Getting the equation from R's lm when using a product

Question

Similar to this question: How to translate the results from lm() to an equation? in which the top voted answer said how to get the form of an equation from lm(y ~ x) and equivalently for lm(z ~ y + x) and other sums, I'm wondering: how can the equation be obtained from the form of lm(z ~ y*x)?

I have:

> summary(lm(log(z) ~ x*y))

Call:
lm(formula = log(z) ~ x * y)

Residuals:
      Min        1Q    Median        3Q       Max 
-0.181142 -0.073755  0.000481  0.082088  0.200902 

Coefficients:
             Estimate Std. Error  t value Pr(>|t|)    
(Intercept)  -9.85368    0.09304 -105.906  < 2e-16 ***
x           -97.41166    6.28269  -15.505  < 2e-16 ***
y            -2.26398    0.14243  -15.895  < 2e-16 ***
x:y          91.69016    9.77390    9.381 6.95e-14 ***

It returns 4 coefficients -- (Intercept), x, y, and x:y -- but I'm not sure how to put them together to get the final equation.

Is it simply that x:y term multiplied by x*y plus the intercept? That is, in this case $\log z = 91.69 xy -9.853$?

By inspecting `model.fit(lm(log(z) ~ x*y))` you will quickly see what `x:y` represents and then you can directly apply the solutions you reference. It's a good idea, though, *not* to transcribe coefficients from the output, because they are rounded. To retain their full precision--which becomes more important with interaction terms like `x:y`--extract the coefficients from the result of `lm` as doubles and use those for subsequent formulas. — whuber, Jul 10 '14 at 20:56

TrynnaDoStat · Accepted Answer · 2014-07-10T21:04:29.600

6

The equation is $\widehat{\log z} = -9.853 -97.41166x -2.26398y + 91.69xy$ where $\widehat{\log z}$ is the estimated value of $\log z$.

edited Jul 10 '14 at 21:04

answered Jul 10 '14 at 20:56

TrynnaDoStat

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+1, this is the answer. I think the confusion might be about manually creating the new `x*y` variable. You'll want to put the 'hat' over the entire "$\log z$", not just the $z$. – gung - Reinstate Monica Jul 10 '14 at 20:58
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I suggest to put an expectation around the left hand side - especially if someone tries to get rid of the log simply by exponentiation. Otherwise the answer is nice. – Michael M Jul 10 '14 at 21:02
I was intending to get rid of the log by simple exponentiation. What's the problem with doing that? – rhombidodecahedron Jul 10 '14 at 21:02
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You might ask that as a new question, @rhombidodecahedron. Briefly, $E(\log z)\ne \log E(z)$. – gung - Reinstate Monica Jul 10 '14 at 21:04
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If you're having trouble with the $\LaTeX$, you can use \$\widehat{\log z}\$ to get $\widehat{\log z}$. – gung - Reinstate Monica Jul 10 '14 at 21:06
The new question has been posted here: http://stats.stackexchange.com/questions/107522/obtaining-an-estimator-for-z-given-an-estimator-for-log-z – rhombidodecahedron Jul 10 '14 at 21:11

Getting the equation from R's lm when using a product

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