Evaluate $\sum_{k=1}^\infty k \alpha^k$

Question

I'm struggling with the derivation of this known result. As a prerequisite, consider the following results. Suppose we have time-series data {$x_i$ for $i = 1, \dots, N$} from a stationary process having mean $\mu$ and variance $\sigma^2$ and theoretical acf $\rho(k)$. The sample mean is $\bar{X} = \sum_{i=1}^N \frac{X_i}{N}$. For correlated observations, it can be shown that the sample mean variance is not $\sigma^2/N$ but is given by:

$$ Var(\bar{X}) = \frac{\sigma^2}{N}\left[ 1 + 2 \sum_{k=1}^{N-1}\left( 1 - \frac{k}{N} \right) \rho(k) \right] $$

I'd like to find the variance of the sample mean for a stationary AR(1) process. The book states that $Var(\bar{X})$ for AR(1) reduces to $ Var(\bar{X}) = \frac{\sigma^2}{N}\left( \frac{1+\alpha}{1-\alpha} \right)$. To show this, I start with the autocorrelation function of an AR(1) process with parameter $\alpha$ which is $\rho(k) = \alpha^{|k|}$ where $\alpha \lt 1$. So the above equation becomes

$$ Var(\bar{X}) = \frac{\sigma^2}{N}\left[ 1 + 2 \sum_{k=1}^{N-1}\left( 1 - \frac{k}{N} \right) \alpha^k \right] $$

Then let $N \rightarrow \infty $. The complication then is ultimately in solving $\sum_{k=1}^\infty (1 - \frac{k}{N}) \alpha^{|k|}$. If I split it then:

$$ \sum_{k=1}^\infty \alpha^k = \sum_{k=0}^\infty \alpha^k - 1 = \frac{\alpha}{1-\alpha} $$

Now I need to solve the second part which is where I get stuck. I don't know how to evaluate:

$$ \sum_{k=1}^\infty k \alpha^k $$

Ultimately, I'd like to show that $ Var(\bar{X}) = \frac{\sigma^2}{N}\left( \frac{1+\alpha}{1-\alpha} \right)$ but I believe by solving the above sum which I'm stuck on I'll be able to show this result from the general equation for $Var(\bar{X})$ above.

@KentaS Yes, because of the limits. I've kept the abs but effectively it is just k. — s5s, Mar 31 '21 at 12:48
Then you should be able to evaluate it by taking the derivative of $1/(1-x)=\sum_{n\ge0}x^n$. — Kenta S, Mar 31 '21 at 12:49
Does this answer your question? Sum of a power series $n x^n$ — Hans Lundmark, Mar 31 '21 at 12:55
No one has mentioned that you seem to have taken the limit $N\to\infty$ wrongly. Note that in $$Var(\bar{X}) = \frac{\sigma^2}{N}\left[ 1 + 2 \sum_{k=1}^{N-1}\left( 1 - \frac{k}{N} \right) \alpha^k \right]$$ the quantity $N$ appears in three places. In two of them, you cannot just "replace" them with $\infty$, but at the same time no instance of $N$ can remain after taking $\lim_{N\to\infty}$. — Calvin Khor, Apr 15 '21 at 09:26
the sum is similar to one that appears here https://math.stackexchange.com/questions/3194201/why-does-this-identity-hold-for-fejér-kernels/3414831#3414831 — Calvin Khor, Apr 15 '21 at 09:51

score 3 · Accepted Answer · answered Mar 31 '21 at 12:56

3

Either take the derivative approach as Kenta S mentioned, or do the following:

$$S=\sum_{1}^{\infty} k\alpha^k \\ \alpha S= \sum_1^{\infty} k\alpha^{k+1} = \sum_2^{\infty} (k-1)\alpha^k =\sum_{2}^{\infty} k\alpha^k -\sum_2^{\infty} \alpha^k \\ =(S-\alpha) -\frac{\alpha^2}{1-\alpha} \\ \implies S= \frac{\alpha}{1-\alpha} +\left ( \frac{\alpha}{1-\alpha} \right)^2$$

answered Mar 31 '21 at 12:56

Vishu

14,469

Thanks, what is the method of differentiating a series called? I believe I've forgotten this trick. – s5s Mar 31 '21 at 13:00
@s5s It’s probably just called ‘differentiating a series’. I may be wrong, though. – Vishu Mar 31 '21 at 13:02
@s5s : Do you perhaps search the term "termwise differentation"? – Gottfried Helms Apr 17 '21 at 06:04

score 1 · Answer 2 · answered Mar 31 '21 at 13:38

$$\sum_{k=1}^{\infty} k\alpha^k$$ $$=\sum_{i=1}^{\infty}\sum_{k=i}^{\infty} \alpha^k$$ $$=\sum_{i=1}^{\infty} \alpha^i \sum_{k=0}^{\infty} \alpha^k$$ $$=\sum_{i=1}^{\infty} \alpha^i \frac{1}{1-\alpha}$$ $$=\frac{1}{1-\alpha} \sum_{i=1}^{\infty} \alpha^i $$ $$=\frac{\alpha}{(1-\alpha)^2}$$

Dole · Answer 3 · 2021-03-31T13:23:47.537

0

Comments for the save.

$$\sum_{k=1}^\infty k \alpha^k=\frac{\partial}{\partial \alpha}\alpha^2\sum_{k=0}^\infty \alpha^{k}-\alpha \sum_{k=0}^\infty\alpha^k$$

which is $$=\frac{\partial}{\partial \alpha}\frac{\alpha^2}{1-\alpha}-\frac{\alpha}{1-\alpha}$$ $$=(\frac{\alpha}{1-\alpha})^2+\frac{\alpha}{1-\alpha}$$

edited Mar 31 '21 at 13:23

answered Mar 31 '21 at 13:10

Dole

2,653

s5s · Answer 4 · 2021-04-15T10:59:24.170

0

So I answer it this way. This answer assumes $|a| < 1$.

$$ \sum_{k=1}^\infty k a^k = \sum_{k=0}^\infty k a^k = \sum_{k=0}^\infty a \frac{d}{d a} [a^k] = a \sum_{k=0}^\infty \frac{d}{da}[a^k] = a \frac{d}{da}[\sum_{k=0}^\infty a^k] = a \frac{d}{da} [\frac{1}{1-a}] = \cdots $$

edited Apr 15 '21 at 10:59

answered Apr 15 '21 at 09:00

s5s

507
4
14

Evaluate $\sum_{k=1}^\infty k \alpha^k$

4 Answers4