```{r}
0.1 * 0.085 + 0.105 * 0.915
```[1] 0.104575Day 08 - Lab : Bayes Rule
For this lab we will be working exercises from the Baye’s Rules Book.
Exercise 1.6 (Applying for an internship)
There are several data scientist openings at a much-ballyhooed company. Having read the job description, you know for a fact that you are qualified for the position: this is your data. Your goal is to ascertain whether you will actually be offered a position: this is your hypothesis.b
a) : From the perspective of someone using frequentist thinking, what question is answered in testing the hypothesis that you’ll be offered the position?
\[P(\text{Qualified} | \text{Offer}^C)\]
b) : Repeat part a from the perspective of someone using Bayesian thinking.
\[P(\text{Offer} | \text{Qualified})\]
c) : Which question would you rather have the answer to: the frequentist or the Bayesian? Explain your reasoning.
Bayesian prospective as that is what I will be interested in knowing.
Exercise 2.1 (Comparing the prior and posterior) [Only parts a-c]
For each scenario below, you’re given a pair of events, \(A\) and \(B\). Explain what you believe to be the relationship between the posterior and prior probabilities of \(B\). \(P(B|A) > P(B)\) or \(P(B|A)<P(B)\) :
a) A = you just finished reading Lambda Literary Award-winning author Nicole Dennis-Benn’s first novel, and you enjoyed it! B = you will also enjoy Benn’s newest novel.
\[P(B | A) > P(B)\]
b) A = it’s 0 degrees Fahrenheit in Minnesota on a January day. B = it will be 60 degrees tomorrow.
\[P(B | A) < P(B)\]
c) B = the authors only got 3 hours of sleep last night. B = the authors make several typos in their writing today.
\[P(B | A) > P(B)\]
Exercise 2.2 (Marginal, conditional, or joint?) [Parts a, b, d]
Define the following events for a resident of a fictional town:
A = drives 10 miles per hour above the speed limit,
B = gets a speeding ticket,
C = took statistics at the local college,
D = has used R,
E = likes the music of Prince F = is a Minnesotan.
Several facts about these events are listed below. Specify each of these facts using probability notation, paying special attention to whether it’s a marginal, conditional, or joint probability.
a) 73% of people that drive 10 miles per hour above the speed limit get a speeding ticket.
\[P(B | A) = 0.73\]
b) 20% of residents drive 10 miles per hour above the speed limit.
\[P(A) = 0.2\]
d) 91% of statistics students at the local college have used R.
\[P(D | C) = 0.91\]
e) 38% of residents are Minnesotans that like the music of Prince.
\[ P(A \text{ and }B) =0.38 \]
Exercise 2.3 (Binomial practice) [Parts c-f]
For each variable Y below, determine whether Y is Binomial. If yes, use notation to specify this model and its parameters. If not, explain why the Binomial model is not appropriate for Y.
c) : Each time they try out for the television show Ru Paul’s Drag Race, Alaska has a 17% probability of succeeding. Let Y be the number of times Alaska has to try out until they’re successful.
No, there is no fixed number of trials. This would be geometric, i.e., number of failures until success.
d) : Y is the amount of time that Henry is late to your lunch date.
No, there are no trials. (no success/failure)
e) : Y is the probability that your friends will throw you a surprise birthday party even though you said you hate being the center of attention and just want to go out to eat.
No, \(Y\) is a probability not an outcome.
f) : You invite 60 people to your “\(\pi\) day” party, none of whom know each other, and each of whom has an 80% chance of showing up. Let Y be the total number of guests at your party.
Yes, \(Y\) is binomial with 60 trials and a success probability of 0.8.
Exercise 2.10 (LGBTQ students: rural and urban)
A recent study of 415,000 Californian public middle school and high school students found that 8.5% live in rural areas and 91.5% in urban areas. Further, 10% of students in rural areas and 10.5% of students in urban areas identified as Lesbian, Gay, Bisexual, Transgender, or Queer (LGBTQ). Consider one student from the study.
a) : What’s the probability they identify as LGBTQ?
Law of total probability \[P(\text{LGBTQ}) = P(\text{LGBTQ} | \text{Rural}) P(\text{Rural}) + P(\text{LGBTQ} | \text{Urban}) P(\text{Urban}) = 0.1 * 0.085 + 0.105 * 0.915\]
b) : If they identify as LGBTQ, what’s the probability that they live in a rural area?
Baye’s rule! \[P(\text{Rural} | \text{LGBTQ}) = \frac{P(\text{Rural} \cap \text{LGBTQ})}{P(\text{LGBTQ})} = \frac{P(\text{LGBTQ} | \text{Rural}) P(\text{Rural})}{P(\text{LGBTQ})} = \frac{0.1 * 0.085}{0.104575}\]
```{r}
0.1 * 0.085 / 0.104575
```[1] 0.08128138c) : If they do not identify as LGBTQ, what’s the probability that they live in a rural area?
\[P(\text{Rural} | \text{Not LGBTQ}) = \frac{P(\text{Not LGBTQ} | \text{Rural}) P(\text{Rural})}{P(\text{Not LGBTQ})} = \frac{(1 - 0.1) * 0.085}{1 - 0.104575}\]
```{r}
(1 - 0.1) * 0.085 / (1 - 0.104575)
```[1] 0.08543429Exercise 2.11 (Internship)
Muhammad applies for six equally competitive data science internships. He has the following prior model for his chances of getting into any given internship, \(\pi\).
