The last post I forget to say that we use Mikownski classes of densities because the MISE is a risk corresponding to the $latex {\mathbb L^2({\mathbb R})}&fg=000000$ norm. Thus, it is natural to assume that $latex {p}&fg=000000$ is smooth with respect to this norm. Another way to describe smoothness in $latex {\mathbb L^{2}({\mathbb R})}&fg=000000$ are …

Photos of Sergey Nikolskii from The Russian Academy of Sciences The MSE gives an error of the estimator $latex {\hat{p}_{n}}&fg=000000$ at an arbitrary point $latex {x_{0}}&fg=000000$, but it is worth to study a global risk for $latex {\hat{p} _{n}}&fg=000000$. The mean integrated squared error (MISE) is an important global measure, $latex \displaystyle \mathrm{MISE}\triangleq\mathop{\mathbb E}_{p}\int\left(\hat{p} _{n}(x)-p(x)\right)^{2}dx &fg=000000$ …

Today we will apply the ideas of the others post by a simple example. Before, we are going to answer the question of the last week. What is exactly the $latex {h_{opt}}&fg=000000$ if we assume that $latex \displaystyle \displaystyle f(x) = \frac{1}{\sqrt{2\pi}} \text{exp}\left(\frac{-x^2}{2}\right)? &fg=000000$ How $latex {f(x)}&fg=000000$ is the density of standard normal distribution. It is …

Before to continue with today’s post we will answer the question of last week, Is it $latex {\hat{f}_{h}(x)}&fg=000000$ a consistent estimator? The answer is yes. Because convergence in mean squared implies convergence in probability.