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## Re: Measure and Integration

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• ... Hi, you have f:R- [0,+oo[ measurable such that int f(u)du
Message 1 of 8 , Oct 1, 2003
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> Let F(x) = Integral [-infinite to x] f(u)du be an absolutely
> continous distribution function. Prove that corresponding
> probability measure m is given by:

Hi,

you have f:R->[0,+oo[ measurable such that int f(u)du<+oo. You define
F(x)=int_oo^x f(u)du and you call m the measure on R such that:
m(]-oo,x])=F(x) for all x in R (the fact that such measure exists is
assumed to be granted) . You want to prove that for all B in B(R) you
have m(B)=int_B f(u)du. Define n(B)=int_B f(u)du. It is not difficult
to show that n(.) is a finite measure on R, and you want to prove
that m=n. Define:

D={B in B(R), n(B)=m(B)}

and:

C={]-oo,x]: x in R}

Then C is a subset of D which is closed under finite intersection. It
is not difficult to show that D is a dynkin sytem on R (see
definition on Tutorial 1 of probability.net, it is important to use
the fact that the mesures m and n are finite). From the dynkin system
theorem (see Tutorial 1), the sigma algebra s(C) on R generated by C
is a subset of D. However, s(C) is nothing but B(R) (exercise). So B
(R) is a subset of D and finally D=B(R). This proves that m=n, and we
are done.

Regards. Noel.
• Hi.... I m lost in some parts of the proof: * Define n(B)=int_B f(u)du [Is it Ok to start assuming this ?] * How can I prove that D is a Dynkin system ? I
Message 2 of 8 , Oct 1, 2003
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Hi.... I'm lost in some parts of the proof:

* Define n(B)=int_B f(u)du [Is it Ok to start assuming this ?]

* How can I prove that D is a Dynkin system ?

I understand the argument when you try to see if the two sets are
equal.... but why is so important that the measures have to be
finite ?

Thanks !!!

Alejandro

--- In probability@yahoogroups.com, "Noel Vaillant" <vaillant@p...>
wrote:
> > Let F(x) = Integral [-infinite to x] f(u)du be an absolutely
> > continous distribution function. Prove that corresponding
> > probability measure m is given by:
>
> Hi,
>
> you have f:R->[0,+oo[ measurable such that int f(u)du<+oo. You
define
> F(x)=int_oo^x f(u)du and you call m the measure on R such that:
> m(]-oo,x])=F(x) for all x in R (the fact that such measure exists
is
> assumed to be granted) . You want to prove that for all B in B(R)
you
> have m(B)=int_B f(u)du. Define n(B)=int_B f(u)du. It is not
difficult
> to show that n(.) is a finite measure on R, and you want to prove
> that m=n. Define:
>
> D={B in B(R), n(B)=m(B)}
>
> and:
>
> C={]-oo,x]: x in R}
>
> Then C is a subset of D which is closed under finite intersection.
It
> is not difficult to show that D is a dynkin sytem on R (see
> definition on Tutorial 1 of probability.net, it is important to use
> the fact that the mesures m and n are finite). From the dynkin
system
> theorem (see Tutorial 1), the sigma algebra s(C) on R generated by
C
> is a subset of D. However, s(C) is nothing but B(R) (exercise). So B
> (R) is a subset of D and finally D=B(R). This proves that m=n, and
we
> are done.
>
> Regards. Noel.
• ... I would not say assuming but just defining . You then have to prove that n is indeed a measure (i.e. that it is countably additive). (see theorem 21 in
Message 3 of 8 , Oct 2, 2003
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--- In probability@yahoogroups.com, "azr124" <azr124@y...> wrote:
> Hi.... I'm lost in some parts of the proof:
>
> * Define n(B)=int_B f(u)du [Is it Ok to start assuming this ?]

I would not say "assuming" but just "defining". You then have to
prove that n is indeed a measure (i.e. that it is countably additive).
(see theorem 21 in Tutorial 5 and exercises preceding it if you are
stuck on the proof)

>
> * How can I prove that D is a Dynkin system ?

If you read the definition in Tutorial 1, I don't think you ll find
it too difficult: you need to show that:

1. R lies in D
2. if A,B are in D and A<B (inclusion) then B\A (set difference) lies
in D
3. if (An) is increasing sequence (i.e. An<A(n+1)) of elements of D,
and A=\/_n An, then A is also an element of D.

> I understand the argument when you try to see if the two sets are
> equal.... but why is so important that the measures have to be
> finite ?

This is important when proving 2. of the dynkin system:
suppose A,B are in D and A<B then:

m(B)=m(B\A)+m(A) (A and B\A are disjoint)

and since m is finite, you can write m(B\A)=m(B)-m(A) (if m wasnt
finite you'd potentially be writing +oo-(+oo) which makes no sense)
Likewise, since n is finite, n(B\A)=n(B)-n(A). Now since A,B are in D
we have m(A)=n(A) and m(B)=n(B). So m(B\A)=n(B\A) and we have proved
that B\A lies in D....

