Friday, 14 February 2014

Lecture 13

The tower property of conditional expectation (also  called the law of iterated expectations) is akin to the law of total probability that we met in Section 6.3. Another name is law of total expectation. A special case is

E[X]=i:P(Ai)>0E[XAi]P(Ai),

where A1,,An are events giving a partition of the sample space Ω. Compare the above to

E[X]=E[E[XY]]=yE[XY=y]P(Y=y).

Sometimes we might write E[E[XY]] as EY[E[XY]], just to remind ourselves that the outer expectation is being applied to the variable Y. Of course it is implicit in all this that E[X] exists.

Remember that P(A|B) is only defined for P(B)>0.  If you want to see something that illustrates this in a cute way, then read this nice description of Borel's paradox. The paradox occurs because of a mistaken step in conditioning on an event of probability 0, which can be reached as a limit of a sequence of events in different ways.

Comments (2)

Loading... Logging you in...
  • Logged in as
Oliver Prior's avatar

Oliver Prior · 580 weeks ago

I'm slightly uncertain in the calculation of the variance of S[N] at the end of Example 13.4, I get it to have, in the second term, [E(X1)]^2 rather than E(X1^2) as a result of the Chain Rule applied when taking the second derivative which is then evaluated at 1 to give E(S[N](S[N]-1)). What am I doing wrong?
1 reply · active 580 weeks ago
Well spotted. As I wrote that in lectures today I was thinking that did not look quite right. It should indeed be

var(S_N)=E[N]var(X) plus (EX)^2 var(N)

This is Examples Sheet 3, #8. Note that with the above formula the variance is dimensionally correct: imagine that X_i are measured in metres, and N is dimensionless. The answer should have units of meters^2.

I have corrected the notes.

p.s. For some odd reason I can't write a plus sign in a comment. I wonder why that is?

Post a new comment

Comments by