Timeline for How can I show that a random walk is not covariance stationary?
Current License: CC BY-SA 3.0
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| May 24, 2014 at 2:37 | comment | added | Dilip Sarwate | @whuber Thanks for the upvote. It occurs to me that even the requirement for identical distribution of the $X_i$ is not needed to prove that the random walk is not covariance stationary. Independence and nonzero variance (or at least one $X_i$ having nonzero variance) might suffice. But at that point we are somewhat far from the usual meaning of random walk. | |
| May 24, 2014 at 2:34 | history | edited | Dilip Sarwate | CC BY-SA 3.0 |
added 289 characters in body
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| May 24, 2014 at 2:16 | comment | added | whuber♦ | This is clean and clear (+1). For the demonstration to be logically complete you need to show (or at least assert the obvious fact) that the variance of $Y_t$ actually does change with $t$. | |
| May 23, 2014 at 20:14 | history | edited | Dilip Sarwate | CC BY-SA 3.0 |
improved and clarified exposition
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| May 23, 2014 at 17:26 | comment | added | Dilip Sarwate | @Charlie $\operatorname{cov}(Y_t+W,Y_t)=\operatorname{cov}(Y_t,Y_t) + \operatorname{cov}(W,Y_t)$ does not require independence: it is a property of covariance that holds generally. $W$ and $Y_t$ are independent because they are sums of two mutually exclusive sets of independent random variables. Generally, if $X$ and $Y$ are independent, so are $g(X)$ and $h(Y)$ for any (measurable) functions $g(\cdot)$ and $h(\cdot)$. If you do not already know these concepts. you are way in over your head when you look at covariance stationarity and random walks. | |
| May 23, 2014 at 15:03 | comment | added | Charlie | Thank you Prof... Where can I find the proof or at least some explanation about the fact that $\operatorname{cov}(Y_t+W,Y_t)=\operatorname{cov}(Y_t,Y_t) + \operatorname{cov}(W,Y_t)$ if $W$ and $Y$ are independent? And why do you say that they are independent? | |
| May 23, 2014 at 13:37 | history | answered | Dilip Sarwate | CC BY-SA 3.0 |