by Let’s say we have a stationary process. That does not guarantee that it is also ergodic. The long-run time average of a single output function of the stationary process may not converge to the expectation of the corresponding variables — and so the long-run time average may not equal the probabilistic (expectational) average. Say we have two coins, where coin A has a probability of 1/2 of coming up heads, and coin B has a probability of 1/4 of coming up heads. We pick either of these coins with a probability of 1/2 and then toss the chosen coin over and over again. Now let H1, H2, … be either one or zero as the coin comes up heads or tales. This process is obviously stationary, but the time averages — [H1 + … + Hn]/n — converges to 1/2 if coin A is chosen, and 1/4 if coin B
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Let’s say we have a stationary process. That does not guarantee that it is also ergodic. The long-run time average of a single output function of the stationary process may not converge to the expectation of the corresponding variables — and so the long-run time average may not equal the probabilistic (expectational) average. 
Say we have two coins, where coin A has a probability of 1/2 of coming up heads, and coin B has a probability of 1/4 of coming up heads. We pick either of these coins with a probability of 1/2 and then toss the chosen coin over and over again. Now let H1, H2, … be either one or zero as the coin comes up heads or tales. This process is obviously stationary, but the time averages — [H1 + … + Hn]/n — converges to 1/2 if coin A is chosen, and 1/4 if coin B is chosen. Both these time averages have a probability of 1/2 and so their expectational average is 1/2 x 1/2 + 1/2 x 1/4 = 3/8, which obviously is not equal to 1/2 or 1/4. The time averages depend on which coin you happen to choose, while the probabilistic (expectational) average is calculated for the whole “system” consisting of both coin A and coin B.