## Monday, 24 February 2014

### Lecture 17

An important special case of Theorem 17.1 is when $Y=aX$, for some constant $a$.
The p.d.f. of $Y$ is then $f_Y(y)=(1/a)f_X(y/a)$.

There are many ways to define transitive ordering relations between random variables. Now you know about $EX\geq EY$ and  $EX\geq_{\text{st}} EY$. Such relationships are important in applications, such as when $X$ and $Y$ are lifetimes of components and we are interested ordering their reliabilities. (Reliability engineering is concerned with determining the ability of a system or component to function under some stated conditions for a specified period of time.)

There are other ordering relations, such as "hazard rate ordering", and "convex ordering". These are described within the Wikipedia article on stochastic ordering. An important order relation in statistics is "likelihood ratio ordering". If $X$ and $Y$ have p.d.fs $f$ and $g$ such that $f(t)/g(t)$ increases with $t$ then $X\geq_{\text{lr}}Y$. It is a stronger ordering than the two we have met, since $X\geq_{\text{lr}}Y\implies X\geq_{\text{st}}Y$.

The treatment of the inspection paradox is this lecture derived from a paper by Sheldon Ross, The inspection paradox, Probability in the Engineering and Informational Sciences, 17, 2003, 47–51. Of course the use of the word "paradox" is somewhat in appropriate, since it is not hard to understand why $X_J\geq_{\text{st}}X_1$.

There are other such inspection paradoxes. Suppose $X_1,X_2,\dotsc$ are i.i.d. $U[0,1]$ and $N$ is the least integer such that $X_1+\cdots+X_N\geq 1$. Then conditional on this,

$X_N\geq_{\text{st}}X_{N-1}\geq_{\text{st}}\cdots \geq_{\text{st}}X_1$.

I decided not to discuss it in the lecture, but at the end of page 69 of today's notes is the puzzle, "Do girls have more brothers than boys do?" The "proof" that the answer is "yes" is false, though initially it looks almost convincing. What is wrong?

What do you think is the truth of the matter? Let $G$ and $B$ be the mean number of brothers possessed by a typical girl and typical boy, respectively.

If every family were to consists of some number of boys and just one girl then $G > B$.

If every family were to consists of only boys or only girls then $G < B$.

What are the minimal conditions under which $G=B$? Must you assume that each child is independently a boy or girl with probability $1/2$?