Fourier Transform of Kernel Density Estimation - Convolution Theorem?

I am reading this paper about density estimation (Appendix A), where the authors apply a Fourier transform to the estimated probability density (the $X_j$ are a sample of $N$ data points drawn independently from an unknown probability density function $f(x)$):

$$\hat{f}(x) = \frac1N \sum^N_{j=1} K(x-X_j)$$

The authors now note that they apply the convolution theorem and the Fourier transform (where the Fourier transform of the function $g(x)$ is defined by $\Phi_g(t) = \int \exp\lbrace\mathrm{i}xt\rbrace g(x) \mathrm{d}x$), and obtain:

$$\Phi_{\hat{f}}(t) = \kappa(t) \frac1N \sum^N_{t=1} \exp\lbrace\mathrm{i}tX_j\rbrace$$

where $\kappa(t)$ is the Fourier transform of the Kernel $K$.

Now, I am a bit stuck on how they obtained this form and how exactly they used the convolution theorem here. In particular, what happened to the argument $X_j$ in the kernel?

EDIT: Ok, I think I got it. Is this the right approach?

We apply the Fourier transform and obtain:

$$\Phi_{\hat{f}}(t) = \frac1N \sum^N_{j=1} \int K(x-X_j) \exp\lbrace\mathrm{i}xt\rbrace\mathrm{d}x$$

We can easily rewrite this equation as an integral over a Dirac delta distribution:

$$\Phi_{\hat{f}}(t) = \frac1N \sum^N_{j=1} \iint K(y) \delta(y-(x-X_j)) \exp\lbrace\mathrm{i}xt\rbrace\mathrm{d}x\mathrm{d}y$$

Integrating over $x$ and using the symmetry of $\delta$ we obtain:

$$\Phi_{\hat{f}}(t) = \frac1N \sum^N_{j=1} \int K(y) \exp\lbrace\mathrm{i}(y+X_j)t\rbrace\mathrm{d}y$$

We see that we can move factor involving $X_j$ outside the integral, and are left with the result:

$$\Phi_{\hat{f}}(t) = \frac1N \sum^N_{j=1} \exp\lbrace\mathrm{i}X_jt\rbrace \int K(y) \exp\lbrace\mathrm{i}yt\rbrace\mathrm{d}y$$

where the first factor is $\Delta(t) = \frac1N \sum^N_{j=1} \exp\lbrace\mathrm{i}X_jt\rbrace$, and the second factor is $\kappa(t) = \int K(y) \exp\lbrace\mathrm{i}yt\rbrace\mathrm{d}y$, exactly as in the paper.

I am still not sure in how far they applied the convolution theorem here?

It's a bit contrived, but observe that by the "sifting property" of Dirac-delta function:

$$K(x-X_j) = K(x) * \delta(x-X_j)$$

where $*$ denotes convolution and $\delta$ is the Dirac-delta function.

Now you can apply Fourier convolution theorem and write:

$$\mathcal{F}\{K(x-X_j)\}(t) = \kappa(t) e^{j t X_j}$$

where $\mathcal{F}\{g(x)\}(t)$ denotes the Fourier transform of a function $g(x)$.

Finally using the linearity property of the Fourier transform:

$$\mathcal{F}\{\hat{f}(x)\}(t) = \frac{1}{N} \sum_{j=1}^N \mathcal{F}\{K(x-X_j)\}(t) = \kappa(t) \sum_{j=1}^N e^{j t X_j}.$$

• Awesome, thanks a lot. Apparently this is known as the “sifting property” of the Dirac-delta.
– mSSM
Feb 9 '17 at 18:30