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If I have a normal distribution $X\sim N(\mu, \sigma)$, but I want to truncate it at 0, so that for all values below 0, the density is equal to 0 (like the blue density on the left plot below): enter image description here

The density would be the truncated normal distribution: https://en.wikipedia.org/wiki/Truncated_normal_distribution

I'm unsure how to calculate the pdf for the situation where $a=0$ and $b=\infty$.

Peter
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    Wouldn't the answer be the PDF of the truncated normal distribution? https://en.wikipedia.org/wiki/Truncated_normal_distribution – Ryan Volpi Mar 17 '21 at 21:38
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    If the underlying density is $f(x)$ and the underlying cumulative distribution function is $F(x)$ then the truncated density is $\dfrac{f(x)}{F(b)-F(a)}$ for $a < x < b$, and in this case $\dfrac{f(x)}{1-F(0)}$ for $0 < x$ – Henry Mar 17 '21 at 22:57
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    Conventionally I would think (a) is right-skewed and (b) left-skewed – Henry Mar 17 '21 at 22:59
  • Please [search our site](https://stats.stackexchange.com/search?q=pdf+trunc*+normal) for solutions. – whuber Mar 17 '21 at 23:26

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