Hello everyone,
I am studying the lecture as the title, in the example of game 4, why var(x4) = sum((x-mu)^2 * P(x))?
I think there is a problem here because var(x) = E[X^2] - E[X ]^2 = sum(X^2 * P(X)) - mu^2.
Anyway, I could not find a way to type the correct math format, I am sorry for the inconvenience.
Hi!
In fact, both formulas are true. You can see it because
E[X] = \sum xP(x) and var(X) = E[(X-\mu)^2]= \sum(X - \mu)^2\cdot P(x)
On the other hand, we know, using expected value properties that var(X) = E[X^2] - E[X]^2.
Was that clear?
Cheers,
Lucas
Dear Mr Lucas,
Could you please guide me the reason to get P(b) = 1/3 and P(c) = 1/6 ?
Thank you.