# Solved – Back Transformation

If I had a response variable that was square-root transformed, and an explanatory variable that is log transformed, and I wished to back transform the model using the summary statistics below, such that Y ~ (X)^2, how would I interpret the meaning of the relationship between X and Y using the estimated Beta coefficient?
I thought it was interpreted as: "If there is a 1% increase in X, there is approximately a change of sqrt(2.1014))/100 units increase in Y."

``    Residuals:         Min      1Q  Median      3Q     Max      -37.051 -12.096  -4.908   9.701  68.071       Coefficients:                 Estimate Std. Error t value Pr(>|t|)         (Intercept)  -3.0147     2.0827  -1.448    0.148         Dose.Back     2.1014     0.1679  12.514   <2e-16 ***     ---     Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  Residual standard error: 16.28 on 1154 degrees of freedom Multiple R-squared:  0.1195,    Adjusted R-squared:  0.1187  F-statistic: 156.6 on 1 and 1154 DF,  p-value: < 2.2e-16 ``
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Your model and its estimates posit that

\$\$sqrt{Y} = 2.1014 D – 3.0147 + varepsilon\$\$

where \$D\$ is `Dose.Back` (or its logarithm) and \$varepsilon\$ is a random variable of zero expectation whose standard deviation is approximately \$16.28.\$ Squaring both sides gives

\$\$Y = (2.1014 D – 3.0147 + varepsilon)^2.\$\$

Adding \$0.01\$ to \$D\$ yields the value

\$\$(2.1014 (D + 0.01) – 3.0147 + varepsilon')^2.\$\$

The difference is

\$\$2(2.1014 D – 3.0147 + varepsilon)(varepsilon' – varepsilon + (0.01)(2.1014)) + (varepsilon' – varepsilon + (0.01)(2.1014))^2.\$\$

This expression, as well as its expectation, are complicated. Let us therefore focus on the simpler question of how the expectation of \$Y\$ varies with \$D\$. Note that

\$\$eqalign{ mathbb{E}(Y) &= mathbb{E}left(2.1014 D – 3.0147 + varepsilonright)^2 \ &= (2.1014D – 3.0147)^2 + 2(2.1014D – 3.0147) mathbb{E}(varepsilon) + mathbb{E}(varepsilon^2) \ &=(2.1014D – 3.0147)^2 + 0 + (16.28)^2. }\$\$

(This result is of considerable interest in its own right because it reveals the role played by the mean squared error in interpreting the relationship between \$D\$ and \$Y\$.)

When \$0.01\$ is added to \$D\$ the value of \$mathbb{E}(Y)\$ increases by

\$\$2(2.1014)(2.1014D – 3.0147)(0.01) + 2.1014(0.01)^2.\$\$

The last term \$2.1014(0.01)^2 approx 0.0002\$ is so small compared to the squared errors (with their typical value of \$16.28\$) that we may neglect it. In this case, to a good approximation, this fitted model associates an (additive) increase in \$D\$ of \$0.01\$ with an increase in \$Y\$ of

\$\$2(2.1014)(2.1014D – 3.0147)(0.01) = 0.0883176 D – 0.126.\$\$

When \$D\$ is the natural logarithm of some quantity \$d\$, a 1% multiplicative increase in \$d\$ causes a value of approximately \$0.01\$ to be added to \$D\$, because

\$\$log(1.01 d) = log(1.01) + log(d) = left(0.01 – (0.01)^2/2 + cdotsright) + D approx 0.01 + D.\$\$

If you used a logarithm to another base \$b\$, entailing \$D = log_b(d) = log(d)/log(b),\$ then a 1% multiplicative increase in \$d\$ causes a value of approximately \$(0.01)/log(b)\$ to be added to \$D\$, so everywhere "\$0.01\$" occurs in the preceding formulas you must use \$(0.01/log(b))\$ instead.

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