Conditional variance calculations (Crypto-currency reward offered)

[ElMacho]

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]

?

 

[MarkFL]

...I'll give 100 DogeCoin (or equivalent) to whomever...

Hello and welcome to MHB! :D

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[ElMacho]

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.

 

[I like Serena]

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...