| \(\pi\) | 0.3 | 0.4 | 0.5 | Total |
|---|---|---|---|---|
| \(f(\pi)\) | 0.25 | 0.60 | 0.15 | 1 |
a): Let Y be the number of internship offers that Muhammad gets. Specify the model for the dependence of Y on \(\pi\) and the corresponding pmf, \(f(y|\pi)\)
\[Y | \pi \sim Binomial(n = 6, \pi)\] \[f(y | \pi) = {6 \choose y} \pi^y (1 - \pi)^{6 - y}\]
b) : Muhammad got some pretty amazing news. He was offered four of the six internships! How likely would this be if \(\pi = 0.3\)?
\[f(y = 4 | \pi = 0.3) = L(\pi = 0.3 | y = 4) = {6 \choose 4} 0.3^4 (1 - 0.3)^{6 - 4}\]
```{r}
choose(6, 4) * 0.3^4 * (1 - 0.3)^(6 - 4)
```[1] 0.059535c) : Construct the posterior model of \(\pi\) in light of Muhammad’s internship news
\[f(\pi | y = 4) = \frac{f(\pi) L(\pi | y = 4)}{f(y = 4)}\]
```{r}
pi_df <- data.frame(
pi = c(0.3, 0.4, 0.5),
prior = c(0.25, 0.6, 0.15),
likelihood = c(
dbinom(4, size = 6, prob = 0.3),
dbinom(4, size = 6, prob = 0.4),
dbinom(4, size = 6, prob = 0.5)
),
posterior = NA
)
```\(f(y = 4) = f(y = 4 | \pi = 0.3)*f(\pi = 0.3) + f(y = 4 | \pi = 0.4)*f(\pi = 0.4) + f(y = 4 | \pi = 0.5)*f(\pi = 0.5)\)
```{r}
(denominator <- dbinom(4, size = 6, prob = 0.3) * 0.25 +
dbinom(4, size = 6, prob = 0.4) * 0.6 +
dbinom(4, size = 6, prob = 0.5) * 0.15)
```[1] 0.132984\[f(\pi = 0.3 | y = 4) = \frac{f(\pi = 0.3) L(\pi = 0.3 | y = 4)}{f(y=4)} = \]
```{r}
(pi_df$posterior[1] <- 0.25 * dbinom(4, size = 6, prob = 0.3) / denominator)
```[1] 0.1119214\[f(\pi = 0.4 | y = 4) = \frac{f(\pi = 0.4) L(\pi = 0.4 | y = 4)}{f(y=4)} = \]
```{r}
(pi_df$posterior[2] <- 0.6 * dbinom(4, size = 6, prob = 0.4) / denominator)
```[1] 0.6237141\[f(\pi = 0.5 | y = 4) = \frac{f(\pi = 0.5) L(\pi = 0.5 | y = 4)}{f(y=4)} = \]
```{r}
(pi_df$posterior[3] <- 0.15 * dbinom(4, size = 6, prob = 0.5) / denominator)
```[1] 0.2643645As a sanity check we know the sum of the posteriors should be 1,
```{r}
sum(pi_df$posterior)
# Equal to doing
0.25 * dbinom(4, size = 6, prob = 0.3) / denominator +
0.6 * dbinom(4, size = 6, prob = 0.4) / denominator +
0.15 * dbinom(4, size = 6, prob = 0.5) / denominator
```[1] 1
[1] 1```{r}
# transpose to get it in the form we are used to
t(pi_df)
library(kableExtra)
library(knitr)
cbind(
c("$f(pi)$", "$L(pi | y)$", "$f(pi | y)$"),
t(pi_df)[-1, ]
) |>
kbl(
digits = 3,
col.names = c("$\\pi$", "0.3", "0.4", "0.5"),
row.names = FALSE
)
``` [,1] [,2] [,3]
pi 0.3000000 0.4000000 0.5000000
prior 0.2500000 0.6000000 0.1500000
likelihood 0.0595350 0.1382400 0.2343750
posterior 0.1119214 0.6237141 0.2643645| $\pi$ | 0.3 | 0.4 | 0.5 |
|---|---|---|---|
| $f(pi)$ | 0.25 | 0.6 | 0.15 |
| $L(pi | y)$ | 0.059535 | 0.13824 | 0.234375 |
| $f(pi | y)$ | 0.111921358960476 | 0.623714131023281 | 0.264364510016243 |
Exercise 2.15 (Cuckoo birds)
Cuckoo birds are brood parasites, meaning that they lay their eggs in the nests of other birds (hosts), so that the host birds will raise the cuckoo bird hatchlings. Lisa is an ornithologist studying the success rate, \(\pi\), of cuckoo bird hatchlings that survive at least one week. She is taking over the project from a previous researcher who speculated in their notes the following prior model for \(\pi\):
| \(\pi\) | 0.6 | 0.65 | 0.7 | 0.75 | Total |
|---|---|---|---|---|---|
| \(f(\pi)\) | 0.3 | 0.4 | 0.2 | 0.1 | 1 |
a) : If the previous researcher had been more sure that a hatchling would survive, how would the prior model be different?
There would be more probability placed on larger values of \(\pi\).
b) : If the previous researcher had been less sure that a hatchling would survive, how would the prior model be different?
There would be less probability placed on larger values of \(\pi\).
c) : Lisa collects some data. Among the 15 hatchlings she studied, 10 survived for at least one week. What is the posterior model for \(\pi\)?
\[f(\pi | y = 10) = \frac{f(\pi) L(\pi | y = 10)}{f(y = 10)}\]
d) : Lisa needs to explain the posterior model for \(\pi\) in a research paper for ornithologists, and can’t assume they understand Bayesian statistics. Briefly summarize the posterior model in context.
The four posterior probabilities would each be the probability of that success rate \(\pi\), based on prior knowledge and data.