What does it mean to say something is independent of something else?
The simplest way to think about it is, “do I learn something more about x by knowing y than not”. If two things are independent, I don’t need to care about y if x is my objective.
It really struggles to spell/typeset in graphics
Representing Probability Distributions
Is of necessity two-dimensional,
We have x and
and Pr(X=x) in one of two types (Pr or f) equating to sums [discrete] or integrals [over continuua].
Probability Distributions of Two Forms
Our core concept is a probability distribution just as above. These come in two forms for two types [discrete (qualitative)] and continuous (quantitative)] and can be either:
If X is normal, then Z = \frac{(X - \mu)}{\sigma} is normal.
The Normal [Plotted]
library(patchwork)Unif <-data.frame(x=seq(0, 1, by =0.005)) %>%mutate(p.x =punif(x), d.x =dunif(x))p1 <-ggplot(Unif) +aes(x=x, y=p.x) +geom_step() +labs(title="Distribution Function [cdf/cmf]") +theme_minimal()p2 <-ggplot(Unif) +aes(x=x, y=d.x) +geom_step() +labs(title="Density Function [pdf/pmf]") +theme_minimal()p2 + p1
The z-transform
The generic z-transformation applied to a variable x centers [mean\approx 0] and scales [std. dev. \approx variance \approx 1] to z_{x} for population parameters.1 In this case, two things are important.
this is the idea behind there only being one normal table in a statistics book.
the \mu and \sigma are presumed known.
z = \frac{x - \mu}{\sigma}
Sample z-scores
The scale command in R does this for a sample.
z = \frac{x - \overline{x}}{s_{x}} where \overline{x} is the sample mean of x and s_{x} is the sample standard deviation of x.
In samples, the 0 and 1 are exact; these are features of the mean and degrees of freedom. If I know the mean and any n-1 observations, the n^{th} observation is exactly the value such that the deviations add up to zero/cancel out.
An Earnings Example
Suppose earnings in a community have mean 55,000 and standard deviation 10,000. This is in dollars. Suppose I earn 75,000 dollars. First, if we take the top part of the fraction in the z equation, we see that I earn 20,000 dollars more than the average (75000 - 55000). Finishing the calculation of z, I would divide that 20,000 dollars by 10,000 dollars per standard deviation. Let’s show that.
I am 2 standard deviations above the average (the +) earnings. All z does is re-scale the original data to standard deviations with zero as the mean. The metric is the standard deviation.
Suppose I earn 35,000. That makes me 20,000 below the average and gives me a z score of -2. I am 2 standard deviations below average (the -) earnings.
Z and symmetry
z is an easy way to assess symmetry.
The mean of z is always zero but the distribution of z to the left and right of zero is informative. If they are roughly even, then symmetry is likely.
If the signs are uneven, then symmetry is unlikely.
In R, z is automated with the scale() command. The last line uses a table and the sign command to show the positive and negative z.
# Generate random normal incomeDataF <-data.frame(Hypo.Income =rnorm(1000, 55000, 10000)) %>%# z-transform income [mean 55000ish, std. dev. 10000ish]mutate(z.Income =scale(Hypo.Income))# Show the data.framehead(DataF)table(sign(DataF$z.Income))
Distributions in R are defined by four core parts:
r: random variables
d: density/probability that Pr(X=x) or f(x)
p: cumulative probability (given q) Pr(X\leq q)=p
q: quantile (given p): x such that Pr(X\leq q)=p
A Grape Escape?
A filling process is supposed to fill jars with 16 ounces of grape jelly, according to the label, and regulations require that each jar contain between 15.95 and 16.05 ounces.
Some Questions
Suppose that the uniform random process of filling in known to fill between 15.9 and 16.1 ounces uniformly.
Plot the histogram of 1000 simulated values. NB: unif is the noun with boundaries a (default 0) and b(default 1).
What is the probability that a random jar contains more than 16.1 ounces?
1-pnorm(16.1, 16.004, 0.028)
[1] 0.0003033834
What is the probability that a random jar contains less than 16.04 ounces?
pnorm(16.04, 16.004, 0.028)
[1] 0.9007286
95% of jars, given a normal, will contain between XXX and XXX ounces of jelly.
qnorm(c(0.025,0.975), 16.004, 0.028)
[1] 15.94912 16.05888
The bottom 5% of jars contain, at most, XXX ounces of jelly.
qnorm(0.05, 16.004, 0.028)
[1] 15.95794
The top 25% of jars contain at least XXX ounces of jelly.
qnorm(0.75, 16.004, 0.028)
[1] 16.02289
Why Normals?
The Central Limit Theorem
They Dominate Ops [6\sigma]
Normal Approximations Abound
Events: The Poisson
Poisson
Take a binomial with p very small and let n \rightarrow \infty. We get the Poisson distribution (y) given an arrival rate \lambda specified in events per period.
f(y|\lambda) = \frac{\lambda^{y}e^{-\lambda}}{y!}
Examples: The Poisson
Walk in customers
Emergency Room Arrivals
Births, deaths, marriages
Prussian Cavalry Deaths by Horse Kick
Fish?
