Good evening.
I got random variables $r_1, r_2 ... r_N$ i.i.d normal with mean $\mu$ and variance $\sigma^2$
I also got $\lambda$ - decay factor.
I got 2 estimators:
$r_N^{*} = \frac{1 - \lambda}{1 - \lambda^N}\sum_{i=1}^{N} \lambda^{i-1} r_i$
and
$v_N = \frac{1 - \lambda^2}{2\lambda(1 - \lambda^{N-1})}\sum_{i=1}^{N} \lambda^{i-1} (r_i - r_N^*)^2$
I need to prove that they are unbiased estimators of mean and variance respectively.
For the mean it is simple. By taking expectation of $r_N$ and using linearity I arrive to this:
$E[r_N^{*}] = \frac{1 - \lambda}{1 - \lambda^N}\sum_{i=1}^{N} \lambda^{i-1} E[r_i] = \mu\frac{1 - \lambda}{1 - \lambda^N}\sum_{i=1}^{N} \lambda^{i-1} = \mu$ using the $(1 - \lambda)^N$ expansion
Now, for the variance for some reason I'm getting stuck:
$E[v_N] = \frac{1 - \lambda^2}{2\lambda(1 - \lambda^{N-1})}\sum_{i=1}^{N} \lambda^{i-1} E[(r_i - r_N^{*})^2]$
The common trick used almost everywhere is to add and subtract the mean, so:
$E[(r_i - r_N^{*})^2] = E[((r_i - \mu) - (r_N^{*} - \mu))^2] = E[(r_i - \mu)^2] + E[(r_N^{*} - \mu)^2] - 2E[(r_i - \mu)(r_N^{*} - \mu)] = 2\sigma - 2E[(r_i - \mu)(r_N^{*} - \mu)]$
How do I proceed? Directly substituing the value for $r_N$ lead me to some disastrous math which I'd rather assume is wrong.