Mr. F
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 Mar19 revised what is the intuition behind Delta method? added 961 characters in body Mar19 answered what is the intuition behind Delta method? Mar19 comment what is the intuition behind Delta method? $f(x) = x(1-x)$, so $f(\mu) = \mu(1-\mu)$. Just follow the exact procedure as in this example, but substituting in the properties of your distribution and your function. Mar19 comment what is the intuition behind Delta method? You're very close to getting it. First, just plug $\mu$ directly into the formula for $f(x) = x(1-x)$. Think about what the function applied to the mean value is (it's not $k\theta$). Secondly, you are correct to think of the CLT for the approximate distribution of $\bar{X}_{n}$.. but why would it have mean $0$? It won't converge to a standard normal unless you subtract the mean and divide by the standard deviation... so what if you don't do these operations? Mar19 comment What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? I've added some Python code that verifies this with eigenvectors and Monte Carlo, and added a bit of discussion. Hopefully it clears things up now. Mar19 revised What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? added 3421 characters in body Mar19 revised What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? added 3421 characters in body Mar19 revised What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? edited body Mar19 comment What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? Ahh, also a friend just pointed out that I had an error. The last row should really be $[5/6, 1/6, 0, 0, 0, 0, 0]$, because you don't automatically go back to start if the next play of the game began with a roll of a $1$. That should fix issues with the eigenvector of my transition matrix. I've edited the matrix to reflect that. Mar19 comment What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? The idea is that by making the transition matrix entry in the bottom left equal to 1.0, we make the chain as a whole recurrent. In the long run, the chain only sits in the 'accept' state for 1 "time unit" before flipping back to the start. Thus, in the long run, the probability of seeing the chain in that accept state should be equal to the reciprocal of the expected number of rolls to get there. So then you need to use either the system of equations method or the eigenvalue/eigenvector method with the transition matrix to get that long-term probability. Mar18 revised What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? edited body Mar18 revised What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? added 207 characters in body Mar18 comment What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? Ah yes, you're right about that. If you roll a 1 at any stage, you go back to the 1 state. I'll modify the probabilities. Mar18 revised What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? edited body Mar18 comment What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? I wanted the top row and the first column to give headers about what is in those rows/columns. It makes it easier to interpret than just putting a matrix. Mar18 comment What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? Any help getting mathjax to understand tabulars? It's not math mode, so how do I signal non-math TeX? Mar18 answered What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6? Mar18 comment Conditional independence Ok, but I'm heading to bed for tonight. I'll pick it up tomorrow. Mar18 comment Conditional independence Ah, yes, you're right. I was mis-reading it.. but it just results in a typo. Where I had written $1/2$ before, it should be $P(D_{j})/2$ because the LTP means the terms add up to $P(D_{j})$, not to 1 as I mistakenly claimed. But I think this still could be fruitful in terms of yielding a system of equations in the $D_{j}$, then applying the given conditions as constraints. Mar18 revised In search of memorable example of “(Pearson-)uncorrelated $\not\Rightarrow$ independent” added 990 characters in body