Talk:Inverse Mills ratio

Possible mistake (turned out not to be a mistake)

I think you have a problem in the conditional mean formula. The corrected formula should be

${\begin{aligned}&\operatorname {E} [\,x\,|\ x>\alpha \,]=\mu +\sigma ^{2}{\frac {\phi {\big (}{\tfrac {\alpha -\mu }{\sigma }}{\big )}}{1-\Phi {\big (}{\tfrac {\alpha -\mu }{\sigma }}{\big )}}},\\&\operatorname {E} [\,x\,|\ x<\alpha \,]=\mu +\sigma ^{2}{\frac {-\phi {\big (}{\tfrac {\alpha -\mu }{\sigma }}{\big )}}{\Phi {\big (}{\tfrac {\alpha -\mu }{\sigma }}{\big )}}},\end{aligned}}$

I have double checked by integration (using Mathematica).--132.64.128.115 (talk) 15:01, 27 October 2011 (UTC)

A later addition (by the same user): this was a mistake. If, instead of using the CDF and PDF of standard normal distribution and standardized bound one simply uses the relevant normal distribution with proper mu and sigma (for example, using "normpdf(x,mu,sigma)" in MATLAB), then one has to multiply the ratio by an additional sigma, simply because PDF(x,mu,sigma)=sigma*phi((x-mu)/sigma). But the equation are right. I am leaving this point because the version history tells me that this is not the first time someone makes this error.--192.114.91.245 (talk) 16:11, 16 November 2011 (UTC)