hi there, not sure on how to approach this problem. I'm a bit lost.
how can I show that a rv y which can be expressed a function of rv x, say g(x), is true if the variance of y given x is 0.
so basically show
y = g(x) iff var(ylx) =0
if y=g(x) then the pmf of y|x is:
y=g(x) with probability 1
everything else with probability 0.
Now find the expected value and variance using the normal definitions.
EDIT: ignore this post, im doing the logic the wrong way round.
If f(*) is a single value function then...
... so that y has mean value and variance 0. But what does it happen if f(*) is a multivalued function, like for example ?...
hi, thanks for the explanation. So just to make sure I am understanding correctly.
say for example f(x) = x. Thus for each unique value for x, there is only one value for f(x).
but if I had say f(x) = ± sqrt x, then I would have 2 values of f(x) for each x.
so then my µ would be E(YlX=xk) = sigma y * P(X=xk n Y = y)/P(X=xk)
what I don't understand is how this will make var(ylx) = 0?
in order for var(ylx) = 0 I must have E(Y^2lX) = E(YlX)^2