# How do I estimate the prediction interval of back transformed log-normal data from Gaussian process?

I have some data that are clearly positively skewed and follow a log-normal distribution, lets assume the initial data is $Z = exp(Y)$, where $Y \sim N(\mu,\sigma^2)$.

A Gaussian process assumes that any subsets from the data is normally distributed. So I transformed the data $Z$ to $Y = ln(Z)$, to make it more Gaussian, and I performed a conventional Gaussian process regression. In the end I predict a future point $Y_*$ where $Y_*\sim N(\mu_*,\sigma_*^2)$.

1- How do I back-transform to the original log scale? i.e., how do I convert $Y_*$ back to $Z_*$? In log-normal kriging they predict the arithmetic mean of the $Z_*$, which is $$E(Z_*) = e^{(\mu_*+\sigma_*^2/2)}.$$ The variance of $Z_*$ will be $$Var(Z_*) = e^{(2\mu_*+\sigma_*^2)}*(e^{\sigma_*^2}-1).$$ Is this correct? or should I calculate the geometric mean of $Z_*$, $E(Z_*) = exp(\mu_*)$, instead?

2- How do I calculate the prediction interval as defined in this link? Normally for Gaussian process the prediction interval of 95% is calculated using the borders $\mu_* \pm 1.96\sigma^2$. How do I calculate the same for the back-transformed data (log-normal scale)? Note you cannot exponentiate the borders from the Gaussian scale to get the prediction interval (or in the reference's case the confidence interval) in the log-normal scale as shown in.

• The argument relating to a confidence interval for the mean doesn't apply to the prediction interval. The assertion in your final sentence is wrong. – Glen_b Jul 27 '17 at 2:22
• If I take the exponent of the borders from the Gaussian scale to the log-normal scale I might have borders for the prediction interval that does not include the mean, i.e., the mean is outside the prediction interval. This was presented in the second reference. – Mahshrp Jul 27 '17 at 12:12
• A prediction interval is not an interval for the mean but an interval for an observation - there are cases where you have an interval that has a 99.99% coverage probability for a future observation (conditional on the model form being right) but has no chance to include the population mean. It's not quite that bad in your situation, but if you include uncertainty in the estimate of $\sigma$, PIs won't include the predicted mean, because that's not finite while the limits on the PI certainly are. ...ctd – Glen_b Jul 27 '17 at 23:37
• ctd... This is because you're dealing with a log-t distribution for both CIs and PIs, which has finite quantiles but no finite moments.] $\,$ I will try to post the explanation of the issue with the premise of the question as an answer as time permits. – Glen_b Jul 27 '17 at 23:43