Oct 15, 2018
Three Axioms of Probability
\[P\left(\bigcup_{i=1}^{\infty}E_i \right) = \sum_{i=1}^{\infty}P(E_i)\]
Suppose that a room contains \(m\) students. What is the probability that at least two of them have the same birthday? Start by assuming every day of the year is equally likely to be a birthday, and disregard leap years. That is, assume there are always \(n = 365\) days to a year.
Let the event \(E\) denote two or more students with the same birthday. In this problem, it is easier to find \(E^c\), as there are a number of ways that \(E\) can take place. There are a total of \(365^m\) possible outcomes in the sample space. \(E^c\) can occur in \(365\times 364 \times \cdots \times (365 - m + 1)\) ways. Consequently,
\[P(E^c) = \frac{365\times 364 \times \cdots \times (365 - m + 1)}{365^m}\] and
\[P(E) = 1 - \frac{365\times 364 \times \cdots \times (365 - m + 1)}{365^m}\]
> m <- 1:60 # vector of number of students
> p <- numeric(60) # initialize vecotr to 0's
> for(i in m){ # index value for loop
+ q = prod((365:(365 - i + 1))/365)
+ p[i] = 1 - q
+ }
> plot(m, p, pch = 19,
+ ylab = "P(at least two students with the same birthday)",
+ xlab = "m = number of students in the room")
> abline(h = 0.5, lty = 2, col = "blue")
> abline(v = 23, lty = 2, col = "blue")
\[P(F|E) = \frac{P(F\cap E)}{P(E)}\]
Suppose two fair dice are rolled where each of the 36 possible outcomes is equally likely to occur. Knowing that the first die shows a 3, what is the probability that the sum of the two dice equals 6?
\[P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/36}{6/36}=\frac{1}{6}\]
Let \(F_1, F_2, \ldots F_n\) be such that \(\bigcup_{i=1}^{n}F_i = \Omega\) and \(F_i \cap F_j = \emptyset\) for all \(i \neq j\), with \(P(F_i) > 0\) for all \(i\). Then, for any event \(E\),
\[P(E) = \sum_{i=1}^{n}P(E \cap F_i) = \sum_{i=1}^{n}P(E|F_i)P(F_i)\]
\[P(F_j|E) = \frac{P(E \cap F_j)}{P(E)} = \frac{P(E|F_i)P(F_i)}{\sum_{i=1}^{n}P(E|F_i)P(F_i)} \]
The television show Let's Make a Deal hosted by Monty Hall, gave contestants the opportunity to choose one of three doors. Contestants hoped to choose the one that concealed the grand prize. Behind the other two doors were much less valuable prizes. After the contestant chose one of the doors, say Door 1, Monty opened one of the other two doors, say Door3, containing a much less valuable prize. The contestant was then asked whether he or she wished to stay with the original choice (Door 1) or switch to the other closed door (Door2). What should the contestant do? Is it better to stay with original choice of switch to the other closed door?
> set.seed(13)
> n <- 10000
> actual <- sample(1:3, size = n, replace = TRUE)
> aguess <- sample(1:3, size = n, replace = TRUE)
> equals <- (actual == aguess)
> not.eq <- (actual != aguess)
> PnoSwitch <- mean(equals)
> PwiSwitch <- mean(not.eq)
> probs <- c(PnoSwitch, PwiSwitch)
> names(probs) <- c("P(Win no Switch)", "P(Win Switch)")
> probs
P(Win no Switch) P(Win Switch)
0.331 0.669
To solve with Bayes' Rule start by assuming the contestant initially guesses Door 1 and that Monty opens Door 3. The the event \(D_i=\) Door \(i\) conceals the prize and \(O_j\) = Monty opens Door \(j\) after the contestant selects Door 1. When a contestant initially selects a door, \(P(D_1)=P(D_2)=P(D_3)=1/3\). Once Monty show the grand prize is not behind Door 3, the probability of winning the grand prize is now one of \(P(D_1|0_3)\) or \(P(D_2|O_3)\). Note that \(P(D_1|O_3)\) corresponds to the strategy of sticking with the initial guess and \(P(D_2|O_3)\) corresponds to the strategy of switching doors.
