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This question already has an answer here:

I'm using a linear model to analyse some data,

    y~N(mu, sigma) where 
    mu[y] <- Intercept + Beta1X + Beta2X1 + Beta3X2
    and Beta2 = Beta1^2
Beta[n] ~ N(mu.b[n], sigma.b[n])

but I have had to log-transform both the predicted and all the predictor variables, because I'm using BUGS, just for efficiency. Gelman alludes to this being called "elasticity" and says the coefficients can be directly interpreted as " a Beta% increase in X is associated with a 1% increase in y".

However, my results :

              Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.1135924  0.1495142   0.760    0.448    
B1           1.4934436  0.0580981  25.706   <2e-16 ***
I(B1^2)     -0.1477196  0.0062205 -23.747   <2e-16 ***
B2           0.0003612  0.0515368   0.007    0.994

suggest that there is a 150% increase in y, with each 1% increase in Beta1 which would be bonkers. Also, how do interpret the "non-linear" or self-interaction term? My suggestion is : A 15% decrease in the effect of B1 on y occurs with each 1% increase in Beta1.

However If I don't transform either the predicted or the predictors I get:

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept)  5.382e+01  3.410e+00  15.782  < 2e-16 ***
B1          -3.026e-02  8.775e-03  -3.449 0.000591 ***
I(B1^2)      8.654e-06  7.828e-06   1.106 0.269264    
B2           2.363e+00  2.490e+00   0.949 0.342789

In which the effect of X1 seems to be reversed, and the effect sizes are miniscule ( certainly not in the order of 150% and 15% respectively)

Hoping for correction!!

Thanks

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marked as duplicate by whuber Jul 29 '14 at 23:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Closely related: stats.stackexchange.com/q/9913. $\endgroup$ – whuber Nov 1 '11 at 17:17
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    $\begingroup$ Where do you get 150% from? The coefficient of 1.49 for B1 turns 1% into 1.49*1% = 1.49%, not 15% and certainly not 150%. $\endgroup$ – whuber Nov 1 '11 at 17:18
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    $\begingroup$ Bingo - I'm an idiot - problem solved. Submit it as an answer and I'll accept it. I'm confusing it, apparently with the interpretation of a logistic regression coefficient, close to zero. $\endgroup$ – rosser Nov 1 '11 at 17:28
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    $\begingroup$ Someone who stumbles across this question because of the title may be interested in reading the following thread: Interpretation of log transformed predictor. $\endgroup$ – gung Nov 2 '12 at 15:07
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(Adding answer from comments so that this question is no longer officially 'unanswered'.)

Where do you get 150% from? The coefficient of 1.49 for B1 turns 1% into 1.49*1% = 1.49%, not 15% and certainly not 150%.

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