Add content for functions/miscellanea

Add actual content to the skeleton of the `functions/miscellanea`
section.

Signed-off-by: Eggert Karl Hafsteinsson <[email protected]>
Signed-off-by: Teodor Dutu <[email protected]>
Signed-off-by: Razvan Deaconescu <[email protected]>

chapters/functions/miscellanea/reading/read.md

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+# Miscellanea
+
+## Simple Probabilities In R
+
+R has functions to compute probabilities based on most common distributions
+
+If $X$ is a random variable with a known distribution, then R can typically compute values of the cumulative distribution function or:
+
+$$F(x)=P[X \leq x]$$
+
+### Examples
+
+:::info Example
+
+If $X \sim Bin(n,p)$ has binomial distribution, i.e.
+
+$$P(X = x) = \displaystyle{n \choose x}p^x(1-p)^{n-x},$$
+
+then cumulative probabilities can be computed with $\verb|pbinom|$, e.g.
+
+```text
+pbinom(5,10,0.5)
+```
+
+gives
+
+$$P[X \leq 5] = 0.623$$
+
+where
+
+$$X \sim Bin(n=10,p= \displaystyle\frac{1}{2})$$
+
+This can also be computed by hand.
+Here we have $n=10$, $p=1/2$ and the probability $P[X \leq 5]$ is obtained by adding up the individual probabilities, $P[X =0]+P[X =1]+P[X =2]+P[X =3]+P[X =4]+P[X =5]$
+
+$$P[X \leq 5] =\displaystyle\sum_{x=0}^5 \displaystyle{10\choose x} \displaystyle\frac{1}{2}^x\displaystyle\frac{1}{2}^{10-x}$$
+
+This becomes
+
+$$P[X \leq 5] = \displaystyle{10 \choose 0} \displaystyle\frac{1}{2}^0\displaystyle\frac{1}{2}^{10-0} + \displaystyle{10 \choose 1} \displaystyle\frac{1}{2}^1\displaystyle\frac{1}{2}^{10-1} + \displaystyle{10 \choose 1} \displaystyle\frac{1}{2}^2\displaystyle\frac{1}{2}^{10-2} + \displaystyle{10 \choose 3} \displaystyle\frac{1}{2}^3\displaystyle\frac{1}{2}^{10-3} + \displaystyle{10 \choose 4} \displaystyle\frac{1}{2}^4\displaystyle\frac{1}{2}^{10-4} + \displaystyle{10 \choose 5} \displaystyle\frac{1}{2}^5\displaystyle\frac{1}{2}^{10-5}$$
+
+or
+
+$$P[X \leq 5] = \displaystyle{10 \choose 0} \displaystyle\frac{1}{2}^{10} + \displaystyle{10 \choose 1} \displaystyle\frac{1}{2}^{10} + \displaystyle{10 \choose 1} \displaystyle\frac{1}{2}^{10} + \displaystyle{10 \choose 3} \displaystyle\frac{1}{2}^{10} + \displaystyle{10 \choose 4} \displaystyle\frac{1}{2}^{10} + \displaystyle{10 \choose 5} \displaystyle\frac{1}{2}^{10}=\displaystyle\frac{1}{2}^{10} \left(1+10+45+\dots \right)$$
+
+Furthermore,
+
+```text
+> pbinom(10,10,0.5)
+[1] 1
+```
+
+and
+
+```text
+> pbinom(0,10,0.5)
+[1] 0.0009765625
+```
+
+It is sometimes of interest to compute $P[X=x]$ in this case, and this is given by the `dbinom` function, e.g.
+
+```text
+> dbinom(1,10,0.5)
+[1] 0.009765625
+```
+
+or $\displaystyle\frac{10}{1024}$
+
+:::
+
+:::info Example
+
+Suppose $X$ has a uniform distribution between `0` and `1`, i.e. $X \sim Unf(0,1)$.
+Then the $punif$ function will return probabilities of the form
+
+$$P[X \leq x]= \int_{-\infty}^{x} f(t)dt= \int_{0}^{x} f(t)dt$$
+
+where $f(t)=1$ if $0 \leq t \leq 1$ and $f(t)=0$.
+For example:
+
+```text
+> punif(0.75)
+[1] 0.75
+```
+
+To obtain $P[a \leq X \leq b],$ we use $punif$ twice, e.g.
+
+```text
+> punif(0.75)-punif(0.25)
+[1] 0.5
+```
+
+:::
+
+## Computing Normal Probabilities In R
+
