协方差函数的绘图应该从 0-shift 开始

Question

下面是老师给我的问题，

Generate a sequence of N = 1000 independent observations of random variable with distribution: (c) Exponential with parameter λ = 1 , by inversion method.

Present graphically obtained sequences(except for those generated in point e) i.e. e.g. (a) i. plot in the coordinates (no. obs., value of the obs) ii. plot in the coordinates (obs no n, obs. no n + i) for i = 1, 2, 3. iii. plot so called covariance function for some values. i.e. and averages:

我写了下面的代码，

(*****************************************************************)
(*Task 01(c) and 02(a)*)
(*****************************************************************)
n = 1000;

taskC = Table[-Log[RandomReal[]], {n}];

ListPlot[taskC, AxesLabel->{"No. obs", "value of the obs"}]

i = 1;
ListPlot[Table[
    {taskC[[k]], taskC[[k+i]]}, 
    {k, 1, n-i,1}],
AxesLabel->{"obs.no.n", "obs.no.n+1"}]

i++;
ListPlot[Table[
    {taskC[[k]], taskC[[k+i]]},
    {k, 1, n-i,1}],
AxesLabel-> {"obs.no.n", "obs.no.n+2"}]

i++;
ListPlot[Table[
    {taskC[[k]], taskC[[k+i]]},
    {k,1,n-i,1}],
AxesLabel->{"obs.no.n", "obs.no.n+3"}]

avg = (1/n)*Sum[taskC[[i]], {i,n}];

ListPlot[Table[1/(n-tau) * Sum[(taskC[[i]]-avg)*(taskC[[i+tau]] - avg), n], {tau, 1,100}], 
    Joined->True, 
    AxesLabel->"Covariance Function"]

他有评论，

The plots of co-variance functions should start from 0-shift. Note that for larger than 0 shifts you are estimating co-variance between independent observations which is zero, while for 0 shift you are estimating variance of observation which is large. Thus the contrast between these two cases is a clear indication that the observations are uncorrelated.

我做错了什么？

如何更正我的代码？

Answer 1

零偏移意味着计算 tau = 0 的协方差，也就是方差。

Labeled[ListPlot[Table[{tau,
    1/(n - tau)*Sum[(taskC[[i]] - avg)*(taskC[[i + tau]] - avg), {i, n - tau}]},
   {tau, 0, 5}], Filling -> Axis, FillingStyle -> Thick, PlotRange -> All,
  Frame -> True, PlotRangePadding -> 0.2, AspectRatio -> 1],
 {"Covariance Function K(n)", "n"}, {{Top, Left}, Bottom}]

Variance[taskC]

0.93484

Covariance[taskC, taskC]

0.93484

(* n = 1 *)
Covariance[Most[taskC], Rest[taskC]]

0.00926913

协方差函数的绘图应该从 0-shift 开始

The plots of co-variance functions should start from 0-shift

random

wolfram-mathematica

stochastic-process