Sample mean

Now let x_t be a time series made up of identically distributed random numbers: m_x and $\sigma_x$ do not depend on time. Let us also suppose that they are independently chosen; this means in particular that for any different times t and s ($t\neq s$):

$$\E(x_t x_s) \eq \E(x_t) \E(x_s)$$ (13)

Suppose we have a sample of n points of x_t and are trying to determine the value of m_x. We could make an estimate $\hat m_x$ of the mean m_x with the formula  

$$\hat m_x \eq {1 \over n} \sum^n_{t = 1} x_t$$ (14)

A somewhat more elaborate method of estimating the mean would be to take a weighted average. Let w_t define a set of weights normalized so that  

$$\sum w_t \eq 1$$ (15)

With these weights, the more elaborate estimate $\hat m$ of the mean is  

$$\hat m_x \eq \sum w_t \, x_t$$ (16)

Actually (14) is just a special case of (16); in (14) the weights are w_t = 1/n.

Further, the weights could be convolved on the random time series, to compute local averages of this time series, thus smoothing it. The weights are simply a filter response where the filter coefficients happen to be positive and cluster together. Figure 6 shows an example: a random walk function with itself smoothed locally.

 

walk
Figure 6 Random walk and itself smoothed (and shifted downward).