Given the following stochastic process:

$$Y_t = t X$$

where: $E(X)=0$ and $VAR(X) = 1 $.

Is it correct to obtain the autocovariance in the following way?

$$COV(Y_t, Y_s) = E [Y_tY_s] ~~~~~~~~~~~~~~~~~~~\text{since, $E[X_t] = 0$.}$$

$$E [Y_tY_s]=E[tXsX] = tsE[X^2] = tsVAR(X) = ts$$

Others say me that the correct way is:

$$E [Y_tY_s]=E[tXsX] = tE[X]sE[X] = 0$$

Which one is correct?


1 Answer 1


Your way is correct. \begin{eqnarray} Cov(Y_t,Y_s) &=& E[Y_t\cdot Y_s]-E[Y_t]\cdot E[Y_s] \\ &=& E[tX\cdot sX]-E[tX]\cdot E[sX] \\ &=& tsE[X^2]-ts(E[X])^2 \\ &=& tsE[X^2] \\ &=& ts(Var[X]-E[X]^2) \\ &=& ts(1-0^2) \\ &=& ts \end{eqnarray}

  • $\begingroup$ Perhaps something to note too is that this process is not stationary, and an auto-covariance function that depends only on the lag (i.e. $t-s$) does not exist. $\endgroup$ Commented Jan 26, 2017 at 2:43

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