Question

Breaking stuff let {xi} n i=1 be a sample of random variables of size n with the property that xi ∼ n (µ, σ2 ) and cov(xi , xj ) = ρσ for all pairs i, j.

a.solve for the mean and variance of x¯.
b.what happens to the variance of x¯ as n → ∞?
c.why do the clt and lln fail in this situation?
d.using the provided code, simulate s = 1000 values of x¯ for n = 50, 100, 500. in all three cases calculate the mean and variance of the resulting distribution. create histograms for each. in this code, σ = 1 and ρ = .4. how does the variance for each n compare to the formula from part (a)?

Answer 1

With

`\bar X=\displaystyle\sum_(i=1)^n\frac{X_i}n`

we find the mean is

`\mathrm E[\bar X]=\displaystyle\frac1n\sum_(i=1)^n\mathrm E[X_i]=\frac{n\mu}n=\mu`

and the variance is

`\mathrm V[\bar X]=\displaystyle\frac1{n^2}\mathrm V\left[\sum_(i=1)^nX_i\right]`

`\mathrm V[\bar X]=\displaystyle\frac1{n^2}\left(\sum_(i=1)^n\mathrm V[X_i]+2\sum_(1\le i<j\le n)\mathrm{Cov}[X_i,X_j]\right)`

`\mathrm V[\bar X]=\displaystyle\frac1{n^2}\left(n\sigma^2+2\frac{n(n-1)}2\rho\sigma\right)`

`\mathrm V[\bar X]=\frac{\sigma^2}n+\frac{n-1}n\rho\sigma`

Then as
`n\to\infty`, we find that
`\mathrm V[\bar X]\to\rho\sigma`. Simply put, the failure of the CLT/LLN has to do with the fact that the
`X_i` are not independent.