# Autocorrelation function from PDF?

1. Mar 24, 2013

### iVenky

What is the exact prodecure for finding out the auto correlation function Rxx(τ) for a given pdf?
Is it possible at all to find out the auto correlation function from the pdf? If not then what is given usually when you find out the auto correlation function Rxx(τ)?

Thanks

2. Mar 24, 2013

### MisterX

The autocorrelation is applied to a stochastic process, which is a family of random variables. A pdf might describe a single random variable. To find the autocorrelation, you would need the joint pdf that relates the random variables.

Some terms that might be worth learning are "stationary process" and "wide-sense stationary". You describe an auto-correlation function Rxx(τ), but in general the autocorrelation will be Rxx(t1, t2). It is only written Rxx(τ) if the processes is a wide-sense stationary process. This is because for a wide-sense stationary process, the autocorrelation only depends on the difference τ between the two times.

With only single pdf for X that was not a joint pdf, you would only be able to find Rxx(0), which is for zero [time] offset.

3. Mar 24, 2013

### iVenky

Ya if it is a "strict sense stationary process" then can we find out Rxx(τ) using the pdf?

4. Mar 24, 2013

### MisterX

You should remember how to find expectation values of functions continuous random variables.
$E[g(X)] = \int _{-\infty}^{\infty} g(x)p_{X}(x)dx$

If you have a joint PDF for two variables X and Y, it is similar, except the integral has to cover all possibilities for X and Y.

$E[g(X, Y)] = \int _{-\infty}^{\infty}\int _{-\infty}^{\infty} g(x, y)p_{XY}(x,y)dxdy$

For example if you wanted to find the auto-covariance of a wide sense stationary stochastic process you'd be finding

$E\left[\left(X_t - E[X_t]\right)\left(X_{t + \tau} - E[X_{t + \tau}]\right)\right]$

For such a process you should have a joint pdf that depends on tau. $p_{XX}(x_1, x_2, \tau)$. This gives the joint PDF for two variables from the process that are seperated by τ. You should not integrate over tau; it does not correspond to one of the random variables.

It's also useful to know

$E\left[\left(X - E[X]\right)\left(Y - E[Y]\right)\right] = E[XY - E[X]Y - E[Y]X + E[Y]E[X] ] = E[XY] - E[X]E[Y] - E[X]E[Y] + E[X]E[Y]$
$= E[XY] - E[X]E[Y]$

So

$E\left[\left(X_t - E[X_t]\right)\left(X_{t + \tau} - E[X_{t + \tau}]\right)\right] = E[X_t X_{t + \tau}] - E[X_t]E[X_{t + \tau}]$

The autocorrelation is the autocovariance divided by the standard deviations of both variables.
$R{xx}(t_1, t_2) = \frac{E\left[\left(X_{t1} - E[X_{t1}]\right)\left(X_{t2} - E[X_{t2}]\right)\right] }{\sigma_{t1} \sigma_{t2}}$

In the problem you are attempting to solve, the standard deviations $\sigma_{t1}$ and $\sigma_{t2}$ might be equal.

Last edited: Mar 24, 2013
5. Mar 24, 2013

### iVenky

So we need to have the joint pdf to find out the Autocorrelation, right?

6. Mar 24, 2013

### rbj

yes. and the assumption that this random process is ergodic. then you can turn any time-average into a probabilistic average.