8
$\begingroup$

I have a vector X of N=900 observations that are best modeled by a global bandwidth Kernel density estimator (parametric models, including dynamic mixture models, turned out not to be good fits):

enter image description here

Now, I want to simulate from this KDE. I know this can be achieved by bootstrapping.

In R, it all comes down to this simple line of code (which is almost pseudo-code): x.sim = mean(X) + { sample(X, replace = TRUE) - mean(X) + bw * rnorm(N) } / sqrt{ 1 + bw^2 * varkern/var(X) } where the smoothed bootstrap with variance correction is implemented and varkern is the variance of the Kernel function selected (e.g., 1 for a Gaussian Kernel).

What we get with 500 repetitions is the following:

enter image description here

It works, but I have a hard time understanding how shuffling observations (with some added noise) is the same thing as simulating from a probability distribution? (the distribution being here the KDE), like with standard Monte Carlo. Additionally, is bootstrapping the only way to simulate from a KDE?

EDIT: please see my answer below for more information about the smoothed bootstrap with variance correction.

$\endgroup$
  • 1
    $\begingroup$ The bootstrap experiment gives you an indication of the variability of the kernel density estimate. This has nothing to do with simulating from the kernel, as better explained by Dougal below. $\endgroup$ – Xi'an Jun 18 '15 at 21:05
  • $\begingroup$ yep, that's quite some variability. Do you think a KDE would be a better approach than a dynamic mixture model here? $\endgroup$ – Antoine Jun 18 '15 at 21:07
  • $\begingroup$ so, I understand that the smooth bootstrap as shown above is not equivalent to simulating from the Kernel. However, it accomplishes the same goal: simulating from the empirical PDF, right? I will try to post the results of the strategy proposed by Douglas below (simulating directly from the KDE) to compare when I have time. $\endgroup$ – Antoine Jun 18 '15 at 21:18
  • $\begingroup$ Simulating from the kernel estimator does not lead to simulations from the empirical cdf and there is no clear definition of an empirical pdf, between histograms and kernel estimates, all of which require calibration of a bandwidth. $\endgroup$ – Xi'an Jun 19 '15 at 6:56
  • $\begingroup$ I disagree with your first comment, please see my answer below. $\endgroup$ – Antoine Jun 19 '15 at 10:53
10
$\begingroup$

Here's an algorithm to sample from an arbitrary mixture $f(x) = \frac1N \sum_{i=1}^N f_i(x)$:

  • Pick a mixture component $i$ uniformly at random.
  • Sample from $f_i$.

It should be clear that this produces an exact sample.

A Gaussian kernel density estimate is a mixture $\frac1N \sum_{i=1}^N \mathcal{N}(x; x_i, h^2)$. So you can take a sample of size $N$ by picking a bunch of $x_i$s and adding normal noise with zero mean and variance $h^2$ to it.

Your code snippet is selecting a bunch of $x_i$s, but then it's doing something slightly different:

  • changing $x_i$ to $ \hat\mu + \frac{x_i - \hat\mu}{\sqrt{1 + h^2 / \hat\sigma^2}} $
  • adding zero-mean normal noise with variance $\frac{h^2}{1 + h^2/\hat\sigma^2} = \frac{1}{\frac{1}{h^2} + \frac{1}{\hat\sigma^2}}$, the harmonic mean of $h^2$ and $\sigma^2$.

We can see that the expected value of a sample according to this procedure is $$ \frac1N \sum_{i=1}^N \frac{x_i}{\sqrt{1 + h^2/\hat\sigma^2}} + \hat\mu - \frac{1}{\sqrt{1 + h^2 /\hat\sigma^2}} \hat\mu = \hat\mu $$ since $\hat\mu = \frac1N \sum_{i=1}^N x_i$.

I don't think the sampling disribution is the same, though.

$\endgroup$
  • $\begingroup$ thank you for this nice answer. I am currently exploring this approach. Would you have a look a this other very recent (and somewhat related) thread please? Thanks in advance. $\endgroup$ – Antoine Jun 7 '15 at 18:08
3
$\begingroup$

To eliminate any confusion about whether it is possible or not to draw values from the KDE using a bootstrap approach, it is possible. The bootstrap is not limited to estimating variability intervals.

Below is a smoothed bootstrap with variance correction algorithm that generates synthetic values $Y_{i}'s$ from a KDE $K$ of window $h$. It comes from this book by Silverman, see page 25 of this document, section 6.4.1 "Simulating from density estimates". As noted in the book, this algorithm allows to find independent realizations from a KDE $\hat{y}$, without requiring to know $\hat{y}$ explicitly:

To generate a synthetic value $Y$ (from a training set $\big\{X_{1},...X_{n}\big\}$):

  • Step 1: Choose $i$ uniformly with replacement from $\big\{1,...,n\big\}$,
  • Step 2: Sample $\epsilon$ from $K$ (i.e., from the Normal distribution if $K$ is Gaussian),
  • Step 3: Set $Y=\bar{X}+(X_{i}-\bar{X}+h.\epsilon)/\sqrt{1+h^{2}{\sigma_{K}}^2/{\sigma_{X}}^2}$.

Where $\bar{X}$ and ${\sigma_{X}}^2$ are the sample mean and variance, and ${\sigma_{K}}^2$ is the variance of $K$ (i.e., 1 for a Gaussian $K$). As explained by Dougal, the expected value of the realizations is $\bar{X}$. Thanks to the variance correction, the variance is ${\sigma_{X}}^2$ (on the other hand, the smoothed bootstrap without variance correction, where step 3 is simply $Y=X_{i}+h.\epsilon$, inflates the variance).

The R code snippet in my question above is strictly following this algorithm.

The advantages of the smoothed bootstrap over the bootstrap are:

  • "spurious features" in the data are not reproduced as different values from the ones in the sample can be generated,
  • values beyond the max/min of the observations can be generated.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.