I have a question regarding quantile regression. Supposing that I have 10000 observations with one response variable and several predictor variables in a dataset collected each year over several years. I run a multivariate quantile regression at 90% percentile. I want to compare the observed 90th percentile response value for each year (a single value) to the 90th percentile predicted response value for that year (a single value) based on the multivariate quantile regression model. How can I do that?
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6$\begingroup$ 'Multivariate' implies that you have more than one dependent variable. I think you meant to say 'multivariable'. $\endgroup$– Frank HarrellCommented Mar 17, 2014 at 12:00
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I am confused. You've done the prediction already? And for each year you are comparing the 90% quantile of the predicted with the 90% quantile of the observed? This is just a linear regression but sounds like all that you really want to see is how well they agree. In which case calculate the mean square error:
$$ MSE(x,y) = \dfrac{1}{N}\sum_{i=1}^N (x-y)^2 $$
Some other metric might also be possible such as the absolute difference.
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$\begingroup$ Thanks pontikos. Basically I have this kind of a model, y = year + x2 + x3 + x4. y, X2 to X4 are continuous variables. There are multiple observations for each year. Now I ran a 90% quantile regression and got the predicted values. I would like to plot the 90% predicted 'response' value versus the 90% observed 'response' value for each year. I want to know the 90% predicted value for each year given the "full model". I think Peter Flom's 2011 NESUG paper answers this question, but I am not sure how he got the single predicted values at different percentiles on page 8. Clear enough? $\endgroup$– KrisCommented Sep 13, 2013 at 23:40
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$\begingroup$ Do you want to extract the 90% quantile of a single, observed value? $\endgroup$ Commented Jul 28, 2014 at 6:01
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$\begingroup$ correlation is not a good measure of agreement. Consider I have two sets of predictions: one is identical to observed and the other is observed/100 + 10^6 (imagine observed have means near 1 say). Both have perfect correlation, but only one has good agreement between observed and predicted. $\endgroup$– Glen_bCommented Aug 29, 2014 at 5:03
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$\begingroup$ @Glen_b yes you are right since the correlation will be the same even if the means of $x$ and $y$ are different. What do you suggest instead? Mean square error? $\endgroup$– pontikosCommented Sep 3, 2014 at 18:31