a syntactically correct way to write the statement above?

Tim
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Nathan
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    Are you familiar with [probability densities](https://en.wikipedia.org/wiki/Probability_density_function) ? – Tim Aug 17 '17 at 12:07
  • @tim that was the page I was reading that made me ask this. In there bacteria example they look at the probability of it dying within a very small time interval. It seemed like it could be written as a limit as to goes to zero. But I'm not sure if my logic and syntax is valid. – Nathan Aug 17 '17 at 12:26
  • @tim are probability densities different to continuous probability distributions? – Nathan Aug 17 '17 at 12:29
  • Discrete probability is easy: the probability to roll four with a six-sided die is simply 1/6 because there are only six possible outcomes. A continuous pdf has an infinite number of possible of outcomes so the probability of any single outcome (not an interval) is 1/infinity = 0. That is the difference. You have to integrate to get the probability of a certain interval. – Louic Aug 17 '17 at 12:47
  • @louic Although many of your statements are correct, they do not logically follow one another as you seem to suggest. For instance, the discrete Poisson distribution has an infinite number of possible outcomes, yet all of them have nonzero probability. Therein lies the danger of relying on intuitive (but meaningless) expressions like "1/infinity". – whuber Aug 17 '17 at 14:16
  • I did say _continuous_ pdf. This makes it correct because the _discrete_ Poisson distribution is not continuous. – Louic Aug 17 '17 at 14:21
  • @louic but there is no discrete pdf's... Moreover, how is Poisson "easy" as compared to dice? With Poisson distribution you also have infinite support. – Tim Aug 17 '17 at 18:18

1 Answers1

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Yes, you are correct, but your notation isn't very clear. You are talking about probability densities. With continuous* random variables there is no point in talking about probabilities, since $\Pr(X=x)=0$ for any $x$. Because of this we use probability densities, i.e. probabilities per foot,

$$ f_X(x) = \lim_{\Delta x \to 0} \frac{ \Pr( x < X \le x + \Delta x ) }{ \Delta x } $$

so when $dx$ is an infinitely small number,

$$ \Pr( x < X \le x + dx ) = f_X(x)\, dx $$

You can also check the Intuition for how the cumulative probability distribution can be derived from probability density function? thread.

* - as noticed by @whuber in the comment, the definition using limits is valid only for absolutely continuous variables.

Tim
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    Because [this issue was recently raised,](https://stats.stackexchange.com/questions/298293) here's a bit of a nit-pick: you are assuming *absolute continuity*. That assumption is tantamount to supposing the limit exists and is finite almost everywhere. For continuous but not absolutely continuous distributions, the limit does not exist (or diverges). – whuber Aug 17 '17 at 13:37
  • Thank you for the edit. Continuous but not absolutely continuous distributions do *not* have density functions, however. Therein lies the distinction. – whuber Aug 17 '17 at 14:10
  • @whuber it seems that the thread you linked begs for an answer :) – Tim Aug 17 '17 at 14:15
  • OK, I posted one. It's hard to do the subject justice in just a screen or two, so constructive comments are (as always) welcome. – whuber Aug 17 '17 at 15:10
  • @Tim Thanks Tim. This is the second question of mine you have answered. You have been a great help in my ongoing journey to try learn stats. – Nathan Aug 17 '17 at 19:32
  • @whuber. Sorry to dig this up again. But from your answer to the question about the difference between absolute continuous and continuous random variables I got the impression that in almost all cases (apart form some weird functions like the cantor function) continuous RVs are the same to absolute continuous RVs. Hence, continuous RVs will also have densities in most cases, no? It sounds like continuous RVs do never have densities in your comment. – guest1 Mar 26 '20 at 15:40