Air Traffic Controllers
FAA Decision: Expend or do not expend scarce resources investigating claimed staffing shortages at the Cleveland Air Route Traffic Control Center.
Essential facts: The Cleveland ARTCC is the US’s busiest in routing cross-country air traffic. In mid-August of 1998, it was reported that the first week of August experienced 3 errors in a one week period; an error occurs when flights come within five miles of one another by horizontal distance or 2000 feet by vertical distance. The Controller’s union claims a staffing shortage though other factors could be responsible. 21 errors per year (21/52 errors per week) has been the norm in Cleveland for over a decade.
Some Questions
Plot a histogram of 1000 random weeks. NB: pois is the noun with no default for \lambda – the arrival rate.
After analyzing the initial data, you discover that the first two weeks of August have experienced 6 errors. What would you now decide? Well, once is 0.0081342. Twice, at random, is that squared. We have a problem.
Deaths by Horse Kick in the Prussian cavalry?
library(vcd)data(VonBort)head(VonBort)
deaths year corps fisher
1 0 1875 G no
2 0 1875 I no
3 0 1875 II yes
4 0 1875 III yes
5 0 1875 IV yes
6 0 1875 V yes
mean(VonBort$deaths)
[1] 0.7
Bernoulli Trials
The Generic Bernoulli Trial
Suppose the variable of interest is discrete and takes only two values: yes and no. For example, is a customer satisfied with the outcomes of a given service visit?
For each individual, because the probability of yes (1) \pi and no (0) 1-\pi must sum to one, we can write:
f(x|\pi) = \pi^{x}(1-\pi)^{1-x}
Binomial Distribution
For multiple identical trials, we have the Binomial:
Informal surveys suggest that 15% of Essex shopkeepers will not accept Scottish pounds. There are approximately 200 shops in the general High Street square.
Draw a plot of the distribution and the cumulative distribution of shopkeepers that do not accept Scottish pounds.
Scots <-data.frame(Potential.Refusers =0:200) %>%mutate(Prob =round(pbinom(Potential.Refusers, size=200, 0.15), digits=4))Scots %>%ggplot() +aes(x=Potential.Refusers, y=Prob) +geom_point() +labs(x="Refusers", y="Prob. of x or Less Refusers") +theme_minimal() -> Plot1Plot1
A Nicer Plot
library(plotly)p <-ggplotly(Plot1)p
More Questions About Scottish Pounds
What is the probability that 24 or fewer will not accept Scottish pounds?
pbinom(24, 200, 0.15)
[1] 0.1368173
What is the probability that 25 or more shopkeepers will not accept Scottish pounds?
1-pbinom(24, 200, 0.15)
[1] 0.8631827
With probability 0.9 [90 percent], XXX or fewer shopkeepers will not accept Scottish pounds.
qbinom(0.9, 200, 0.15)
[1] 37
Application: The Median is a Binomial with p=0.5
Interestingly, any given observation has a 50-50 chance of being over or under the median. Suppose that I have five datum.
What is the probability that all are under?
pbinom(0,size=5, p=0.5)
[1] 0.03125
What is the probability that all are over?
dbinom(5,size=5, p=0.5)
[1] 0.03125
What is the probability that the median is somewhere in between our smallest and largest sampled values?
Everything else.
The Rule of Five
This is called the Rule of Five by Hubbard in his How to Measure Anything.
Geometric Distributions
How many failures before the first success? Now defined exclusively by p. In each case, (1-p) happens k times. Then, on the k+1^{th} try, p. Note 0 failures can happen…
Pr(y=k) = (1-p)^{k}p
Example: Entrepreneurs
Suppose any startup has a p=0.1 chance of success. How many failures?
Example: Entrepreneurs
Suppose any startup has a p=0.1 chance of success. How many failures for the average/median person?
qgeom(0.5,0.1)
[1] 6
[Geometric] Plot 1000 random draws of “How many vendors until one refuses my Scottish pounds?”
How many failures before the r^{th} success? In each case, (1-p) happens k times. Then, on the k+1^{th} try, we get our r^{th} p. Note 0 failures can happen…
Pr(y=k) = {k+r-1 \choose r-1}(1-p)^{k}p^{r}
Needed Sales
I need to make 5 sales to close for the day. How many potential customers will I have to have to get those five sales when each customer purchases with probability 0.2.
In this last example, I was concerned with sales. I might also want to generate revenues because I know the rough mean and standard deviation of sales. Combining such things together forms the basis of a Monte Carlo simulation.
Some of the basics are covered in a swirl on simulation.
An Example
Customers arrive at a rate of 7 per hour. You convert customers to buyers at a rate of 85%. Buyers spend, on average 600 dollars with a standard deviation of 150 dollars.
Distributions are how variables and probability relate. They are a graph that we can enter in two ways. From the probability side to solve for values or from values to solve for probability. It is always a function of the graph.
Distributions generally have to be sentences.
The name is a noun but it also has
parameters – verbs – that makes the noun tangible.
