I have computed a truncated normal distribution, which total probability density (i.e. area under the curve) is equal to 0.92.

The distribution represents well the reality of the phenomenon I am investigating, i.e. I am expecting that when x = 0 the probability will be positive (non-zero). The mode is also where I expected it to be. However, any non represented probabilities for x < 0 do not make physical sense, thus I would like to "rescale" my truncated distribution so that the total area under the density is 1 (not 0.92).

The distribution being asymetrical, how can I do this?

  • $\begingroup$ There's no point in a truncated normal where there's positive probability for that point. Either you don't mean 'probability' or you don't mean 'truncated'. $\endgroup$
    – Glen_b
    Aug 5, 2014 at 12:01

1 Answer 1


Divide the constant term out the front by the area. That rescaled density has area 1.

So if the area under $c_0 g_X(x)$* over the truncated domain is exactly 0.92, then the area under $\frac{c_0}{0.92}g_X(x)$ on the same domain is 1.

*(where $g$ is some function normalized by $c_0$ to integrate to 1 on the untruncated domain)


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