David Pilling |
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Time Seriesstarted 13th October 2016
The autocorrelation at lag k is defined as: ` rho_k=(E[(z_t-mu)(z_(t+k)-mu)])/sigma_z^2 `
where ` z_t ` is the time series, μ is its mean value, ` σ_z^2 ` is its variance, and E denotes the expected value. For a stationary process ` E[z_t-mu]=0`
` sigma_z^2=E[(z_t-mu)^2] `
for all `t`. By the definition of `sigma` `rho_0` equals 1. If ` z_t` is a random variable, then `rho_k` equals 0 because the values of `z_t` and `z_(t+1)` are independent: ` rho_k=(E[(z_t-mu)]E[(z_(t+k)-mu)])/sigma_z^2 `
and ` E[(z_t-mu)]` is zero. ` E[(z_t-mu)(z_s-mu)]=0` for `t!=s`
If ` z_t=r_t-r_(t-1)+mu ` with ` r_t ` a random variable with mean 0 then ` rho_1=(E[(r_t-r_(t-1))(r_(t+1)-r_t)])/sigma_z^2 `
Expanding the argument of E and removing the terms which are zero by the independence assumption ` rho_1=(-E[r_t^2])/sigma_z^2 `
` sigma_z^2=E[(r_t-r_(t-1))^2] `
` sigma_z^2=2E[r_t^2] `
` rho_1=-1/2 `
The best intuitive way of looking at this I can think of is to imagine three data values, the first and last will average out to their mean positions, the correlation between the difference between the middle point and the first and the difference between the middle point and last will always consist of a value less than the mean multiplied by a value greater than the mean. Put it another way average values with a correlation of 1 with those with a correlation of 0 get a correlation of 0.5. Credits References |
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