# 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

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