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Let $(x_i,y_i)_{1\leq i\leq n}$ some dataset. I want to estimate the conditional expectation $E[Y\mid X=x]$ and the conditional variance $V[Y\mid X=x]$.

I used Nadaraya-Watson's estimator to estimate the conditional expectation: $$\hat{E}[Y\mid X=x]=\frac{\sum_{i=1}^ny_iK\left(\frac{x-x_i}{h}\right)}{\sum_{i=1}^nK\left(\frac{x-x_i}{h}\right)}$$ In fact I just use the ksmooth function in R.

Now let $z_i$ the squared residual: $z_i = \left(y_i - \hat{E}[Y\mid X=x_i]\right)^2$. Then I use Nadaraya-Watson's estimator once again to get an estimation of the conditional variance: $$\hat{V}[Y\mid X=x]=\frac{\sum_{i=1}^nz_iK\left(\frac{x-x_i}{h}\right)}{\sum_{i=1}^nK\left(\frac{x-x_i}{h}\right)}$$

So this is some kind of two-step estimation, which bothers me a little. Is this a good way to estimate the conditional variance? If not, how to do this properly (and not paramatrically)?

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  • $\begingroup$ Use a bootstrap? $\endgroup$ – Repmat Sep 5 '16 at 18:01
  • $\begingroup$ It seems reasonable to me. Consider, are you bothered by the comparable two-step estimation in the simple case of the overall (marginal) $y$ variance? (That is simply the limit of uniform weights, i.e. as your bandwidth $h$ goes to infinity.) $\endgroup$ – GeoMatt22 Sep 5 '16 at 18:43
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You could simply use the fact that $V(Y) = \mathbb{E}[Y^2] - \mathbb{E}[Y]^2$, estimate the conditional expectations of $Y^2$ and $Y$ separately, then estimate the conditional variance by: $$\hat{V}(Y|X) = \hat{\mathbb{E}}(Y^2|X) - \hat{\mathbb{E}}(Y|X)^2$$ No double smoothing involved this way.

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