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
Topics:
Lars Pålsson Syll considers the following as important: Economics
This could be interesting, too:
Lars Pålsson Syll writes Andreas Cervenka och den svenska bostadsbubblan
Lars Pålsson Syll writes Debunking the balanced budget superstition
Lars Pålsson Syll writes How inequality causes financial crises
Lars Pålsson Syll writes Income inequality and the saving glut of the rich
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.