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

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.

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,

> pbinom(10,10,0.5)
[1] 1

and

> 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.

> dbinom(1,10,0.5)
[1] 0.009765625

or $\displaystyle\frac{10}{1024}$

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:

> punif(0.75)
[1] 0.75

To obtain $P[a \leq X \leq b],$ we use $punif$ twice, e.g.

> 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

Example

If $Z \sim N(0,1)$ then we can e.g. obtain $P[Z\leq1.96]$ with

> 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.

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

> pnorm(2)
[1] 0.9772499

or specify $\mu$ and $\sigma$

> 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 $1.96$.