# Math Help - conditional variance problem

1. ## conditional 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

2. ## Re: conditional variance problem

Originally Posted by Kuma
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
The statement is equivalent so say that g(x) must be a single value function...

Kind regards

$\chi$ $\sigma$

3. ## Re: conditional variance problem

Originally Posted by chisigma
The statement is equivalent so say that g(x) must be a single value function...

Kind regards

$\chi$ $\sigma$
I'm still a bit confused. Can you elaborate a bit more?
So do you mean for every Y value there must only be one X value?

I'm not sure how I can apply this to the question.

4. ## Re: 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.

5. ## Re: conditional variance problem

Originally Posted by Kuma
I'm still a bit confused. Can you elaborate a bit more?
So do you mean for every Y value there must only be one X value?

I'm not sure how I can apply this to the question.
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 $x_{n}$ and You know its probability function $p_{n}=P \{x=x_{n}\}$ and that we want to find the mean value of the random variable y=f(x) ...

$E \{y| x=x_{k}\}= \mu$ (1)

If f(*) is a single value function then...

$P\{y=f(x_{n})|x=x_{k}\} =\begin{cases}1 &\text{if }n=k\\ 0 &\text{if }n \ne k\end{cases}$ (2)

... so that y has mean value $\mu= f(x_{k})$ and variance 0. But what does it happen if f(*) is a multivalued function, like for example $f(x)= \sqrt{x}$?...

Kind regards

$\chi$ $\sigma$

6. ## Re: 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

7. ## Re: conditional variance problem

Originally Posted by Kuma
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...

$P\{y=f(x_{n})|x=x_{k}\} =\begin{cases}1 &\text{if }n=k\\ 0 &\text{if }n \ne k\end{cases}$ (1)

... it means that $f(x_{k})$ is the only possible value for y, so that is $\mu_{y}=x_{k}$ and $\sigma_{y}^{2}=0$...

Kind regards

$\chi$ $\sigma$