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I run a log linear model $$\log(Y)=\alpha + \beta X + \epsilon$$

and wonder how to calculate the mean of predicted values, in the same dimension as the initial (untransformed) variable Y. I would like to find the same value than the variable $Y$, calculated on the data.

I applied the formula that I have found on this link : https://davegiles.blogspot.com/2013/08/forecasting-from-log-linear-regressions.html, but with the formula mean of $$y_t^* = \exp\{\log(y_t)^* + ( s^2 / 2)\}$$

where

$$\log(y_t)^* := \widehat{log Y}=\hat \alpha + \hat \beta X$$

I can't land on my feet.

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  • $\begingroup$ What is the question? Is the question how to land on your feet? ... No, seriously it would be nice if you could us math type setting and state a little more clearly what it is you would like help with. $\endgroup$
    – Jesper for President
    Oct 17, 2019 at 12:57
  • $\begingroup$ Here is a piece of code to illustrate what I would like : in this code I would like to find the same value as mean(Y), but from the predicted values (what I tried to do at he last line) eps<-rnorm(1000,0,0.11) X<-rnorm(1000,1.8,0.5) a<-0.5 b<-1.2 Y=exp(a+bX+eps) log_Y<-log(Y) mean(Y) res<-summary(lm(log_Y~X)) achap<-coef(res)[1,1] bchap<-coef(res)[2,1] Ypred<-achap+bchapX mean(exp(Ypred)) var_res<-sigma(lm(log_Y~X))**2 mean(exp(Ypred+var_res/2)) $\endgroup$
    – thogs
    Oct 17, 2019 at 13:13
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    $\begingroup$ stats.stackexchange.com/search?q=duan+smear $\endgroup$
    – whuber ♦
    Oct 17, 2019 at 13:42
  • $\begingroup$ You seem to be requesting two contradictory things: with this model, there's no assurance the mean predicted value of the $y_i$ will equal the original mean. You could impose that constraint, but only by (a) complicating the model; (b) making it worse in most respects; and (c) rendering any hypothesis tests suspect. Which, then is more important: having a good fit to an appropriate model or reproducing the mean of (explicitly random) responses exactly? $\endgroup$
    – whuber ♦
    Oct 17, 2019 at 18:29

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