Talk:Truncated normal distribution

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[edit] Regarding the pdf

I am concerned that: f(x;\mu,\sigma, a,b) = \frac{\frac{1}{\sigma}\phi(\frac{X - \mu}{\sigma})}{\Phi(\frac{b - \mu}{\sigma}) - \Phi(\frac{a - \mu}{\sigma}) },

cannot be the formula for a PDF of a truncated random normal variable. Say there is a left truncated (a = 0) normal random variable with positive mean.

If we choose X to be a negative value, then \phi(\frac{X - \mu}{\sigma}) is positive, \Phi(\frac{b - \mu}{\sigma}) is 1 and \Phi(\frac{a - \mu}{\sigma}) is positive. Altogether, the PDF cannot be zero as it should be.

Perhaps defining it piecewise is the most logical idea because I cannot think of an explicit formula.

Well, the article already mentioned that the domain of X is [a,b]. Thus f(x = ;μ,σ,a,b) is zero outside a and b. So, in your example, if a = 0, then f(x) is zero. Robbyjo (talk) 20:04, 20 February 2008 (UTC)

In my opinion the formula for is incorrect. It should be: f(x;\mu,\sigma, a,b) = \frac{\phi(\frac{X - \mu}{\sigma})}{\Phi(\frac{b - \mu}{\sigma}) - \Phi(\frac{a - \mu}{\sigma}) }. In the current version, if you truncate at a=-inf, b=+inf you will not get Normal distribution Compare also: http://rss.acs.unt.edu/Rdoc/library/msm/html/tnorm.html

—Preceding unsigned comment added by 128.143.16.201 (talk) 20:31, 20 February 2008 (UTC)

It's not a typo; note that \phi(\cdot) is the standard normal pdf. So \frac{1}{\sigma}\phi(\frac{X - \mu}{\sigma}) gives you the pdf for X˜N(μ,σ). Write that out and you'll see why. Josuechan (talk) 23:11, 20 February 2008 (UTC)