This example shows how to estimate conditional quantiles of a response given predictor data using quantile random forest and by estimating the conditional distribution function of the response using kernel smoothing.
For quantile-estimation speed,
oobQuantileError use linear interpolation to predict quantiles in the conditional distribution of the response. However, you can obtain response weights, which comprise the distribution function, and then pass them to
ksdensity to possibly gain accuracy at the cost of computation speed.
Generate 2000 observations from the model
is uniformly distributed between 0 and 1, and . Store the data in a table.
n = 2000; rng('default'); % For reproducibility t = randsample(linspace(0,1,1e2),n,true)'; epsilon = randn(n,1).*sqrt(t.^2/2 + 0.01); y = 0.5 + t + epsilon; Tbl = table(t,y);
Train an ensemble of bagged regression trees using the entire data set. Specify 200 weak learners and save the out-of-bag indices.
rng('default'); % For reproducibility Mdl = TreeBagger(200,Tbl,'y','Method','regression',... 'OOBPrediction','on');
Mdl is a
Predict out-of-bag, conditional 0.05 and 0.95 quantiles (90% confidence intervals) for all training-sample observations using
oobQuantilePredict, that is, by interpolation. Request response weights. Record the execution time.
tau = [0.05 0.95]; tic [quantInterp,yw] = oobQuantilePredict(Mdl,'Quantile',tau); timeInterp = toc;
quantInterp is a 94-by-2 matrix of predicted quantiles; rows correspond to the observations in
Mdl.X and columns correspond to the quantile probabilities in
yw is a 94-by-94 sparse matrix of response weights; rows correspond to training-sample observations and columns correspond to the observations in
Mdl.X. Response weights are independent of
Predict out-of-bag, conditional 0.05 and 0.95 quantiles using kernel smoothing and record the execution time.
n = numel(Tbl.y); quantKS = zeros(n,numel(tau)); % Preallocation tic for j = 1:n quantKS(j,:) = ksdensity(Tbl.y,tau,'Function','icdf','Weights',yw(:,j)); end timeKS = toc;
quantKS is commensurate with
Evaluate the ratio of execution times between kernel smoothing estimation and interpolation.
ans = 4.3864
It takes much more time to execute kernel smoothing than interpolation. This ratio is dependent on the memory of your machine, so your results will vary.
Plot the data with both sets of predicted quantiles.
[sT,idx] = sort(t); figure; h1 = plot(t,y,'.'); hold on h2 = plot(sT,quantInterp(idx,:),'b'); h3 = plot(sT,quantKS(idx,:),'r'); legend([h1 h2(1) h3(1)],'Data','Interpolation','Kernel Smoothing'); title('Quantile Estimates') hold off
Both sets of estimated quantiles agree fairly well. However, the quantile intervals from interpolation appear slightly tighter for smaller values of
t than the ones from kernel smoothing.