- Thread starter vaisman
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- The expected values (aka, means) are each/all the sum of [joint probability * value]. For example, in the case of variable (G), its mean = (20% * 3.0) + (60% * 9.0) + (20% * 12.0) = 8.4%. In case it helps, below I represent the assumption situation in a
*probability matrix;*this assumption set is "simpler" in the sense that non-diagonals in the matrix are zeros.

- Variance applies, σ^2(X) = E(X^2) - [E(X)]^2. For example σ^2(B) = E(B^2) - [E(B)]^2 = 50.60 - 7.0^2 = 1.60. I hope that clarifies!

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#3

- The expected values (aka, means) are each/all the sum of [joint probability * value]. For example, in the case of variable (G), its mean = (20% * 3.0) + (60% * 9.0) + (20% * 12.0) = 8.4%. In case it helps, below I represent the assumption situation in a
*probability matrix;*this assumption set is "simpler" in the sense that non-diagonals in the matrix are zeros.

- Variance applies, σ^2(X) = E(X^2) - [E(X)]^2. For example σ^2(B) = E(B^2) - [E(B)]^2 = 50.60 - 7.0^2 = 1.60. I hope that clarifies!

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#4

Hi David. I have another question- just making sure I am getting it correct. This is Miller Chapter 6 Bayesian Analysis Page 101

In the attached screenshot, I see that the grid of the previous question (page 100) is being used for the question given above.

Just want to confirm if this is a typo or there is something that I am missing?

In the attached screenshot, I see that the grid of the previous question (page 100) is being used for the question given above.

Just want to confirm if this is a typo or there is something that I am missing?

And then I notice a second problem, I don't know how this happened but the problem statement omits the following emphasized sentence: "Stars are rare. Of a given pool of managers, only 16% turn out to be stars.

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#7

Thanks for your amazing real time feedback David ! What I heard about you in many online forums about being prompt and very helpful is cent percent true !

Apologies but I am here to trouble you one more time :

Bayesian Analysis Page 108 (Calculation of P(6B) using this function:

I can't quite understand this calculation as I haven't studied this previously. May I request you to put a full solution or show the full method of solving this?

I interpreted this as (0.75)^6 x (0.25)^4= 0.0007.

Maybe I am missing something but don't know where to search for this. All your help is greatly appreciated and sorry for asking too many questions David !

Apologies but I am here to trouble you one more time :

Bayesian Analysis Page 108 (Calculation of P(6B) using this function:

I can't quite understand this calculation as I haven't studied this previously. May I request you to put a full solution or show the full method of solving this?

I interpreted this as (0.75)^6 x (0.25)^4= 0.0007.

Maybe I am missing something but don't know where to search for this. All your help is greatly appreciated and sorry for asking too many questions David !

Sure, frankly I try to respond quickly especially if I think we have an error (as above). Plus, we consider it a contribution (to help us fix mistakes or otherwise improve). In regard to the follow-up: We are given the assumption that "outperformers beat the market 75% of the time" which statistically can be written as a

- Critical (lookup) values will be provided; e.g., GARP's recent practice paper includes a normal Z lookup. But if student's t lookup values are needed, they will be provided
**However**, you would be expected to know to reject a large computed t (e.g., t = 7 or t = 12), as such a value represents too many standard deviations (away from the hypothesized mean) to accept the null. We generally need the lookup for the zone around t = 2.0; i.e. if the t = 2.0 +/- x, it depends on the d.f. But if t = 5.0 or 15.0, we don't need a table to reject. I hope that's helpful!

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