tag:blogger.com,1999:blog-5418994588621362010.post2235854441564397140..comments2023-11-15T10:08:56.796-08:00Comments on the blog formerly known as The Statistical Mechanic: mean meanWolfganghttp://www.blogger.com/profile/07086991199438418163noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-5418994588621362010.post-79787080834041277632010-08-20T14:05:07.058-07:002010-08-20T14:05:07.058-07:00in your second example, E[y(t)] grows linearly in ...in your second example, E[y(t)] grows linearly in t, so I guess it is intuitive that the sample mean doesn't necessarily converge. BTW, the process Y(t) with Y(0) \sim N(0,1) and Y(t) = Y(0), t >= 1 is a trivial example of a process with sample mean <> expectation valueMattnoreply@blogger.comtag:blogger.com,1999:blog-5418994588621362010.post-46090253460394870942010-03-22T09:30:41.152-07:002010-03-22T09:30:41.152-07:00Very nice. I think there are many such processes, ...Very nice. <br>I think there are many such processes, but I wanted to find one where the sequence of sample means is bounded, while in this example it obviously bounces between positive and negative values without bounds.wolfganghttp://tsm2.blogspot.comnoreply@blogger.comtag:blogger.com,1999:blog-5418994588621362010.post-51846100754463963862010-03-22T08:25:01.525-07:002010-03-22T08:25:01.525-07:00Hi, I believe you can achieve the same effect much...Hi, I believe you can achieve the same effect much easier.<br>Use a process for y,N where N(t+1) = 10*N(t) and N(1) = 1 , y(t) = N(t)*rand and rand is either +1 or -1<br>Write down the first three or four terms to see what I am talking about.<br><br>This gives you E[y] = 0 but the sample mean will not converge.Anonymousnoreply@blogger.com