Uncertainty - June 20

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Uncertainty - June 20

Amit Goyal
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Re: Uncertainty - June 20

Vasudha
Could u give a small hint?

On Tue, Jun 19, 2012 at 9:23 PM, Amit Goyal [via Discussion forum] <[hidden email]> wrote:
June20__Uncertainty.png


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Re: Uncertainty - June 20

Vasudha
E(Y+Z)=E(Y)+E(E(Z|Y))=E(Y)+0. so the expected values of both r equal. every risk averse individual would prefer Y to Y+Z coz they wont take additional risk if the expected value isnt increasing. risk neutral people would be indifferent. and risk loving people would prefer Y+Z. this means only statement 2 is correct.
i know this is wrong. in fact i think the first step is wrong.
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Re: Uncertainty - June 20

Amit Goyal
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Sure. For understanding, lets do the problem for the finite state space or sample space i.e. S = {s(1), s(2), ..., s(n)}. Y takes only two values y(1) and y(2) i.e. Y: S --> {y(1), y(2)}. And its given that Z is another random variable Z: S --> R. It is also given that Pr(Y = y(1)) = pi(1) and Pr(Y = y(2)) = pi(2). And we know that E(Z|Y) = 0.
We can think of Y and Y + Z as two lotteries or gambles. The question asks you that if an individual has to choose between the two which one will he pick depending on his attitude towards risk?
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Re: Uncertainty - June 20

Vasudha
Hmm..so is what I did completely wrong?
r
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Re: Uncertainty - June 20

r
In reply to this post by Amit Goyal
(iii) and (iv)
AJ
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Re: Uncertainty - June 20

AJ
In reply to this post by Amit Goyal
I am getting same answers as vasudha..
expected returns of both are same.
and risk in Y+Z is more..

So, just 2nd is true.
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Re: Uncertainty - June 20

Amit Goyal
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This post was updated on .
In reply to this post by Vasudha
Hi Vasudha and AJ,

Your conclusion is correct and I could also see that you have right reasons in mind. The only thing that is missing is the convincing expressions. Here is how you prove it:
Given that the person is risk averse, his utility function for money will be a concave function.
Eu(Y + Z)
= E(Eu(Y + Z)|Y)
= pi(1) Eu((Y + Z)|Y = y(1)) + pi(2) Eu((Y + Z)|Y = y(2))
= pi(1) Eu((y(1) + Z)|Y = y(1)) + pi(2) Eu((y(2) + Z)|Y = y(2))
≤ pi(1) u(E(y(1) + Z)|Y = y(1)) + pi(2) u(E(y(2) + Z)|Y = y(2))                       [By concavity of u(.)]
= pi(1) u(y(1)) + pi(2) u(y(2))                                        [E(Z|Y = y(1)) = 0 and E(Z|Y = y(2)) = 0]
= Eu(Y)

This automatically disproves (iii), (iv) and (v). For (i), just construct an example. As you said a risk loving person with appropriate choice of state space, pi, lotteries and utilities would do.
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Re: Uncertainty - June 20

Vasudha
I see :-)

On Wed, Jun 20, 2012 at 5:53 AM, Amit Goyal [via Discussion forum] <[hidden email]> wrote:
Hi Vasudha,

Your conclusion is correct and I could also see that you have right reasons is mind. The only thing that is missing is the convincing expressions. Here is how you prove it:
Given that the person is risk averse, his utility function for money will be a concave function.
Eu(Y + Z)
= E(Eu(Y + Z)|Y)
= pi(1) E(Eu(Y + Z)|Y = y(1)) + pi(2) E(Eu(Y + Z)|Y = y(2))
= pi(1) E(Eu(y(1) + Z)|Y = y(1)) + pi(2) E(Eu(y(2) + Z)|Y = y(2))
≤ pi(1) E(u(E(y(1) + Z)|Y = y(1))) + pi(2) E(u(E(y(2) + Z)|Y = y(2)))  [By concavity of u(.)]
= pi(1) E(u(y(1))) + pi(2) E(u(y(2)))
= Eu(Y)

This automatically disproves (iii), (iv) and (v). For (i), just construct an example. As you said a risk loving person with appropriate choice of state space, pi, lotteries and utilities would do.


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