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.
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.
