# If I know the density I'm estimating is symmetric about 0, how to impose this restriction in my kernel density estimator?

Suppose I'm interested in estimating the unknown smooth density of $$X$$ denoted by $$f(\cdot)$$ using data $$\{X_i\}_{i=1}^{n}$$. Suppose I also know that $$f(\cdot)$$ is symmetric about 0 in the sense that $$f(-x)=f(x)$$ for any $$x$$ in the support. My questions are

1.How to impose or incorporate this symmetry restriction in the usual kernel density estimator defined as

$$\widehat{f}(x)=\frac{1}{nh}\sum_{i=1}^{n}k(\frac{X_i-x}{h})$$, where $$k(\cdot)$$ is the kernel function.

2.How does the symmetry-restricted kernel density estimator improve upon the naive kernel estimator defined above?

Intuitively, the symmetry-restricted kernel density estimator should be better because it used more information, But I don't know how to show or quantify such improvement. For example, does it converge faster?

$$\hat f(x) = \frac{1}{2nh}\sum_i k\left(\frac{X_i-x}{h} \right)+k\left(\frac{-X_i-x}{h} \right)$$
If you used the same bandwidth as for an ordinary kernel estimator, you would expect that the variance component of error would be halved, and the bias component not changed. Presumably you could (in principle) get a smaller $$h$$ and smaller bias, but less variance reduction. You won't get an improved rate of convergence, just a constant factor.
This paper actually has the details, both for when the centre of symmetry is known (your case) and when it's unknown. If it's unknown you need to estimate it, and you have to be careful that your estimator isn't too bad. The paper shows that (for large enough $$n$$ and under weak assumptions about smoothness) you can always get an improvement even if the centre of symmetry has to be estimated.