What is the interpretation/logic of this formula:

It's probability density function $f$, for $a<x<b$, is given by

$$ f(x;\mu,\sigma^2,a,b) = P(x|a<x<b) = \frac{p(x, a<x<b)}{P(a<x<b)}=\frac{p(x)}{P(a<x<b)} \\ = \frac{\tfrac{1}{\sqrt{2\pi}\sigma} \exp\{-\tfrac{(x-\mu)^2}{2\sigma^2}\}}{\Phi(\tfrac{b-\mu}{\sigma}) -\Phi(\tfrac{a-\mu}{\sigma})} = \frac{\tfrac{1}{\sigma} \phi(\tfrac{x-\mu}{\sigma})}{\Phi(\tfrac{b-\mu}{\sigma}) -\Phi(\tfrac{a-\mu}{\sigma})} $$

EDIT: I was confused with what goes in the numerator and denominator and the rationale behind it in order to be able to explaint it to myself. I was reading about conditional probability P(B|A) formula and I was confused because I thought that numerator equals Probability of X being in the range [a, b] times probability of X given its in the range.

Since I am learning this all by myself there are lots of things I am confused with and have no one to ask, except you guys here. As a result, this was the cause of a vague question at first.

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    $\begingroup$ What exactly is unclear for you in this formula? $\endgroup$ – Tim Jun 17 '16 at 14:45
  • $\begingroup$ @Tim What is p(x)? Is it PDF of original distribution? $\endgroup$ – Quirik Jun 17 '16 at 15:18
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    $\begingroup$ Going on no information at all, one would guess that $f$ is the density of a truncated distribution, $P$ is a probability, and $p$ is a probability density for the untruncated version. Those guesses at least would make the equations sort of correct (the derivation seems less than rigorous because it appears to confound densities with probabilities). Presumably, though, the source of that formula explained what they meant by all these terms. $\endgroup$ – whuber Jun 17 '16 at 16:59
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    $\begingroup$ Can you provide more context? Where are you getting this from? What did they say the constituent terms stood for? Can you link to the source? $\endgroup$ – gung - Reinstate Monica Jun 18 '16 at 12:06
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    $\begingroup$ @whuber, it doesn't seem the original did define the terms. Although it is possible to figure them out from the larger context (eg, the post shows an example calculation for the expectation). $\endgroup$ – gung - Reinstate Monica Jun 18 '16 at 12:25

From the axioms of probability, total probability is $ P(\Omega) = P(-\infty < X < \infty) = 1 $. Let's denote probability of $X$ being in some subset $P(a < X \le b)$ as $\pi$. If $a > -\infty$ and $b < \infty$, then obviously $\pi < 1$. If your variable is truncated, i.e. it has restricted range, then it's total probability is some $\pi$, so we have to normalize it that it's total probability is equal to $1$, that is why we divide density function, or probability function, by $\pi$. It is a property of any truncated distribution, not just truncated normal.

I was confused because I thought that numerator equals Probability of X being in the range [a, b] times probability of X given its in the range.

You are correct! We are talking here about conditional probability of $X = x$ given that distribution of $X$ is truncated to the $(a,b]$ range since conditioning is about restricting sample space. That is what normalization of the probability by dividing by $\pi$ is about. So your formula can be re-writed as:

$$ \underbrace{f(x ~ | ~ a < X \leq b)}_\text{truncated density} = \frac{\overbrace{f(x)}^\text{non-truncated density}}{\underbrace{F(b)-F(a)}_{P(a < X \leq b)}} ~~ \text{for } a < x \leq b $$

  • $\begingroup$ Just to make one thing clear, if I understood correctly, the probability function of original distribution from the numerator is restrictred only to values of the random variable in the interval [a, b]? $\endgroup$ – Quirik Jun 17 '16 at 18:06
  • $\begingroup$ That is what truncation is about... check the linked Wikipedia page. $\endgroup$ – Tim Jun 17 '16 at 18:18
  • $\begingroup$ Thank you for your effort. I was confused with the facts I explained in my edited questions. In addition, could you please see my doubt I posted in comment to my post above. $\endgroup$ – Quirik Jun 19 '16 at 7:58
  • $\begingroup$ In a nutshell, this is basically the formula P(B|A) = P(A and B)/P(A), where P(B|A) is truncated (renormalized) density; P(A) is probability of X being in the truncated area which is equal to the area under the curve from a to b; P(A and B) is (untruncated) density of X=x and if a<x<b. Is my understanding correct? $\endgroup$ – Quirik Jun 30 '16 at 7:13
  • $\begingroup$ @quirik yes, you are $\endgroup$ – Tim Jun 30 '16 at 7:15

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