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\begin{fcdefn}[Modified upper shadow]
  \glssymboldefn{modifieduppershadow}%
  Let $n \in \Nbb$, let $J \subset [n]$, let $r < s$ and let $\mathcal{A} \subset
  [n]^{(r)}$.
  Define the \emph{modified upper shadow} $\partial_J^s \mathcal{A}$ to be $\{B
  \in [n]^{(s)} : A \subset B, (B \setminus A) \cap J = \emptyset\}$.
\end{fcdefn}

\begin{fclemma}[]
  % Lemma 7.7
  \label{lemma:7.7}
  Let $n, r, s, J, \mathcal{A}$ be as above.
  Let $\alpha = \frac{|\mathcal{A}|}{{n \choose r}}$ and $\beta =
  \frac{|\mshad_J^s \mathcal{A}|}{{n \choose s}}$.
  Assume that $r \le \frac{n}{2}$.
  Then
  \[
    \beta
    \ge \alpha \left( 1 - \frac{2|J|}{n} \right)^{s - r}
  .\]
\end{fclemma}

\begin{proof}
  Pick a random pair $(A, B)$ with $A \in [n]^{(r)}$, $B \in [n]^{(s)}$, $A
  \subset B$.
  Then
  \[
    \Pbb[(A, B) \in \mathcal{A} \times \mshad_J^s \mathcal{A}]
    \ge \alpha \left( 1 - \frac{2|J|}{n} \right)^{s - r}
  .\]
  Also, the probability is at most $\beta$.
  The result follows.
\end{proof}

\begin{fccoro}[]
  % Corollary 7.8
  \label{coro:7.8}
  For every $p \in (0, \half)$, $\eps > 0$, there exists $m$ such that for every
  $n \in \Nbb$ and every intersecting family $\mathcal{A} \subset [n]^{(r)}$
  where $r = pn$, there exists $J \subset [n]$, $|J| \le m$, and an intersecting
  family $\mathcal{B}$ of subsets of $J$ such that $|\mathcal{A} \setminus
  \ol{\mathcal{B}}| \le \eps {n \choose r}$.
\end{fccoro}

\begin{proof}
  Suppose not.
  Let $\mathcal{C} = hacal \setminus \ol{\mathcal{B}}$.
  Then $\mathcal{C}$ has density $\ge \eps$.
  Apply the \nameref{thm:7.5} Theorem to $\mathcal{A}$ to obtain $J$ and an
  intersecting family $\mathcal{B}$ of subsets of $J$ with
  $\mup(\ol{\mathcal{A}} \setminus \ol{\mathcal{B}}) \le \frac{\eps}{4}$.

  Note that since $\mathcal{C} \cap \ol{\mathcal{B}} = \emptyset$, $\mshad_J^s
  \mathcal{C} \cap \ol{\mathcal{B}} = \emptyset$ for every $s > r$.
  Also, if $s \le r + n^{2 / 3}$, then
  \[
    \frac{|\mshad_J^s \mathcal{C}|}{{n \choose s}}
    \ge \eps\left(1 - \frac{2|J|}{n}\right)^{n^{2 / 3}}
    \ge \eps \left( 1 - \frac{2|J|}{n^{1 / 3}} \right)
    \ge \frac{3\eps}{4}
  ,\]
  for $n$ sufficiently large.

  But (by law of total probability),
  \[
    \mup(\ol{\mathcal{C}})
    = \sum_{s \ge r} \mup([n]^{(s)}) \cdot \left[ \frac{|\mshad_J^s
    \mathcal{C}|}{{n \choose s}} \right]
    \ge \frac{3\eps}{4} \sum_{r \le s \le r + n^{2 / 3}} \mup([n]^{(s)})
    \ge \frac{3\eps}{4} (1 - o(1))
    \ge \frac{5\eps}{16}
  .\]
  But $\ol{\mathcal{C}} \subset \ol{\mathcal{A}} \setminus \ol{\mathcal{B}}$, so
  this is a contradiction.
\end{proof}

\newpage
\section{Invariance principles}

\begin{example*}
  Let $X_i$ uniform on $\{-1, 1\}$, $Y_i \sim N(0, 1)$.
  Then $\frac{1}{\sqrt{n}} (X_1 + \cdots + X_n) \approx \frac{1}{\sqrt{n}} (Y_1
  + \cdots + Y_n) \sim N(0, 1)$.
  Here, $\approx$ is meant to mean ``has approximately the same distribution
  as''.

  Here, we saw that replacing $X_i$ by another variable $Y_i$ with roughly
  similar properties (e.g. same mean and variance) didn't affect the
  distribution of the sum by much.

