Lecture 6

Here is some interesting reading for you about the law courts and conditional probability.

• A formula for justice
  Bayes' theorem is a mathematical equation used in court cases to analyse statistical evidence. But a judge has ruled it can no longer be used. Will it result in more miscarriages of justice?
• Court of Appeal bans Bayesian probability (and Sherlock Holmes)
  In a recent judgement the English Court of Appeal has not only rejected the Sherlock Holmes doctrine shown above, but also denied that probability can be used as an expression of uncertainty for events that have either happened or not.
• Prosecutor's fallacy
  The prosecutor's fallacy is a fallacy of statistical reasoning, typically used by the prosecution to argue for the guilt of a defendant during a criminal trial. 

I mentioned David Aldous's blog post Presenting probability via math puzzles is harmful.
Do you think it has harmed you to see the problem "I have two children one of whom is a boy (born on a Thursday)"?

If you read the Wikipedia article on Simpson's paradox you will find that an alternative name is the Yule-Simpson effect. Udny Yule (a Fellow of St John's College, and a lecturer in statistics at Cambridge) appears to be the first one to have commented on this phenomenon (in 1903).

Our Example 6.7 was artificial, so it would be easy to see what was happening (women predominated amongst independent school applicants and women were more likely to gain admission.) In the Wikipedia article you can read about more practical examples. One of the best-known real-life examples of Simpson's paradox occurred when the University of California, Berkeley was sued for bias against women who had applied for admission to graduate schools there.

I remember once extracting data from Cambridge University Tripos results to show that in each of certain set of carefully selected subjects (including Engineering, English, ...) women were more likely to obtain Firsts than men, but that men were more likely to obtain Firsts when results over all subjects were combined. This sounds paradoxical. Can you figure out how it can happen? Hint: some subjects award greater percentages of Firsts than do others. Some subjects are relatively more popular with women. 

Lecture 5

I have updated the notes to go as far as next Wednesday, Lecture 9.  By the way, if you are printing these notes, you can choose the format you most prefer. I use Acrobat to print with "Fit" which gives me a large typeface on A4. But "Actual size" (to give you wide margins) or "multiple pages" 2 per sheet can look nice also.

I have made some small changes to Example 5.2 in the notes for Lecture 5, and one important correction to a typo in the formula for the Poisson distribution. By the way, I mentioned horse kicks. I was imperfectly recalling that a famous example of data that is Poisson distributed is due to von Bortkiewicz (1898), who studied the numbers of Prussian cavalryman being killed by the kick of a horse in each of 20 years. This is an example with $n$ (number of cavalryman) large, and $p$ (chance a particular cavalryman being killed) small. You can read more about this famous example here.

You might compare the answer to Examples Sheet 1 #10 to a Bonferroni inequality working in the reverse direction, $\geq$.

We saw by examples today that it is possible to have three events $A_1,A_2,A_3$ such that every pair of events is independent, but the three events are not mutually independent. Here is another example. Suppose the sample sample consists of 4 equally likely outcomes $\{w_1,w_2,w_3,w_4\}$. Let

$A_1=\{w_1,w_2\}$,
$A_2=\{w_1,w_3\}$,
$A_3=\{w_1,w_4\}$.

Then $P(A_i\cap A_j)=P(A_i)P(A_j)=1/4$, but $1/4=P(A_1\cap A_2\cap A_3)\neq P(A_1)P(A_2)P(A_3)=1/8$.

I mentioned in the lecture that we might generalize this to a puzzle. Can you find four events $A_1,A_2,A_3,A_4$ such that every subset of either two or three events is a set of mutually independent events, but the four events are not mutually independent?

At the end of the lecture a student told me this nice answer. Let $A_i$ be the probability the $i$th dice roll is odd, $i=1,2,3$ and $A_4$ be the event the sum of the three rolls is even. That works! And it looks like we could generalize this example to give $A_1,\dotsc,A_{k+1}$ such that every  $k$ or fewer events are independent, but the $k+1$ events are not independent.

You can find another answer at the end of this page, but I think the above is nicer.

limsup of a sequence of sets (or events)

When seen for the first time it can be difficult to immediately grasp the meaning of  $\bigcap_{i=1}^\infty \bigcup_{k=i}^\infty A_k$.

I found that when I was a student and encountering this for the first time. This is one reason I included #7 on Examples Sheet 1. My advice is to pick it apart slowly.

$A_k$ is the event that the $k$th toss is a head.

First: The event $\bigcup_{k=i}^\infty A_k$ is true (occurs) if and only if there is at least one head somewhere amongst the tosses $i,i+1,i+2,\dots$ .

Second: When we put $\bigcap_{i=1}^\infty$ in front we create an event that is true if and only if the above is true for all $i$. So how can the result be true? Well, if and only if there are an infinite number of heads. (There might also be an infinite number of tails.)

In general, $\bigcap_{i=1}^\infty \bigcup_{k=i}^\infty E_k$ is occurs (or is true) if and only if an infinite number of the events $E_1,E_2,\dotsc$ occur.

The idea is similar to the "$\limsup$" of a sequence $\{a_1,a_2,\dots\}$, which is defined as

$\lim_{i\to\infty} \sup_{k\geq i} a_k$.

How about $\bigcup_{i=1}^\infty \bigcap_{k=i}^\infty A_k$? That is the event that eventually we see only heads. Now perhaps you can see that

$\bigcup_{i=1}^\infty \bigcap_{k=i}^\infty E_k \subseteq\bigcap_{i=1}^\infty \bigcup_{k=i}^\infty E_k $.

Here is an article on it that might also help.