Regards. Noel.
• Hello: I m really stuck on the below as far as setting up the problem. Since I don t know how much is bet in total I don t see how can I work out the below
Message 4 of 8 , Oct 5, 2003
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Hello:

I'm really stuck on the below as far as setting up the
problem. Since I don't know how much is bet in total
I don't see how can I work out the below problem? Is
there an algebraic way to determine this?

--------------------------------------------------
I have the following 3 horse race with the following
quoted odds by the bookmaker:

horse #1: 6-1 for
horse #2: 7-2 against
horse #3: 50-1 against

So:
\$6 bet on #1 would return \$7 if he won, 0 otherwise.
\$2 bet on #2 would return \$9 if he won, 0 otherwise.
\$1 bet on #3 would return \$51 if he won, 0 otherwise.

We can interpret this as a one period market with 3
assets: \$1 bet on #1, \$1 bet on #2, and \$1 bet on #3.

What are the associated "risk-neutral" probabilities?
What is the bookmaker taking on every bet?

__________________________________
Do you Yahoo!?
The New Yahoo! Shopping - with improved product search
http://shopping.yahoo.com
• Hi Group, I sent this out before, but haven t heard back from anyone. Can anyone help me with this? Thanks, M.W. __________________________________ Do you
Message 5 of 8 , Oct 9, 2003
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Hi Group,

I sent this out before, but haven't heard back from
anyone. Can anyone help me with this?

Thanks,
M.W.

__________________________________
Do you Yahoo!?
The New Yahoo! Shopping - with improved product search
http://shopping.yahoo.com

[Non-text portions of this message have been removed]
• Can anyone help me regarding the below? Thanks. -M.W. ... __________________________________ Do you Yahoo!? The New Yahoo! Shopping - with improved product
Message 6 of 8 , Oct 9, 2003
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Can anyone help me regarding the below?
Thanks.

-M.W.

>
> Hello:
>
> I'm really stuck on the below as far as setting up
> the
> problem. Since I don't know how much is bet in
> total
> I don't see how can I work out the below problem?
> Is
> there an algebraic way to determine this?
>
>
> --------------------------------------------------
> I have the following 3 horse race with the following
> quoted odds by the bookmaker:
>
> horse #1: 6-1 for
> horse #2: 7-2 against
> horse #3: 50-1 against
>
> So:
> \$6 bet on #1 would return \$7 if he won, 0 otherwise.
> \$2 bet on #2 would return \$9 if he won, 0 otherwise.
> \$1 bet on #3 would return \$51 if he won, 0
> otherwise.
>
> We can interpret this as a one period market with 3
> assets: \$1 bet on #1, \$1 bet on #2, and \$1 bet on
> #3.
>
> What are the associated "risk-neutral"
> probabilities?
> What is the bookmaker taking on every bet?
>
>
> __________________________________
> Do you Yahoo!?
> The New Yahoo! Shopping - with improved product
> search
> http://shopping.yahoo.com
>
>

__________________________________
Do you Yahoo!?
The New Yahoo! Shopping - with improved product search
http://shopping.yahoo.com
• ... Hi M.W. you can view the economy as having one time period and three possible states of the world in the future. State 1: horse 1 wins. State 2: horse 2
Message 7 of 8 , Oct 9, 2003
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> > \$6 bet on #1 would return \$7 if he won, 0 otherwise.
> > \$2 bet on #2 would return \$9 if he won, 0 otherwise.
> > \$1 bet on #3 would return \$51 if he won, 0
> > otherwise.

Hi M.W.

you can view the economy as having one time period and three possible
states of the world in the future. State 1: horse 1 wins. State 2:
horse 2 wins etc... A triplet (a,b,c) can represent a financial asset:

you receive a in state 1
you receive b in state 2
you receive c in state 3

Betting on horse 1, is just like buying the asset (7,0,0), the price
of which is 6. The price of (0,9,0) is 2. The price of (0,0,51) is 1.
It follows that the price of (1,0,0) is 6/7, that of (0,1,0) is 2/9
and that of (0,0,1) is 1/51. These (I believe) is what we call the
risk neutral probabilities. Indeed the price of (a,b,c) is (by
linearity):

Price(a,b,c)=a*6/7+b*2/9+c*1/51

So formally, 6/7, 2/9 and 1/51 look like probabilities and Price
(a,b,c) is the expected payoff with respect to these probabilitities.
Now in practice, the bookmaker will sell (1,0,0), (0,1,0) and (0,0,1)
to different people, so as to be as unexposed as possible: so the
bookmaker is trying to sell (1,1,1). Their liability is 1 whatever
happens. However, they have received:

6/7+2/9+1/51=1.09897 ...

So the bookmaker is making a profit of about 10% on its turnover...

Hope this helps.

Regards. Noel
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