  How can we define ``approximately same distribution''?
  You may have seen before that we can define it as $|\Pbb[X \le t] - \Pbb[Y \le
  t]| \le \eps$ holding for all $t$.
  This can be rephrased in terms of $\Ebb \indicator{X \le t}$.
  We will instead use a notion of similar distribution where we use continuous
  functions (in fact we will even require stronger conditions than this).
\end{example*}

\begin{fcthm}[Generalisation / modification of the Berry--Esseen Theorem]
  % Theorem 8.1
  \label{thm:8.1}
  Let $X_1, \ldots, X_n$ and $Y_1, \ldots, Y_n$ be sequences of independent
  random variables.
  Suppose that $\Ebb X_i = \Ebb Y_i$ and $\Ebb X_i^2 = \Ebb Y_i^2$ for each $i$.
  Let $\psi : \Rbb \to \Rbb$ be such that $\|\psi'''\|_\infty \le C$ (bounded
  third derivative).
  Then
  \[
    |\Ebb \psi(X_1 + \cdots + X_n - \Ebb \psi(Y_1 + \cdots + Y_n)|
    \le \frac{1}{6} C \left( \sum_{i = 1}^{n} (\|X_i\|_3^3 + \|Y_i\|_3^3)
    \right)
  .\]
\end{fcthm}

\begin{note*}
  For $Y \sim N(0, 1)$, we have $\|Y\|_3^3 = 2\sqrt{\frac{2}{\pi}}$ and
  $\|Y\|_4^4 = 3$ (will be an exercise on the example sheet).
\end{note*}

\begin{proof}
  By the triangle inequality, the quantity we wish to bound is at most
  \[
    \sum_{i = 1}^{n} |\Ebb \psi(Y_1 + \cdots + Y_{i - 1} + X_i + \cdots + X_n)
    - \Ebb \psi(Y_1 + \cdots + Y_{i - 1} + Y_i + X_{i + 1} + \cdots + X_n)|
    Y_|
  .\]
  Write $U_i$ for $Y_1 + \cdots + Y_{i - 1} + X_{i + 1} + \cdots + X_n$.
  So the above is
  \[
    \sum_{i = 1}^{n} |\Ebb \psi(U_i + X_i) - \Ebb \psi(U_i + Y_i)|
  .\]
  By Taylor's Theorem,
  \begin{align*}
    \psi(U_i + X_i)
    &= \psi(U_i) + X_i \psi'(U_i) + \frac{X_i^2}{2} \psi''(U_i) +
    \frac{X_i^3}{6} \psi'''(V_i) \\
    \psi(U_i + Y_i)
    &= \psi(U_i) + Y_i \psi'(U_i) + \frac{Y_i^2}{2} \psi''(U_i) +
    \frac{Y_i^3}{6} \psi'''(W_i)
  \end{align*}
  where $V_i$ is between $U_i$ and $U_i + X_i$, $W_i$ is between $U_i$ and $U_i
  + Y_i$.

  Taking expectations and subtracting, and using the fact that $\Ebb X_i = \Ebb
  Y_i$ and $\Ebb X_i^2 = \Ebb Y_i^2$ and also that $X_i$ and $Y_i$ are
  independent of $U_i$, we get
  \[
    \Ebb\left(
      \frac{X_i^3}{6} \psi'''(V_i) - \frac{Y_i^3}{6} \psi'''(W_i)
    \right)
  ,\]
  which has size at most $\frac{C}{6}(\Ebb|X_i|^3 + \Ebb |Y_i|^3) = \frac{C}{6}
  (\|X_i\|^3 + \|Y_i\|_3^3)$.

  Summing gives the result.
\end{proof}

\begin{fccoro}[]
  % Corollary 7.2
  \label{coro:7.2}
  Let $X_1, \ldots, X_n$ be independent with $\Ebb X_i = 0$ and $\Ebb X_i^2 =
  \sigma_i^2$ with $\sum \sigma_i^2 = 1$.
  Let $\psi$ be such that $\|\psi'''\|_\infty \le C$.
  Then
  \[
    \left|
      \Ebb \psi \left( \sum_{i = 1}^{n} X_i \right)
      - \Ebb \psi (Y)
    \right|
    \le \frac{C}{6} \left( \sum_{i = 1}^n \|X_i\|_3^3 + 2 \sum_{i = 1}^n
    \sigma_i^3 \sqrt{\frac{2}{\pi}} \right)
  \]
  where $Y \sim N(0, 1)$.
\end{fccoro}

\begin{proof}
  Let $Y_1, \ldots, Y_n$ be normal with mean zero and $\Ebb Y_i^2 = \sigma_i^2$.
  Then $\sum_{i = 1}^n Y_i \sim N(0, 1)$.
  By the previous theorem, we get a bound of
  \[
    \frac{C}{6} \left( \sum_{i = 1}^n \|X_i\|_3^3 + \sum_{i = 1}^n \sigma_i^3 2
    \sqrt{\frac{2}{\pi}} \right)
    .
    \qedhere
  \]
\end{proof}