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
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.
In order to understand the 'core of problem' there is an example of discrete probability function. Let's suppose to have a discrete random variable x that can assume a countable set of values $\displaystyle x_{n}$ and You know its probability function $\displaystyle p_{n}=P \{x=x_{n}\} $ and that we want to find the mean value of the random variable y=f(x) ...
$\displaystyle E \{y| x=x_{k}\}= \mu $ (1)
If f(*) is a single value function then...
$\displaystyle P\{y=f(x_{n})|x=x_{k}\} =\begin{cases}1 &\text{if }n=k\\ 0 &\text{if }n \ne k\end{cases}$ (2)
... so that y has mean value $\displaystyle \mu= f(x_{k})$ and variance 0. But what does it happen if f(*) is a multivalued function, like for example $\displaystyle f(x)= \sqrt{x}$?...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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
That is quite obvious that if a 'random' variable y has probability...
$\displaystyle P\{y=f(x_{n})|x=x_{k}\} =\begin{cases}1 &\text{if }n=k\\ 0 &\text{if }n \ne k\end{cases}$ (1)
... it means that $\displaystyle f(x_{k})$ is the only possible value for y, so that is $\displaystyle \mu_{y}=x_{k}$ and $\displaystyle \sigma_{y}^{2}=0$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$