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

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- Sep 30th 2011, 03:52 PMKumaconditional variance problem
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 - Sep 30th 2011, 07:48 PMchisigmaRe: conditional variance problem
- Oct 1st 2011, 01:44 AMKumaRe: conditional variance problem
- Oct 1st 2011, 06:06 AMSpringFan25Re: conditional variance problem
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.** - Oct 1st 2011, 06:16 AMchisigmaRe: conditional variance problem
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

$\displaystyle \chi$ $\displaystyle \sigma$ - Oct 1st 2011, 11:46 AMKumaRe: conditional variance problem
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 - Oct 1st 2011, 07:49 PMchisigmaRe: conditional variance problem
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

$\displaystyle \chi$ $\displaystyle \sigma$