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Find mean and variance of a random vector

  1. Oct 27, 2014 #1
    1. The problem statement, all variables and given/known data
    The components of a random vector ##\mathbf{X} = [X_1, X_2, \ldots, X_N]^{\intercal}## all have the same mean ##E_X[X]## and the same variance ##var(X)##. The "sample mean" random variable
    $$
    \bar{X} = \frac{1}{N}\sum_{i = 1}^NX_i
    $$
    is formed. If the ##X_i##'s are independent, find the mean and variance of ##\hat{X}##. What happens to the variance as ##N\to\infty##? Does this tell you anything about the PMF of ##\bar{X}## as ##N\to\infty##?

    2. Relevant equations


    3. The attempt at a solution
    Since the ##X_i##'s are independent,
    $$
    \hat{X} = \frac{1}{N}\sum_{i = 1}^NE_{X_i}[X_i]
    $$
    but the \(X_i\)'s have the same means, thus, all are the same so
    \begin{align*}
    &= \frac{1}{N}NE[X]\\
    &= E[X]
    \end{align*}
    Additionally, independence implies that ##cov(X_i, X_j) = 0## for ##i\neq j## so the variance is
    \begin{align*}
    var\Big(\sum var(X_i)\Big)
    &= \sum_{i = 1}^Nvar(X_i)\\
    &= \sum_{i = 1}^N\big(E[X_i^2] - E^2[X_i]\big)\\
    &= \sum_{i = 1}^N\big(E[X_i^2]\big) - NE^2[X]
    \end{align*}
    Can the variance be simplified any more? I am not sure what happens to the variance as N tends to infinity either.
     
  2. jcsd
  3. Oct 27, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You are complicating things unnecessarily. If ##\mu = E\, X_i## and ##\sigma^2 = \text{Var}\, X_i## ##\forall i##, then
    [tex] E\, \sum_{i=1}^n X_i = n \mu, \; \text{Var} \, \sum_{i=1}^n X_i = n \sigma^2\\
    \text{so}\\
    E\, \bar{X} = \mu, \; \text{Var}\, \bar{X} = \frac{\sigma^2}{n}.[/tex]
    Leaving the variance in this last form is the usual way these things are done, and yields the most insight into the problems's structure.
     
  4. Oct 28, 2014 #3

    Ok so that means the variance goes to zero then but what does that mean for the PMF?
     
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