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\fancyhead[LO,LE]{Unit 3 Methods Statistics}
\fancyhead[CO,CE]{Andrew Lorimer}

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  \title{Statistics}
  \author{}
  \date{}
  \maketitle

  \section{Probability}

  \[ \Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B) \]
  \[ \Pr(A \cup B) = 0 \tag{mutually exclusive} \]

  \section{Conditional probability}

  \[ \Pr(A|B) = \frac{\Pr(A \cap B)}{\Pr(B)} \quad \text{where } \Pr(B) \ne 0 \]
  
  \[ \Pr(A) = \Pr(A|B) \cdot \Pr(B) + \Pr(A|B^{\prime}) \cdot \Pr(B^{\prime}) \tag{law of total probability} \]
  
  \[ \Pr(A \cap B) = \Pr(A|B) \times \Pr(B) \tag{multiplication theorem} \]

  For independent events:
  
  \begin{itemize}
    \item \(\Pr(A \cap B) = \Pr(A) \times \Pr(B)\)
    \item \(\Pr(A|B) = \Pr(A)\)
    \item \(\Pr(B|A) = \Pr(B)\)
  \end{itemize}

  \subsection{Discrete random distributions}

  Any experiment or activity involving chance will have a probability associated with each result or \textit{outcome}. If the outcomes have a reference to \textbf{discrete numeric values} (outcomes that can be counted), and the result is unknown, then the activity is a \textit{discrete random probability distribution}.

  \subsubsection{Discrete probability distributions}
  
  If an activity has outcomes whose probability values are all positive and less than one ($\implies 0 \le p(x) \le 1$), and for which the sum of all outcome probabilities is unity ($\implies \sum p(x) = 1$), then it is called a \textit{probability distribution} or \textit{probability mass} function.

  \begin{itemize}
    \item \textbf{Probability distribution graph} - a series of points on a cartesian axis representing results of outcomes. $\Pr(X=x)$ is on $y$-axis, $x$ is on $x$ axis.
    \item \textbf{Mean $\mu$} or \textbf{expected value} \(E(X)\) - measure of central tendency. Also known as \textit{balance point}. Centre of a symmetrical distribution.
      \begin{align*}
        \overline{x} = \mu = E(X) &= \frac{\Sigma(xf)}{\Sigma(f)} \\
        &= \sum_{i=1}^n (x_i \cdot P(X=x_i)) \\
        &= \int_{-\infty}^{\infty} x\cdot f(x) \> dx \quad \text{(for pdf } f \text{)}
        &= \sum_{-\infty}^{\infty} 
      \end{align*}
    \item \textbf{Mode} - most popular value (has highest probability of \(X\) values). Multiple modes can exist if \(>1 \> X\) value have equal-highest probability. Number must exist in distribution.
    \item \textbf{Median \(m\)} - the value of \(x\) such that \(\Pr(X \le m) = \Pr(X \ge m) = 0.5\). If \(m > 0.5\), then value of \(X\) that is reached is the median of \(X\). If \(m = 0.5 = 0.5\), then \(m\) is halfway between this value and the next.
      \[ m = X \> \text{such that} \> \int_{-\infty}^{m} f(x) dx = 0.5 \]
    \item \textbf{Variance $\sigma^2$} - measure of spread of data around the mean. Not the same magnitude as the original data. For distribution \(x_1 \mapsto p_1, x_2 \mapsto p_2, \dots, x_n \mapsto p_n\):
      \begin{align*}
        \sigma^2=\operatorname{Var}(x) &= \sum_{i=1}^n p_i (x_i-\mu)^2 \\
        &= \sum (x-\mu)^2 \times \Pr(X=x) \\
        &= \sum x^2 \times p(x) - \mu^2
      \end{align*}
    \item \textbf{Standard deviation $\sigma$} - measure of spread in the original magnitude of the data. Found by taking square root of the variance: $\sigma =\operatorname{sd}(X)=\sqrt{\operatorname{Var}(X)}$
  \end{itemize}