+To compute probabilities $X\sim N(\mu,\sigma^2)$ is usually transformed, since we know that
+
+$$Z:=\displaystyle\frac{X-\mu}{\sigma} \sim(0,1)$$
+
+The probabilities can then be computed for either $X$ or $Z$ with the `pnorm` function in R.
+
+### Details
+
+Suppose $X$ has a normal distribution with mean $\mu$ and variance
+
+$$X\sim N(\mu,\sigma^2)$$
+
+then to compute probabilities, $X$ is usually transformed, since we know that
+
+$$Z=\displaystyle\frac{X-\mu}{\sigma} \sim(0,1)$$
+
+and the probabilities can be computed for either $X$ or $Z$ with the `pnorm` function.
+
+### Examples
+
+:::info Example
+
+If $Z \sim N(0,1)$ then we can e.g. obtain $P[Z\leq1.96]$ with
+
+```text
+> pnorm(1.96)
+[1] 0.9750021
+
+> pnorm(0)
+[1] 0.5
+
+> pnorm(1.96)-pnorm(1.96)
+[1] 0
+
+> pnorm(1.96)-pnorm(-1.96)
+[1] 0.9500042
+```
+
+The last one gives the area between `-1.96` and `1.96`.
+
+:::
+
+:::info Example
+
+If $X \sim N(42,3^2)$ then we can compute probabilities either by transforming
+
+$$
+\begin{aligned}
+  P[X\leq x] &= P\left[\displaystyle\frac{X-\mu}{\sigma} \leq \displaystyle\frac{x-\mu}{\sigma}\right] \\
+  &= P\left[Z \leq \displaystyle\frac{x-\mu}{\sigma}\right]
+\end{aligned}
+$$
+
+and calling `pnorm` with the computed value $z=\displaystyle\frac{x-\mu}{\sigma}$, or call `pnorm` with $x$ and specify $\mu$ and $\sigma$.
+
+To compute $P[X\leq 48]$, either set $z=(48-42)/3=2$ and obtain
+
+```text
+> pnorm(2)
+[1] 0.9772499
+```
+
+or specify $\mu$ and $\sigma$
+
+```text
+> pnorm(42,42,3)
+[1] 0.5
+```
+
+:::
+
+## Introduction to Hypothesis Testing
+
+### Details
+
+If we have a random sample $x_1, \ldots, x_n$ from a normal distribution, then we consider them to be outcomes of independent random variables $X_1, \ldots, X_n$ where $X_i \sim N(\mu, \sigma^2)$.
+Typically, $\mu$ and $\sigma^2$ are unknown but assume for now that $\sigma^2$ is known
+
+Consider the hypothesis
+
+$$H_0: \mu = \mu_0 \text{ vs. } H_1: \mu > \mu_0$$
+
+where
+
+$$\mu_0$$
+
+is a specified number.
+
+Under the assumption of independence, the sample mean
+
+$$\overline{x} = \displaystyle\frac{1}{n}\displaystyle\sum^n_{i=1}x_i$$
+
+is also an observation from a normal distribution, with mean $\mu$ but a smaller variance.Specifically, $\overline{x}$ is the outcome of
+
+$$\overline{X} = \displaystyle\frac{1}{n}\displaystyle\sum^n_{i=1}X_i$$
+
+and
+
+$$X \sim N(\mu, \displaystyle\frac{ \sigma^2}{n})$$
+
+so the standard deviation of $X$ is $\displaystyle\frac{\sigma}{\sqrt{n}}$, so the appropriate error measure for $\overline{x}$ is $\displaystyle\frac{\sigma}{\sqrt{n}}$, when $\sigma$ is unknown.
+
+If $H_0$ is true, then
+
+$$z:= \displaystyle\frac{\overline{x}-\mu_0}{\sigma / \sqrt{n}}$$
+
+is an observation from an $n \sim N (0,1)$ distribution, i.e. an outcome of
+
+$$Z= \displaystyle\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}$$
+
+where $Z \sim N(0,1)$ when $H_0$ is correct.
+It follows that e.g. $P[\vert Z \vert > 1.96] = 0.05$ and if we observe $\vert Z \vert > 1.96$ then we reject the null hypothesis.
+
+Note that the value $z^\ast = 1.96$ is a quantile of the normal distribution and we can obtain other quantiles with the `pnorm` function, e.g. `pnorm` gives $1.96$.