# Limit of Sample ACF for X_t = cos(wt)

• SpringPhysics
In summary, the problem discusses showing that the sample autocorrelation function at lag h for n observations of xt = cos(wt) converges to cos(wh) as n tends to infinity. This is done by substituting the given expression for the sample autocovariance function at lag h and simplifying using formulas for summing integers and squares. The cosine sums are then evaluated using Euler's formula to convert them to exponentials, leading to the desired convergence.
SpringPhysics

## Homework Statement

Show that the sample acf at lag h for n observations of xt = cos(wt) converges to cos(wh) as n tends to infinity.

## Homework Equations

The sample acf at lag h is defined as the sample autocovariance function at lag h divided by the sample autocovariance function at lag 0, where the sample autocovariance function at lag h is defined as:

acvf(h) = $\sum$t=1n-|h|(xt+|h| - $\overline{x}$)(xt - $\overline{x}$)

Here, $\overline{x}$ is the sample mean of the n observations.

## The Attempt at a Solution

A previous question asked to show something similar for xt = a + bt, so what I did there was just substitute everything in (everything works out nicely because there are formulas for summing integers and squares from 1 to n).

I attempted to do the same:

xt - $\overline{x}$ = cos(wt) - 1/n $\sum$s=1n cos(ws)

xt+h - $\overline{x}$ = cos(wt + wh) - 1/n $\sum$s=1n cos(ws)

So then the numerator of acvf(h) becomes:
$\sum$t=1n-|h|(cos(wt)cos(wt+wh) - 1/n cos(wt+wh)$\sum$s=1n cos(ws) - 1/n cos(wt) $\sum$s=1n cos(ws) + 1/n2 $\sum$s=1n $\sum$r=1n cos(ws)cos(wr))

And the denominator of acvf(h) becomes:
$\sum$t=1ncos(wt)cos(wt) - 1/n $\sum$s=1n $\sum$r=1n cos(ws)cos(wr)

And at this point, I have no idea how to evaluate the cosine sums to find the limit as n goes to infinity of the fraction. I attempted to convert the cosines to exponentials by Euler's formula but I ended up with more terms that were even messier.

Any help would be appreciated, even just some starting point or confirmation that some particular method works, and I'll try to grind through it again. Thanks.

UPDATE: I solved the problem using exponentials (a bit tedious but it worked). This thread can be closed now.

## What is the limit of the sample ACF for X_t = cos(wt)?

The limit of the sample ACF for X_t = cos(wt) is 0. This means that as the lag increases, the correlation between the values of X_t decreases and approaches 0.

## How is the sample ACF calculated for X_t = cos(wt)?

The sample ACF for X_t = cos(wt) is calculated by taking the correlation between X_t and X_t+h, where h is the lag. This is repeated for different values of h to create the ACF plot.

## What does a sample ACF plot for X_t = cos(wt) look like?

A sample ACF plot for X_t = cos(wt) will have a peak at lag 0 with decreasing values as the lag increases. It will also have a sinusoidal shape due to the periodic nature of the cosine function.

## What does the limit of the sample ACF tell us about X_t = cos(wt)?

The limit of the sample ACF tells us that there is no correlation between the values of X_t at different lags. This means that the values of X_t are not dependent on each other and are independent.

## Can the limit of the sample ACF for X_t = cos(wt) be negative?

No, the limit of the sample ACF for X_t = cos(wt) cannot be negative. The correlation coefficient can range from -1 to 1, but since the sample ACF is calculated using the cosine function, the values will always be positive and approach 0 as the lag increases.

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