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Joint Distribution of Changing Mean

  1. Sep 29, 2006 #1
    Let X be a random variable with mean [tex]\mu[/tex] and standard deviation 1.

    Let's add a twist.

    Suppose [tex]\mu[/tex] is randomly distributed about 0 with standard deviation 1.

    At each iteration, we select a new [tex]\mu[/tex] according to its distributuion. This mean is then used in the distribution for X. Then we pick an X according to its distribution.

    My question: What is the resulting joint distribution? Given this joint distribution, I should be able to calculate the mean and standard deviation. Clearly, the mean X will be 0, but what will be the standard deviation of X? It seems that it should, at a minimum, be greater than 1.

    Thanks!
     
    Last edited: Sep 29, 2006
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  3. Sep 29, 2006 #2

    mathman

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    let m=mean

    E(X2|E(X)=m)=1+m2

    E(m2)=1, since std dev(m)=1 and mean=0

    Therefore E(X2)=2 or

    std. dev.(X)=sqrt(2)
     
  4. Sep 29, 2006 #3
    Could you elaborate a bit more? I can confirm that the std deviation is indeed sqrt(2), however, I don't understand where the following formula comes from:

    E(X^2 | E(x) = m) = 1 + E(m^2)

    From the definition,

    [tex]\sigma_x^2 = E(x^2) - E(x)^2 = E(x^2) - m^2[/tex]

    Presumably, I stick your formula into the formula I just wrote above...but I'm still confused where your formula comes from. Also, m (in my original post [tex]\mu[/tex]) is not the same...it is determined by a normal distribution.
     
  5. Sep 29, 2006 #4

    Hurkyl

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    I think the easiest way to do this is to simplify the description -- X is the sum of two normal distributions. (admittedly, it's good to be able to do it different ways, though)
     
  6. Sep 29, 2006 #5
    Interesting, I had wondered if that was okay to do...as the variance of X would then be the sum of the variances of the two normal distributions...and this is, in fact, sqrt(2). Could you explain how these are equivalent pictures?

    In general, I would like to consider a set of distributions

    [tex]\mu_1 \sigma_1[/tex]

    [tex]\mu_2 \sigma_2[/tex]

    [tex]\mu_3 \sigma_3[/tex]

    ...

    where the [tex]\mu_i[/tex] are distributed normally with mean [tex]\mu[/tex] and std deviation [tex]\Delta \mu[/tex]

    where the [tex]\sigma_i[/tex] are distributed normally with mean [tex]\sigma[/tex] and std deviation [tex]\Delta \sigma[/tex]

    For each distribution, we pick x once. What is the expected value of x and what is the standard deviation?
     
    Last edited: Sep 29, 2006
  7. Sep 30, 2006 #6

    Hurkyl

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    X is a normal distribution centered about u.

    X - u is a normal distribution centered about 0.

    X - u is, presumably independent from u. (You never actually specified in your problem that the only dependence of X on u is that u is the mean of the distribution on X, but I assume it was meant)

    X = (X - u) + u
     
  8. Sep 30, 2006 #7

    mathman

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    note: (You wrote: E(X^2 | E(x) = m) = 1 + E(m^2). If you look carefully at what I wrote I had m^2 on the right in that line and then took average over m on both sides to get E(X^2).)

    In your expression for sig2(x), you implicitly defined it for a specific value of m. In the expression I wrote, I just made it explicit and rearranged terms, while using the fact that the variance of x is 1.

    All I assumed about m is that it was random with first and second moments 0 and 1; normal distribution is unnecessary.
     
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