  \subsubsection{Expectation theorems}

  \begin{align*}
    E(aX \pm b) &= aE(X) \pm b \\
    E(z) &= z \\
    E(X+Y) &= E(X) + E(Y) \\
    E(X)^n &= \Sigma x^n \cdot p(x) \\
    &\ne [E(X)]^2
  \end{align*}


  \section{Binomial Theorem}

  \begin{align*}
    (x+y)^n &= {n \choose 0} x^n y^0 + {n \choose 1} x^{n-1}y^1 + {n \choose 2} x^{n-2}y^2 + \dots + {n \choose n-1}x^1 y^{n-1} + {n \choose n} x^0 y^n \\
    &= \sum_{k=0}^n {n \choose k} x^{n-k} y^k \\
    &= \sum_{k=0}^n {n \choose k} x^k y^{n-k}
  \end{align*}

  \begin{enumerate}
    \item powers of \(x\) decrease \(n \rightarrow 0\)
    \item powers of \(y\) increase \(0 \rightarrow n\)
    \item coefficients are given by \(n\)th row of Pascal's Triangle where \(n=0\) has one term
    \item Number of terms in \((x+a)^n\) expanded \& simplified is \(n+1\)
  \end{enumerate}

  Combinations: \(^n\text{C}_r = {N\choose k}\) (binomial coefficient) 
  \begin{itemize}
    \item Arrangements \({n \choose k} = \frac{n!}{(n-r)}\)
    \item Combinations \({n \choose k} = \frac{n!}{r!(n-r)!}\)
    \item Note \({n \choose k} = {n \choose k-1}\)
  \end{itemize}

  \subsubsection{Pascal's Triangle}

  \begin{tabular}{>{$}l<{$\hspace{12pt}}*{13}{c}}
    n=\cr0&&&&&&&1&&&&&&\\
    1&&&&&&1&&1&&&&&\\
    2&&&&&1&&2&&1&&&&\\
    3&&&&1&&3&&3&&1&&&\\
    4&&&1&&4&&6&&4&&1&&\\
    5&&1&&5&&10&&10&&5&&1&\\
    6&1&&6&&15&&20&&15&&6&&1
  \end{tabular}

  \colorbox{cas}{On CAS:} (soft keys) \keystroke{\(\downarrow\)} \(\rightarrow\) \keystroke{Advanced} \(\rightarrow\) \verb;nCr(n,cr);

  \section{Binomial distributions}

  (aka Bernoulli distributions)

  \begin{align*}
    \Pr(X=x) &= {n \choose x} p^x (1-p)^{n-x} \\
    &= {n \choose x} p^x q^{n-x}
  \end{align*}

  \begin{enumerate}
    \item Two possible outcomes: \textbf{success} or \textbf{failure}
    \item \(\Pr(\text{success})\) is constant across trials (also denoted \(p\))
    \item Finite number \(n\) of independent trials
  \end{enumerate}

  If these conditions are met, then it is a Binomial Random Variable. This variable is said to have a \textit{binomial probability distribution}.

  \begin{itemize}
    \item \(n\) is the number of trials
    \item There are two possible outcomes: \(S\) or \(F\)
    \item \(\Pr(\text{success}) = p\)
    \item \(\Pr(\text{failure}) = 1-p = q\)
    \item Shorthand notation: \(X \sim \operatorname{Bi}(n,p)\)
  \end{itemize}

  \colorbox{cas}{On CAS:} Main \(\rightarrow\) Interactive \(\rightarrow\) Distribution \(\rightarrow\) \verb;binomialPDf; \\
  Input \verb;x; (no. of successes), \verb;numtrial; (no. of trials), \verb;pos; (probbability of success)

  \subsection{Applications of binomial distributions}

  \[ \Pr(X \ge a) = 1 - \Pr(X < a) \]

  \subsection{Expected value of a binomial distribution}

  \[ E(X \sim \operatorname{Bi}(n,p))=np \]

  \subsection{Variance}

  \[ \sigma^2(X) = np(1-p) \]

  \subsection{Standard deviation}

  \[ \sigma(X) = \sqrt{np(1-p)} \]

\end{document}