Example for updating prior from posterior

Bayes Theorem \[ p(A|B)=\frac {p(B|A)~p(A)} {p(B)} \]

1. Breast Cancer prevelance = .35%, \(p(\ddot\frown)=.0035\)
   • What is the probability of not having breast cancer \(p(\ddot\smile) = 1 -.0035=.9965\)
2. Mammogram false negative = 11%, \(p(-|\ddot\frown)=.11\)
3. Mammogram false positive = 7%, \(p(+|\ddot\smile)=.07\)

Finding the true positive, getting a + mammogram conditional on having breast cancer. This is also called sensitivity of a test, how well a test can detect a disease. \[ p(+|\ddot\frown) = 1 - p(-|\ddot\frown) = 1 - .11=.89 \] This is because the true positive and false negative must = 1.00. Meaning that we can take the inverse of what we already know to be the false negative.

Finding the true negative, getting a - mammogram conditional on not having breast cancer. This is also called specificity of a test, how well a test can discriminate between not having the disease. \[ p(-|\ddot\smile) = 1 - p(-|\ddot\smile) = 1 - .07=.93 \]

We now have 6 pieces of information from our original 3 pieces. \(p(\ddot\frown)=.0035\) Probability of having cancer \(p(\ddot\smile) =.9965\) Probability of not having cancer \(p(+|\ddot\frown)=.89\) True positive \(p(-|\ddot\frown)=.11\) False negative \(p(-|\ddot\smile)=.93\) True Negative \(p(+|\ddot\smile)=.07\) False positive

We want to know what is the probability of having breast cancer conditional on getting a + mammogram. Question: \(p(\ddot\frown|+)=?\)

Using bayes theorem we plug in the numbers.

\[ p(\ddot\frown|+)=\frac {p(+|\ddot\frown)~p(\ddot\frown)} {p(+|\ddot\frown) \times p(\ddot\frown) + p(+|\ddot\smile) \times p(\ddot\smile)} \]

\[ p(\ddot\frown|+)=\frac{.89\times .0035} {.89 \times .0035 + .07 \times .9965} = .0427 \] A 4% chance that a women over the age of 40 selected at random and tests positive on mammogram has breast cancer.

What if there is a second test? Rather than using the prevalence we now use the result from the first test. \(p(\ddot\frown)=.0427\) \(p(\ddot\smile)= 1 - .0427 = .9573\) \[ p(\ddot\frown|2nd +)=\frac{.89\times .0427} {.89 \times .0427 + .07 \times .9573} =.3619 \] There is a 36% chance that a women who tested positive twice on a mammogram has breast cancer.

Using a binomial distribution we can examine the probability of having 10 women with breast cancer from a sample of 1,000.

dbinom(10, size = 1000, prob=.0035)

[1] 0.002258614

What about at least 1 woman with breast cancer from a sample of 1,000?

dbinom(1, size = 1000, prob=.0035)

[1] 0.1054129

An 11% chance.

In a sample of 10,000, what is the cumulative probability of seeing at least 100 or less cases of breast cancer?

library(RcmdrMisc)
x <- 1:100
plotDistr(x, dbinom(x, size = 10000, prob=.0035), discrete=TRUE)

With this plot we can see that the seeing 33-38 cases of breast cancer has a probability of around .30, let’s check this hypothesis.

sum(dbinom(33:37, size=10000, prob=.0035))

[1] 0.3277125

The cumulative probability of seeing between 33 and 37 cases of breast cancer in a sample of 10,000 is .327, or a 32.8% chance.