Based on how the show is designed, the following are know:
\[P(D_1|O_3) = \frac{P(O_3|D_1)P(D_1)}{P(O_3|D_1)P(D_1) + P(O_3|D_2)P(D_2) + P(O_3|D_3)P(D_3)}\] \[P(D_1|O_3) = \frac{1/2\cdot 1/3}{1/2 \cdot 1/3 + 1 \cdot 1/3 + 0 \cdot 1/3}=\frac{1}{3}\]
\[P(D_2|O_3) = \frac{P(O_3|D_2)P(D_2)}{P(O_3|D_1)P(D_1) + P(O_3|D_2)P(D_2) + P(O_3|D_3)P(D_3)}\]
\[P(D_2|O_3) = \frac{1\cdot 1/3}{1/2 \cdot 1/3 + 1 \cdot 1/3 + 0 \cdot 1/3}=\frac{2}{3}\]
Consider the case with \(n\) Doors and just one grand prize. The probability of winning the grand prize on the first choice is \(\frac{1}{n}\). This is the same probability of winning if you were to use the "stay" strategy. You will win the grand prize if both of the following events occur:
You choose a door that does not have the grand prize \(\left(P(G) = \frac{n-1}{n}\right)\)
You choose the door with the grand prize by switching doors \(\left(P(C|G) = \frac{1}{n-2}\right)\)
In other words, the probability of winning using the "switch" strategy is
\[P(G \cap C) = P(G) \cdot P(C|G) = \frac{n-1}{n} \cdot \frac{1}{n-2}\] For \(n=3\) doors, the probability of winning with the "switch" strategy is:
\[\frac{n-1}{n}\cdot \frac{1}{n-2} =\frac{3-1}{3}\cdot\frac{1}{3-2} = \frac{2}{3}\] Note: \(\frac{n-1}{n(n-2)} > \frac{1}{n} \rightarrow\) switching is always better!
A discrete random variable \(X\) is a function from \(\Omega\) into the real numbers \(\bf{R}\) with a range that is finite or countably infinite. That is, \(X: \Omega \rightarrow {x_1, x_2, \ldots, x_m}\), or \(X:\Omega \rightarrow {x_1, x_2, \dots}\).
A function \(X\) from \(\Omega\) into the real numbers \(\bf{R}\) is a continuous random variable if there exists a nonnegative function \(f\) such that for every subset \(C\) of \(\bf{R}\), \(P(X \in C) = \int_C\;f(x)\,dx\). In particular, for \(a \leq b\), \(P(a < X \leq b) = \int_a^b\; f(x)\,dx.\)
The function that assigns probability to the values of the random variable is called the probability density function (pdf). Many authors will refer to the pdf as the probability mass function pmf when working with discrete random variables. We will denote the pdf as \(p(x)=P(X=x)\) for each \(x\). All pdfs must satisfy the following two conditions:
\(p(x)\geq0\) for all \(x\).
\(\sum_{\forall x}p(x) = 1\).
The function \(f\) is called the probability density function of \(X\) and satisfies the following conditions:
\[\mu_X = E(X) = \sum_{x}x\cdot p(x)\]
\[\mu_X = E(X) = \int_{-\infty}^{\infty}\;x\,f(x)\,dx\]
\[\sigma^2 = \text{Var}(X) = E[(X - \mu_X)^2] = \sum_{x}(x- \mu_X)^2 \cdot p(x)\]
\[\sigma^2 = \text{Var}(X) = E[(X - \mu_X)^2] = \int_{-\infty}^{\infty}(x - \mu_X)^2\;f(x)\,dx\]
> S <- expand.grid(roll1 = 1:6, roll2 = 1:6) > S$Sum <- apply(S, 1, sum) > head(S)
roll1 roll2 Sum 1 1 1 2 2 2 1 3 3 3 1 4 4 4 1 5 5 5 1 6 6 6 1 7
Find the expected value and variance of \(X\)
> table(S$Sum)
2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 5 4 3 2 1
> T1 <- MASS::fractions(table(S$Sum)/sum(!is.na(S$Sum))) > T1
2 3 4 5 6 7 8 9 10 11 12 1/36 1/18 1/12 1/9 5/36 1/6 5/36 1/9 1/12 1/18 1/36
> x <- as.numeric(names(T1)) > x
[1] 2 3 4 5 6 7 8 9 10 11 12
> px <- T1 > EX <- sum(x*px) > EX
[1] 7
> VX <- sum((x - EX)^2*px) > VX
[1] 35/6
Let \(X\) be a continuous random variable with pdf \(f(x) = 2x, 0 < x < 1\). Find the mean and variance of \(X\).
\[\mu_x = E(X) = \int_{0}^{1} x \cdot 2x\, dx = \frac{2}{3}x^3 \mid_0^1 = \frac{2}{3}\]
\[\sigma^2 = \text{Var}(X) = E(X^2) - \mu_X^2\] \[\sigma^2 = \int_0^1 x^2(2x)\,dx - \left(\frac{2}{3}\right)^2 = \left[\left(\frac{1}{2}y^4 \right)\right]_0^1 - \frac{4}{9}= \frac{1}{18}\]
> f <- function(x){2*x}
> integrate(f, 0, 1)
1 with absolute error < 1.1e-14
> ex <- function(x){x*f(x)}
> EX <- integrate(ex, 0, 1)
> EX
0.6666667 with absolute error < 7.4e-15
> EX <- MASS::fractions(EX$value) > EX
[1] 2/3
> g <- function(x){(x - 2/3)^2*f(x)}
> VX <- integrate(g, 0, 1)
> VX
0.05555556 with absolute error < 6.2e-16
> VX <- MASS::fractions(VX$value) > VX
[1] 1/18
What is the area under \(f\)?
