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How to prove the claim that the conditional distribution of a normal random variable is also normal random? And how to think it intuitively?

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  • $\begingroup$ The duplicate addresses the question about intuition. It also shows one way one might prove this claim generally. However, a much simpler way is available: simply inspect the density function. (This is a one-line proof.) $\endgroup$
    – whuber
    Apr 8 '21 at 14:18
  • $\begingroup$ @whuber The link is about bivariate normal distribution. Is there an easier way to think about the simplest case that conditional larger or smaller than a constant? I.e. why is a one-sided truncated normal distribution still "normal"? $\endgroup$ Apr 8 '21 at 21:35
  • $\begingroup$ I don't follow, because truncated Normal distributions are not Normal, nor are they the same as conditional distributions. $\endgroup$
    – whuber
    Apr 8 '21 at 21:57
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    $\begingroup$ @whuber I find I misunderstand the very basic definition of conditional distribution. Thank you very much for your reply to make me realize that. $\endgroup$ Apr 9 '21 at 13:08