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Let $T \sim Unif(0, 1)$. Then, $f_T(t) = 1 \text{ for } t \text{ in [0, 1] (0 elsewhere)}$.

How do we formally compute $f_{T \mid T > 0.5}$?

Intuitively, $f_{T \mid T > 0.5}(t) = 2 \text{ for } t \text{ in [0.5, 1] (0 elsewhere)}$, because we know that $\int_{0.5}^1 f_{T \mid T > 0.5}(t)dt$ must be 1, because T can't be greater than 1, and we are given that T is greater than 0.5, and so T must be between 0.5 and 1. Thus, the integral of the PDF over that range must add up to 1. In addition, we also know that $T \sim Unif(0, 1)$, which is just a flat line. Thus, it makes sense that $f_{T \mid T > 0.5}$ must also be a flat line. The only flat line that integrates to 1 over the range of 0.5 to 1 is $f_{T \mid T > 0.5}(t) = 2$.

This distribution is simple enough that I can just figure this out in my head. However, I'm struggling to figure out the formal math behind how to do this in general.

For example, how would you compute $f_{X \mid X > 1}(x)$ if $X \sim \mathcal{N}(0, 1)$? I can't as easily reason about the normal distribution as I can about the uniform distribution, so I do not know how to approach this.

In general, how do you compute the PDF of a conditional distribution?

Pro Q
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    Consider how much of the density you discard when you condition. Since the conditional PDF must integrate to $1$, how would you use that to adjust your conditional PDF? – Dave Jul 26 '21 at 21:40
  • Do you always multiply by some constant to rescale? If so, what is the name of the formal rule for doing that? – Pro Q Jul 26 '21 at 21:41
  • @ProQ Bayes' rule. – mhdadk Jul 26 '21 at 21:52
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    You describe *truncated* distributions. At https://stats.stackexchange.com/a/525918/919 I posted a general, formal account of truncation, including a derivation of the CDF and PDF of a truncated distribution. The specific case of a truncated Normal distribution is treated at https://stats.stackexchange.com/questions/166273 (univariate) and https://stats.stackexchange.com/a/158008/919 (multivariate). – whuber Jul 26 '21 at 22:38
  • Thank you! The [Wikipedia page for truncated distributions](https://en.wikipedia.org/wiki/Truncated_distribution) contained the answer I needed. In general, you always multiply the previous PDF by a constant $\frac{1}{F(b) - F(a)}$, where $F$ is the CDF for the function. For the uniform distribution, $F(x) = x$, so $F(1) = 1$ and $F(0.5) = 0.5$ and the PDF is $f(x) = 1$. So, the new PDF of the truncated function is indeed $\frac{1}{F(b) - F(a)} f(x) = (2) * (1) = 2$ over the range, and $0$ otherwise. (Since this question was marked as a duplicate, I cannot post this as an answer.) – Pro Q Jul 27 '21 at 20:54
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    Re "always multiply:" only in the special case of truncation in the tails. (This is commonest.) – whuber Jul 28 '21 at 16:22

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