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Math Help - conditional variance problem

  1. #1
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    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
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    MHF Contributor chisigma's Avatar
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    Re: conditional variance problem

    Quote Originally Posted by Kuma View Post
    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
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    Re: conditional variance problem

    Quote Originally Posted by chisigma View Post
    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.
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  4. #4
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    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.
    Last edited by SpringFan25; October 1st 2011 at 07:14 AM.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: conditional variance problem

    Quote Originally Posted by Kuma View Post
    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
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    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 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
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  7. #7
    MHF Contributor chisigma's Avatar
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    Re: conditional variance problem

    Quote Originally Posted by Kuma View Post
    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 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
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