# Transformation of confidence intervals

I'm using Matlab to perform a linear regression. In order to prevent the prediction of negative values I used a box-cox-transformation of the dependent variable ($=y_t$) with $\lambda = 0.5$.

$y^{(\lambda)} = \frac{y_t^{\lambda} - 1}{\lambda}$

After that I perform the linear regression with $y^{(\lambda)}$ as dependent variable. To get the result in my original form I transform $y^{(\lambda)}$ back into $y_t$.

$y_t = (\lambda(\frac{1}{\lambda} + y^{(\lambda)}))^{\frac{1}{\lambda}}$

My question now is, can I transform the confidence intervals in the same way as I transform my dependent variable and how do I prove it or disprove it?

-
The confidence interval for which estimator? –  Learner Feb 23 '13 at 11:52
I mean the confidence interval for the dependent variable. Sorry for not clarifying that. –  Portbane Feb 23 '13 at 12:05

## 1 Answer

Let's say the transformed regression equation is $E \left[ y^{\left( \lambda \right)} |x \right] = x \beta$ and let's call $\hat{y}^{\left( \lambda \right)} = x \hat{\beta}$ where $\hat{\beta}$ is the estimated parameters vector. Let's call $H$ the covariance matrix of $\hat{\beta}$. Finally let's create this new function $g \left( \hat{y}^{\left( \lambda \right)} \right)$ such that $$g \left( \hat{y}^{\left( \lambda \right)} \right) = \left( \lambda \left( \frac{1}{\lambda} + \hat{y}^{\left( \lambda \right)} \right) \right)^{\frac{1}{\lambda}}$$ Now, in order to construct the confidence interval, you need to approximate the variance of your new estimator $g \left( \hat{y}^{\left( \lambda \right)} \right)$. It is possible to do so using the delta method $$\frac{\partial g}{\partial \hat{\beta}^T} H \frac{\partial g}{\partial \hat{\beta}}$$ where $\frac{\partial g}{\partial \hat{\beta}}$ is a vector of derivatives of $g$ with each of $\hat{\beta}_1, \ldots, \hat{\beta}_k$ the different parameters you estimated and $\hat{\beta}^T$ is obviously the transpose of $\hat{\beta}$.

-
Thanks for your answer. But what you are saying is that I can't just simply use the confidence interval calculated by Matlab and transform each boundary with $y_t = (\lambda(\frac{1}{\lambda} + y^{(\lambda)}))^{\frac{1}{\lambda}}$? –  Portbane Feb 23 '13 at 15:12
That is what I am saying. –  Learner Feb 23 '13 at 23:37
Thanks so much for your help. –  Portbane Feb 24 '13 at 14:25
Are you sure about that, @Learner? The delta method gives us an asymptotic variance for any asymptotically linear estimate, sure. However, @Portbane has said that this is a linear regression, so we know that this is not just asymptotically linear. Moreover, since the confidence interval is based on quantiles of the prediction distribution, it seems that we can not only do better than the delta method here, we can also do less work we CAN simply map the linear confidence interval through the function g. Or have I missed something obvious? –  dan mackinlay Jan 26 '14 at 14:54