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I am about to compute the variance of the following two variables: $u_n=2.\frac{X_1+X_2+ \ldots +X_n}{n}-1$ and $w_n=max\{X_1,X_2, \ldots ,X_n\}$ which almost surely converge to $m$ I evaluated before.

I have done $E(u_n)=m$ and $E(w_n)=m-\frac{1^n+\ldots+(m-1)^n}{m^n}$

How can I evaluate the variance? through the expectation or $D(u_n)=D(\frac{X_1+X_2+ \ldots +X_n}{n}-1)$ (how to eliminate the brackets?)? I have no idea with that especially $w_n$. Hope any help!

$D(u_n)$ is the variance of $u_n$ i mean.

$(X_i)_{i=1,\ldots,n}$ is an i.i.d sequence that has distribution $1/m$ obtaining value $1,2,\ldots \quad or \quad m$ with probability $1/m$.

Am I right that: $$D(u_n)=D(\frac{X_1+X_2+ \ldots +X_n}{n}-1)= \frac{1}{n^2}D(X_1)=\frac{1}{n^2}[E(X_1^2)-(E(X_1))^2]$$ then apply that $$E(X_1^2)=\frac{1}{m}(1^2+\ldots+m^2)$$

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  • $\begingroup$ The $X_i$ are i.i.d Uniform on some range? $\endgroup$
    – Glen_b
    Aug 19, 2016 at 6:04
  • $\begingroup$ I am still not clear what is m. By the way $var(u_n)=\frac{4}{n} var(x)$ which is obviously because x_i are iid. $w_n$ is a bit more tricky as you need to find the density which is not difficult. $\endgroup$
    – TPArrow
    Aug 19, 2016 at 8:28
  • $\begingroup$ the original is: n of m balls which are numbered from 1 to m are drawn with replacement. $(X_i)_{i=1,\ldots n}$ are the number on the ith selected ball, estimate m ? $\endgroup$
    – Godgog Arsenal
    Aug 20, 2016 at 3:53

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