What if explained deviance is greater than 1.0 (or 100%)?

Question

I am using explained deviance (sometimes referred to as percent deviance, or deviance explained by the model) as a goodness-of-fit measure for my species distribution model. Explained deviance is calculated as: (Null Deviance - Residual Deviance) / Null Deviance , and the greater the explained deviance, the greater the explanatory power of the model. One of my deviance values is greater than 1.0 (when multiplied, greater than 100%)....why is that?

Edit: Here is my R code & output for the boosted regression tree:

CODE

spurge10.tc5.lr001 <- gbm.step(data=model.data10, gbm.x = 6:34, gbm.y = 4, family = "bernoulli", tree.complexity = 5, learning.rate = 0.001, bag.fraction = 0.5)

OUTPUT: I am using "estimated cv deviance"

fitting final gbm model with a fixed number of 1400 trees for PresOrAbs

mean total deviance = 1.386 mean residual deviance = 0.718

estimated cv deviance = 1.079 ; se = 0.046

training data correlation = 0.861 cv correlation = 0.58 ; se = 0.038

training data ROC score = 0.982 cv ROC score = 0.819 ; se = 0.021

Thanks!

Answer 1

My understanding of gmb.step is that the deviance values in the output are actual deviance (simliar to "variance") values and not percentages. Therefore, you yourself will need to actually calculate the percent of deviance explained relative to the null model.

Typically, this calculation would be (null deviance - residual deviance) / null deviance. Using your values, that would be 1.386 (mean total deviance) - 0.718 (mean residual deviance) / 1.386, or 48.2% deviance explained.

I am less familiar with using the cv deviance and am trying to find a definition of it now. I am speculating that "cv deviance" is the model residual deviance from the cross validation procedure. If so, % deviance explained should be reduced relative to the first calculation because it would reflect deviance explained on test sets withled from training data. If I am correct about the cv deviance definiion, the calculation would then be (1.386-1.079)/1.386 or 22.2% explained.

Hope this is helpful.