MHB Conditional variance calculations (Crypto-currency reward offered)

AI Thread Summary
The discussion centers on a mathematical problem involving conditional variance calculations in the context of cryptocurrency rewards. A user is seeking clarification on how to derive a specific formula for V(C) given certain conditions about V(A|B), V(A), and the independence of A and C. Another participant provides a derivation that leads to a different result, suggesting a potential typo in the original problem statement. The conversation highlights the complexities of variance calculations in probability theory, particularly in financial contexts like cryptocurrencies. The thread emphasizes the need for accurate problem formulation to achieve correct mathematical conclusions.
ElMacho
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I'm reading a journal article that implies the following but I can't see how it is done. I'll give 100 DogeCoin (or equivalent) to whomever can explain this in full.

Given that
V(A|B) = s
V(A) = r*s + w
B = A + C

and A & C are independent
so V(B) = V(A) + V(C) & V(C) = V(B) - V(A)

Then how can it be shown that

V(C) = [s*(r*s + w)] / [(r-1)s+w]

?
 
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ElMacho said:
...I'll give 100 DogeCoin (or equivalent) to whomever...

Hello and welcome to MHB! :D

We are a free math help site, so it is not our policy as MHB Math Helpers in general to personally expect or accept payment for help given on our site. However, if you feel like you want to make a genuine http://mathhelpboards.com/misc.php?do=donate to MHB (in USD) for whatever reason, then we certainly would not discourage that. (Sun)
 
MarkFL said:
Hello and welcome to MHB! :D

We are a free math help site, so it is not our policy as MHB Math Helpers in general to personally expect or accept payment for help given on our site. However, if you feel like you want to make a genuine http://mathhelpboards.com/misc.php?do=donate to MHB (in USD) for whatever reason, then we certainly would not discourage that. (Sun)

Crypto-currencies can be transferred much more easily than USD. They can easily be transferred for cash online.
 
ElMacho said:
I'm reading a journal article that implies the following but I can't see how it is done. I'll give 100 DogeCoin (or equivalent) to whomever can explain this in full.

Given that
V(A|B) = s
V(A) = r*s + w
B = A + C

and A & C are independent
so V(B) = V(A) + V(C) & V(C) = V(B) - V(A)

Then how can it be shown that

V(C) = [s*(r*s + w)] / [(r-1)s+w]

?

Welcome to MHB, ElMacho!

By definition we have that:
$$V(A|B) = \frac{V(AB)}{V(B)} = \frac{V(A(A+C))}{V(A+C)}$$
Since $A(A+C)=A$ and $A,C$ independent, we get that:
$$V(A|B)= \frac{V(A)}{V(A)+V(C)} = \frac{rs+w}{rs+w+V(C)} = s$$
Therefore:
$$V(C)=(rs+w)\frac{1-s}s$$
This is different from the expected result, so I suspect there is some kind of typo in your problem statement...
 
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