> curve(f, 0, 1)
The standard deviation of \(X\) is \(\sigma_{X} = \sqrt{\text{Var}[X]}\)
If \(X\) is a random variable and \(a\) and \(b\) are constants, then
\[E[a \pm bX] = a \pm bE[X]\] \[\text{Var}[a \pm bX] = b^2\text{Var}[X] \] \[\text{Var}[aX \pm bY] = a^2 \text{Var}[X] + b^2\text{Var}[Y]\]
\[\bar{X} = \frac{1}{n}\sum_{i=1}^{n}X_i\] is called the (sample) mean.
\[E[\bar{X}] = \mu_{\bar{X}} = \mu_{X} = \mu\]
\[\text{Var}[\bar{X}] = \frac{\sigma^2_X}{n} = \frac{\sigma^2}{n}\]
\[Z = \frac{\text{statistic} - \mu_{\text{statistic}}}{\sigma_{\text{statistic}}}\]
\[Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{14 - 10}{\frac{2}{\sqrt{4}}} = 4\]
\[T = \frac{\text{statistic} - \mu_{\text{statistic}}}{\hat{\sigma}_{\text{statistic}}}\]
> rs <- c(12, 14, 2, 20) > xbar <- mean(rs) > SD <- sd(rs) > Tscore <- (xbar - 10)/(SD/sqrt(4)) > c(xbar, SD, Tscore)
[1] 12.0000000 7.4833148 0.5345225
\[T = \frac{\bar{X} - \mu_{\bar{X}}}{\hat{\sigma}_{\bar{X}}} = \frac{12 - 10}{\frac{7.483315}{\sqrt{}4}} = 0.5345\]
Let X and Y be random variables. Then X and Y are independent if for any sets A and B,
\[P(X \in A , Y \in B) = P(X \in A)P(Y \in B).\]
Two events A and B are independent if
\[P(A \cap B) = P(A)P(B)\]
Two events A and B are independent if (and only if)
\[P(B|A) = P(B)\]
Suppose we take a random sample of \(n = 100\) online adults and find 71 use Facebook, 18 use Twitter, and 15 use both.
Facebook
Twitter Yes No Sum
Yes 15 3 18
No 56 26 82
Sum 71 29 100
Let A = respondent uses Facebook and B = respondent uses Twitter.
We know that \(P(A) = \frac{71}{100}=0.71\), \(P(B) = \frac{18}{100}=0.18\), and \(P(A\cap B)= \frac{15}{100}=0.15\).
Since \(P(A \cap B)\neq 0\), the events A and B are not mutually exclusive.
Recall that A and B are independent if \(P(A \cap B)=P(A)P(B)\). Since \(P(A \cap B) = 0.15 \neq P(A)P(B)=0.71\times 0.18 = 0.1278,\) the events A and B are not independent.
What do we expect in each cell?
Facebook
Twitter Yes No
Yes 15 3
No 56 26
Facebook
Twitter Yes No
Yes 12.78 5.22
No 58.22 23.78
\[TS = \sum_{\text{all cells}}\frac{(O - E)^2}{E}\]
> on <- c(15, 3, 56, 26)
> ONL <- matrix(data = on, nrow = 2, byrow = TRUE)
> dimnames(ONL) <- list(Twitter = c("Yes", "No"), Facebook = c("Yes", "No"))
> ONLT <- as.table(ONL)
> ONLTDF <- as.data.frame(ONLT)
> TDF <- vcdExtra::expand.dft(ONLTDF)
> dim(TDF)
[1] 100 2
> T1 <- xtabs(~Twitter + Facebook, data = TDF) > T1
Facebook
Twitter No Yes
No 26 56
Yes 3 15
> chisq.test(T1, correct = FALSE)
Pearson's Chi-squared test
data: T1
X-squared = 1.6217, df = 1, p-value = 0.2029
> library(tidyverse) > TDF %>% + ggplot(aes(x = Twitter, fill = Facebook)) + + geom_bar(position = "fill") + + labs(y = "Fraction")
> obs.stat <- chisq.test(T1, correct = FALSE)$stat > obs.stat
X-squared 1.621673
> set.seed(123)
> sims <- 10^4 - 1
> ts <- numeric(sims)
> for(i in 1:sims){
+ ts[i] <- chisq.test(xtabs(~Twitter + sample(Facebook), data = TDF),
+ correct = FALSE)$stat
+ }
> pvalue <- (sum(ts >= obs.stat) + 1)/(sims + 1)
> pvalue
[1] 0.2546
> library(ggplot2) > ggplot(data = data.frame(x = ts), aes(x = x)) + + geom_density(fill = "peachpuff", adjust = 3) + + theme_bw() + + stat_function(fun = dchisq, args = list(df = 1), color = "red") + + geom_vline(xintercept = obs.stat, linetype = "dashed", + color = "lightblue") + + xlim(0, 8)
