Expectation - May 20

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AJ
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Re: Expectation - May 20

AJ
How are we defining sample space here...?

Can the sample space just have 9 elements....? all Xs

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Re: Expectation - May 20

Amit Goyal
Administrator
In reply to this post by anon_econ
Vasudha,

Its better to write the sample space over those 9 terms because we know the probabilities associated with those 9 terms. Each term is true with half probability and the terms are independent. Hence there are 2^9 = 512 elements in the sample space. An example is TTTTTFFFF i.e. terms 1 to 5 are true and 6 to 9 are false.
Now consider propositions 1. Let E1 be the event that proposition 1 is true.
Give an example of an element from E1.

Define I[E1]: S --> R as
I[E1](x) = 1 if x ∈ E1
           = 0 if x ∉ E1
Like wise define I[E2], ....., I[E9].
Interpret N = I[E1] + I[E2] + ... + I[E9]
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Re: Expectation - May 20

anon_econ
An example of an element of E1 is x1^~x3^x7 where '^' denotes 'and'.
0=<N<=9 and N=n means that n out of 9 propositions are true. Can't think beyond this.. <smiley image="smiley_unhappy.gif"/>
AJ
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Re: Expectation - May 20

AJ
In reply to this post by Amit Goyal
ummmm...
Example from E1 is
TFFFFFFFF

and N is event that all propositions are true.

Please say this one is correct!!!
AJ
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Re: Expectation - May 20

AJ
No. of propositions is 7 .. right..??

Not 9... so, this should be till E7

I should probably leave and see this discussion kal.. :/
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Re: Expectation - May 20

anon_econ
Oh sorry! it should be 7!
AJ
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Re: Expectation - May 20

AJ
But in sir's reply... its till 9 ....
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Re: Expectation - May 20

Amit Goyal
Administrator
In reply to this post by anon_econ
You can write the sample space as true false values for 7 propositions also and you will have 2^7 elements. But as you pointed out it will be harder (not impossible) and very inconvenient to assign the probability function to the events of the sample space. [Events can be assigned zero probability. So the count will remain 2^7 if you write your sample space this way. In the example you said that proposition 1, 2 and 7 cannot be false at the same time. So the event that all 1, 2 and  7 are false is a zero probability event ].
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Re: Expectation - May 20

Amit Goyal
Administrator
This post was updated on .
In reply to this post by anon_econ
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Re: Expectation - May 20

anon_econ
ok. got it :)
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Re: Expectation - May 20

Amit Goyal
Administrator
In reply to this post by AJ
Let me ask more specific questions:
True/ False
(i) TFFFFFFFF ∈ E1
(ii) FFFFFFFFF ∈ E1
(iii) FFFFFFTFF ∈ E1

I hope you have figured out that N is a random variable. Also answer these:
What is N (TFFFFFFFF)?
What is N (FFFFFFFFF)?
What is N (FFFFFFTFF)?


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Re: Expectation - May 20

anon_econ
true, true, false.
7,7,6.
AJ
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Re: Expectation - May 20

AJ
In reply to this post by Amit Goyal
(i) true
(ii) true
(iii) false

N=9
N=9
N=8
AJ
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Re: Expectation - May 20

AJ
sorry typo.... 7,7, 6

as in all propositions are true for first two.. and for third 1st is false.. so N=6
AJ
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Re: Expectation - May 20

AJ
So, now we have to find E(N) ...??
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Re: Expectation - May 20

anon_econ
In reply to this post by AJ
so we have to find E(N) which will be P(E1)+P(E2)+...P(E7).
AJ
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Re: Expectation - May 20

AJ
ya... it shud be sum of these probabilities....

and P(E1)= 7/8 .. by any chance...?????
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Re: Expectation - May 20

anon_econ
In reply to this post by anon_econ
which is 7*7/8 i think..6.125.
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Re: Expectation - May 20

anon_econ
lol we are posting at the same time. AJ if this is correct then u got the answer right in the first go..if sir says it's right u've got to tell me how u did it :D
AJ
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Re: Expectation - May 20

AJ
Ya, we are posting at same time.. pretty much excited that world has started to make sense once again...
AND

I got that answer.. by "stupidity"...
I used binomial .. n=7
p=7/8

and I got that 7/8 by .. (1/2+1/2+1/2-1/4-1/4-1/4+1/8)

so .. expectation=np= 7*7/8

:p
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