TEST 01

Section A

Section A1

Assume

• $n$ sample data points $X_i$, where $i\in \{0,1,\dots ,(n-1)\}$,
• from an underlying probability density $f(x)$ (and cumulative distribution function $F(x)$).

$$\begin{array}{rl}\int k(x)dx& =1\\ k(-x)& =k(x)\\ \int x^{2}k(x)dx& =1\end{array}$$

CC++

#include <stdio.h>

int main(void) {
  printf("Hello world!\n");
  return 0;
}

#include <iostream>

int main(void) {
  std::cout << "Hello world!" << std::endl;
  return 0;
}

Section A2

MORE TEXT

CC++C#

#include <stdio.h>

int main(void) {
  printf("Hello world!\n");
  return 0;
}

#include <iostream>

int main(void) {
  std::cout << "Hello world!" << std::endl;
  return 0;
}

using System.IO;

static class Program
{
    static Main()
    {
        Console.WriteLine("Hello world!");
    }
}

$$\mathrm{MISE}(h)=\mathbb{E}\{\int {[{\hat{f}}_h(x)-f(x)]}^{2}dx\}$$

Section A3

Issue	Solution / mitigation	Notes
1. KDE doesn’t characterise the distribution (because it requires that we store the original samples).	Resample the smoothed distribution.	Need the relevant C# classto distinguish whether the data is original sample or smoothed.
2. KDE distorts the variance, which is a key risk measure.	Explicitly correct for the increase in variance – see below.	Implicit assumption that this does not distort the distribution. I think we are assuming at least symmetry (which would not be true e.g. for log-normals).
3. KDE tails are asymptotically $\exp (-(x/h{)}^2)$, i.e. the tail depends on $h$, not the actual tail shape.	Not sure – some thoughts set out below.	It is dangerous to make assumptions about the tails (including whether the tails on either side are similar.

Section B

Assume

• $n$ sample data points $X_i$, where $i\in \{0,1,\dots ,(n-1)\}$,
• from an underlying probability density $f(x)$ (and cumulative distribution function $F(x)$).

$$\begin{array}{rl}\int k(x)dx& =1\\ k(-x)& =k(x)\\ \int x^{2}k(x)dx& =1\end{array}$$

MORE TEXT

$$\mathrm{MISE}(h)=\mathbb{E}\{\int {[{\hat{f}}_h(x)-f(x)]}^{2}dx\}$$

Issue	Solution / mitigation	Notes
1. KDE doesn’t characterise the distribution (because it requires that we store the original samples).	Resample the smoothed distribution.	Need the relevant C# classto distinguish whether the data is original sample or smoothed.
2. KDE distorts the variance, which is a key risk measure.	Explicitly correct for the increase in variance – see below.	Implicit assumption that this does not distort the distribution. I think we are assuming at least symmetry (which would not be true e.g. for log-normals).
3. KDE tails are asymptotically $\exp (-(x/h{)}^2)$, i.e. the tail depends on $h$, not the actual tail shape.	Not sure – some thoughts set out below.	It is dangerous to make assumptions about the tails (including whether the tails on either side are similar.

Section C

Assume

• $n$ sample data points $X_i$, where $i\in \{0,1,\dots ,(n-1)\}$,
• from an underlying probability density $f(x)$ (and cumulative distribution function $F(x)$).

$$\begin{array}{rl}\int k(x)dx& =1\\ k(-x)& =k(x)\\ \int x^{2}k(x)dx& =1\end{array}$$

MORE TEXT

$$\mathrm{MISE}(h)=\mathbb{E}\{\int {[{\hat{f}}_h(x)-f(x)]}^{2}dx\}$$

Issue	Solution / mitigation	Notes
1. KDE doesn’t characterise the distribution (because it requires that we store the original samples).	Resample the smoothed distribution.	Need the relevant C# classto distinguish whether the data is original sample or smoothed.
2. KDE distorts the variance, which is a key risk measure.	Explicitly correct for the increase in variance – see below.	Implicit assumption that this does not distort the distribution. I think we are assuming at least symmetry (which would not be true e.g. for log-normals).
3. KDE tails are asymptotically $\exp (-(x/h{)}^2)$, i.e. the tail depends on $h$, not the actual tail shape.	Not sure – some thoughts set out below.	It is dangerous to make assumptions about the tails (including whether the tails on either side are similar.