Talk:Truncated normal distribution

Mathematics Start‑class Low‑priority

	Mathematics portal This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics articles
Start	This article has been rated as Start-class on Wikipedia's content assessment scale.
Low	This article has been rated as Low-priority on the project's priority scale.

Statistics Unassessed

	This article is within the scope of WikiProject Statistics, a collaborative effort to improve the coverage of statistics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.StatisticsWikipedia:WikiProject StatisticsTemplate:WikiProject StatisticsStatistics articles
???	This article has not yet received a rating on Wikipedia's content assessment scale.
???	This article has not yet received a rating on the importance scale.

Mean

The expression for the mean is given as: $\mu +{\frac {\phi (\alpha )-\phi (\beta )}{Z}}\sigma$ . This must be incorrect, because it sometimes gives mean values outside the truncation bounds. For example, $\mu =0$ , $\sigma =2$ , $a=2$ , $b=10$ gives a mean of 1.53. I believe the correct expression is $\mu +{\frac {\phi (\alpha )-\phi (\beta )}{Z}}\sigma ^{2}$ . Agreed? Jtmcg1128 (talk) 18:03, 13 October 2011 (UTC)[reply]

The expression $\mu +{\frac {\phi (\alpha )-\phi (\beta )}{Z}}\sigma$ is correct. It's an issue of $\phi (\cdot )$ being defined as the standard normal pdf: $\phi (\xi )={\frac {1}{\sqrt {2\pi }}}\exp {(-{\frac {1}{2}}\xi ^{2}})$ , see discussion on the pdf definition below. The pdf of a distribution with arbitrary mean and standard deviation is ${\frac {1}{\sigma }}\phi ({\frac {x-\mu }{\sigma }})$ . (Karekafi (talk) 17:25, 14 November 2011 (UTC))[reply]

Regarding the pdf

I am concerned that $f(x;\mu ,\sigma ,a,b)={\frac {{\frac {1}{\sigma }}\phi ({\frac {X-\mu }{\sigma }})}{\Phi ({\frac {b-\mu }{\sigma }})-\Phi ({\frac {a-\mu }{\sigma }})}}$ $f(x;\mu ,\sigma ,a,b)={\frac {{\frac {1}{\sigma }}\phi ({\frac {X-\mu }{\sigma }})}{\Phi ({\frac {b-\mu }{\sigma }})-\Phi ({\frac {a-\mu }{\sigma }})}}$ , 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 $\phi ({\frac {X-\mu }{\sigma }})$ $\phi ({\frac {X-\mu }{\sigma }})$ is positive, $\Phi ({\frac {b-\mu }{\sigma }})$ $\Phi ({\frac {b-\mu }{\sigma }})$ is 1 and $\Phi ({\frac {a-\mu }{\sigma }})$ $\Phi ({\frac {a-\mu }{\sigma }})$ 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 $f(x=;\mu ,\sigma ,a,b)$ 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)[reply]

In my opinion the formula for is incorrect. It should be: $f(x;\mu ,\sigma ,a,b)={\frac {\phi ({\frac {X-\mu }{\sigma }})}{\Phi ({\frac {b-\mu }{\sigma }})-\Phi ({\frac {a-\mu }{\sigma }})}}$ $f(x;\mu ,\sigma ,a,b)={\frac {\phi ({\frac {X-\mu }{\sigma }})}{\Phi ({\frac {b-\mu }{\sigma }})-\Phi ({\frac {a-\mu }{\sigma }})}}$ . 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)[reply]
- 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 $X\sim N(\mu ,\sigma )$ . Write that out and you'll see why. Josuechan (talk) 23:11, 20 February 2008 (UTC)[reply]

I reverted the numerator for the PDF back to ${\frac {1}{\sigma }}\phi \left({\frac {X-\mu }{\sigma }}\right)$ because the edit by IP: 152.78.63.13 appears incorrect as Josuechan pointed out above. I also cleaned up the formatting of the discussion a little. --V madhu (talk) 18:17, 4 August 2009 (UTC)[reply]

The correct Density function is this one

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 }})}}

. This function is both positive and integrates to 1 (because of change of variables,

dz={\frac {dx}{\sigma }}

). The incorrect version, listed above, integrates to

{\sigma }

.Iwaterpolo (talk) 01:57, 3 June 2010 (UTC)[reply]

I agree on the PDF formulation. For an easier understanding I added the exact definition of the standard normal pdf in the text below. This may help people not completely familiar with the exact notation to more quickly understand the problem. (Karekafi (talk) 16:40, 14 November 2011 (UTC))[reply]

Interactive calculators

The Statistics Online Computational Resource provides an interactive truncated normal distribution calculator (Java applet). If may be useful to learners, practitioners and instructors. Iwaterpolo (talk) 17:40, 29 November 2010 (UTC)[reply]

Entropy formula

Appears to be wrong. The values from the formula don't agree with numerical computation of of the entropy. I derived the case for a one-sided truncated normal, and that differs from this case, but I haven't had time to go back and derive the two-sided case. Would be nice if someone can track down a reference for this or find the correct formula. --Jpillow (talk) 17:20, 12 January 2012 (UTC)[reply]

Simulation

The simulation section appears to be wrong/misleading. I understand the formula as basically simulating rejecting sampling by basically drawing uniformly from the lower and upper bounds of the CDF of the non-truncated normal (with appropriate parameters), then inverting to obtain the actual value. However, in this case the use of $\Phi$ is misleading because it is referring to the CDF of the normal distribution with parameters $\mu ,\sigma$ instead of the standard normal.

There are several ways to resolve, this, but I feel that the following would be easiest. Note that using $\alpha ,\beta$ with the standard normal CDF instead of $a,b$ would require the result to be multiplied by $\sigma$ then added to $\mu$ .

A random variate x defined as $x=\Phi '^{-1}(\Phi '(a)+U*(\Phi '(b)-\Phi '(a)))$ with $\Phi '$ the CDF of a normal distribution with mean $\mu$ and variance $\sigma ^{2}$ , and $\Phi '^{-1}$ its inverse, $U$ a uniform random number on $(0,1)$ , follows the distribution truncated to the range $(a,b)$ .