## Re: relationship % conditional probability and conditional expectation

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• Thank you for not only addressing my question, but also clearing many of the other (mis)conceptions about elementary probability theory. One of the
Message 1 of 5 , Apr 30, 2004
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Thank you for not only addressing my question, but also clearing many
of the other (mis)conceptions about elementary probability theory.
One of the difficulties I've found on the subject is notations are
highly overloaded with meanings. I'm in the middle of transition
migrating my previous elementary probability theory knowledge to
measure-theoric one and I think your reply clarifies much of my cloud.

> Rmk2: a conditional expectation is always a
> conditional expectation "knowing some sigma-
> algebra G". For example E[X|Y] is in fact E[X|s(Y)]
> where s(Y) is the sigma algebra on W generated by Y
> i.e. the set {Y^{-1}(B) , B in B(R)}. E[X|Y,Z] is
> in fact E[X|s(X,Y)] where s(X,Y) is the sigma
> algebra on W generated by X and Y. E[X|A] is in
> fact E[X|s(1_A)], where s(1_A)={0,W,A,A^c} etc...

Is 1_A the indicator function for event A defined as:
1_A = 1 if A occurs, and 1_A=0 otherwise?
So the second parameter to the expectation notation should be of type
sigma algebra, not of random variable? Right?

> > If
> > P{Y1+...+Yk<tk, k=1...n | Y1+...+Yn-1, tn=u, Y1+...+Yn=y}
> > = 1 - (Y1+...+Yn-1)/tn (1)
>
> For this equation to make sense, I believe you should have
> written.
>
> P{Y1+...+Yk<tk, k=1...n | Y1+...+Yn-1, tn=u, Y1+...+Yn=y}
> = 1 - (Y1+...+Yn-1)/u (1*)
>
> with "u" instead of "tn". So I assume (1*) is the correct
(1) is the exact quote from the textbook. By the way, why using tn
at the RHS doesn't make sense? Remember tn is just an RV.

Thank you again.
Casey
• ... Hi Casey, you re very welcome. ... Strictly speaking yes, the second argument to the *conditional* expectation should always be a sigma-algebra, e.g.
Message 2 of 5 , May 1, 2004
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> Thank you for not only addressing my question,
> but also clearing many of the other (mis)conceptions

Hi Casey, you re very welcome.

> Is 1_A the indicator function for event A defined as:
> 1_A = 1 if A occurs, and 1_A=0 otherwise?
> So the second parameter to the expectation notation
> should be of type sigma algebra, not of random variable?
> Right?

Strictly speaking yes, the "second argument" to the
*conditional* expectation should always be a
sigma-algebra, e.g. E[X|G]. However, as very often
in mathematics, notations get a little bit sloppy,
when there is no ambiguity, when it looks nicer or
when it is more convenient. So if Y is a random variable,
and s(Y) is the sigma-algebra generated by Y, *no one*
will ever write E[X|s(Y)], but E[X|Y] instead. It is
more convenient to write E[X|Y] and there is no risk
in doing so, since everyone knows it means exactly
E[X|s(Y)]... Similarly, if A is some event (an element of F)
everyone will write E[X|A] instead of E[X|1_A] or E[X|s(1_A)]
or E[X|{0,W,A,A^c}]... Also, if B is another event,
everyone will write P(B|A) instead of E[1_B|A]... Note
that strictly speaking P(B|A) is a *random variable* and
not a number. In all elementary textbooks, you will
see the formula:

P(B|A)=P(B/\A)/P(A) (*)

where in fact, strictly speaking, P(B|A) is the random
variable:

P(B|A)=(P(B/\A)/P(A))*1_A + (P(B/\A^c)/P(A^c))*1_A^c

For all outcome w in A, (i.e. such that 1_A(w)=1), we have:

P(B|A)(w)=P(B/\A)/P(A)

I believe equation (*) is damaging because it is imcompatible
with the true definition of E[X|G] (which means that beginners
get very confused I would think)

> at the RHS doesn't make sense? Remember tn is just an RV.

Yes, tn is a random variable. If you write something like
E[X|tn], then this is also a random variable, which is measurable
with respect to s(tn), i.e. which can be expressed as a
(measurable) function of tn, i.e. E[X|tn]=g(tn) for some
measurable function g:R->R. So an equation like E[X|tn]=g(tn)
makes perfect sense. Now, by definition, E[X|tn=u] is just
another notation for g(u). More precisely, E[X|tn=u] is
supposed to refer to the "value at u" of "the" (not quite
unique) function g such that if you take the composition
of g and tn, i.e. (g o tn) you obtain E[X|tn], or in other
words the function g such that E[X|tn]=g(tn). Hence an equation
like E[X|tn=u]=g(u) makes perfect sense. But writing something
like E[X|tn=u]=g(tn) makes no sense in my opinion... Neither would
E[X|tn]=g(u) be meaningful...

Noel.
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