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Let $(x_i,y_i,z_i)_{i=1,\dots,n}$ be an i.i.d. sample of $(X,Y,Z)$. How one can estimate the following object $$\int_{-\infty}^xf(\bar x,y|z)\mathrm{d}\bar x$$ where $f(x,y|z)$ is a density of $X,Y$ conditional on $Z$.

We know that $F(x) = \int_{-\infty}^xf(\bar x)\mathrm{d}x$ can be estimated using $\hat F(x) = \frac{1}{n}\sum_{i=1}^n1[x_i\leq x]$. We also know that $\hat f(x,y|z) = \frac{\frac{1}{h_{xn}h_{yn}}\sum_{i=1}^nK\left(\frac{x-x_i}{h_{xn}}\right)K\left(\frac{y-y_i}{h_{yn}}\right)K\left(\frac{z-z_i}{h_{zn}}\right)}{\sum_{i=1}^nK\left(\frac{z-z_i}{h_{zn}}\right)}$

Can I construct the estimator for the above mentioned object like this? $$\frac{\frac{1}{h_{yn}}\sum_{i=1}^n1[x_i\leq x]K\left(\frac{y-y_i}{h_{yn}}\right)K\left(\frac{z-z_i}{h_{zn}}\right)}{\sum_{i=1}^nK\left(\frac{z-z_i}{h_{zn}}\right)}$$ where $K$ is a kernel function $1$ is an indicator variable, $h$ is a bandwidth and $n$ is a sample size.

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I guess you have a sample from $(X,Y,Z)$. By definition, we have that

$$f(x,y\vert z)=\dfrac{f(x,y,z)}{f(z)}.$$

Using the joint sample from $(X,Y,Z)$, construct a nonparametric estimator (e.g. KDE) of their corresponding density $\varphi_{X,Y,Z}$. Using the marginal sample from $Z$ construct a nonparametric (KDE) estimator of the density of $Z$, $\varphi_Z$. Plug these estimators in the in integral

$$\int_{-\infty}^xf(\bar x,y|z)\mathrm{d}\bar x \approx \int_{-\infty}^x \dfrac{\varphi_{X,Y,Z}(\bar{x},y,z)}{\varphi_Z(z)}\mathrm{d}\bar x,$$

and voilà. The integral can be calculated using quadrature or Monte Carlo methods.

Note that the bandwidth parameter might not be the same for all the entries as you are suggesting. In addition, possible dependencies of $X,Y,$ and $Z$ must be taken into account.

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  • $\begingroup$ @Bayes What you are using here is the well-known Empirical Cumulative Distribution Function (ECDF). You are also mixing two types of nonparametric estimators whose use might require a theoretical justification. Moreover, the proposed estimator does not consider dependencies between the entries. $\endgroup$
    – Pigeon
    Apr 22 '13 at 14:05
  • $\begingroup$ OK, I have deleted that comment. The empirical distribution function may be extremely slow in more than one dimension. Your estimator is not taking into account the possible dependencies between the entries (it uses a (strange) diagonal bandwidth matrix for $Y$ and $Z$). That is all I can contribute, cheers. $\endgroup$
    – Pigeon
    Apr 22 '13 at 14:08
  • $\begingroup$ I think you had better be careful about bandwidths, lest you create situations where $\phi_Z(z)=0$ but $\phi_{X,Y,Z}(\bar{x},y,z)$ is nonzero. This casts doubt on the numerical stability of the entire strategy: it would be good to see it carried out on some actual data to verify it actually works. $\endgroup$
    – whuber
    Apr 22 '13 at 14:45
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Bayes,

Actually, you don't need to estimate separate KDEs for f(x,y,z) and f(z) to get f(x,y,z). You can just estimate the joint $f(x,y,z)$ by KDE that uses the Gaussain kernels and then analytically compute $f(x,y|z)$ for any selected value of z. Then, as Pigeon already explained, just integrate over the x. But, if you are using the Gaussian kernels, you can do this part analytically as well. So you are in effect completely avoiding Monte Carlo. You will, however, require samples from the joint pdf f(x,y,z) and I assume you can get those. Then you'll need a good multivariate KDE that uses Gaussian kernels. Perhaps you can try this one: http://www.mathworks.com/matlabcentral/fileexchange/41187-fast-kernel-density-estimator-multivariate

Note that your equation for $\hat f(x,y|z)$ assumes independance of x,y,z, but I don't think that your initial definition makes that assumption. Anyway, if you're able to get the i.i.d. samples from the joint $f(x,y,z)$, then the above KDE will take into acount the possible correlation accross x,y and z .

You can find the neccesary equations on how to compute conditionals and marginals of multivariate Gaussians in textbooks or check the net.

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