1,641 reputation
718
bio website
location
age
visits member for 2 years, 9 months
seen yesterday

?- love(math) is unrequited. true.


Mar
19
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.
Mar
19
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.
Mar
18
revised What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6?
edited body
Mar
18
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
Mar
18
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.
Mar
18
revised What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6?
edited body
Mar
18
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.
Mar
18
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?
Mar
18
answered What is the expected number of dice one needs to roll to get any monotonically increasing series of 1 to 6?
Mar
18
comment Conditional independence
Ok, but I'm heading to bed for tonight. I'll pick it up tomorrow.
Mar
18
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.
Mar
18
revised In search of memorable example of “(Pearson-)uncorrelated $\not\Rightarrow$ independent”
added 990 characters in body
Mar
18
comment Conditional independence
Because the whole sample space $\Omega_{i} = H_{i}\cup\bar{H_{i}}$ by definition in that case, and because $H_{i}$ is disjoint from its negation... again, unless you're using 'negation' in some non-obvious way, in which case my interpretation is off.
Mar
18
answered In search of memorable example of “(Pearson-)uncorrelated $\not\Rightarrow$ independent”
Mar
18
comment Conditional independence
If I am understanding the notation (which maybe I am not), then $\bar{H_{i}}$ is the negation of $H_{i}$, basically like the complement if you want to think of $H_{i}$ as a set/event. Then the law of total probability gives my statement. That may not be what the symbol $\bar{H_{i}}$ is actually meant for though... Does it explain that in any more detail?
Mar
18
awarded  Commentator
Mar
18
comment Conditional independence
If I am understanding the notation correctly, then in order for those fractions to be unity, we must have $P(D_{j}|H_{i}) = P(D_{j}|\bar{H_{i}}) = 1/2$, because $P(D_{j}|H_{i}) + P(D_{j}|\bar{H_{i}}) = 1$. So given that it's this specific, I'm guessing you can derive a system of equations for determining the different probabilities by assuming each fraction equals some value $f_{ij}$, possibly not unity. Then that system probably will only have non-trivial solutions under the stated result. At least, this is the line of thinking I'd look down first.
Mar
17
comment Computing the derivative of a quadratic form and matrix chain rule
If it makes you feel better, most of the work I do is Bayesian computational statistics, so.. functional analysis and Markov chain sampling theory. But I have to pay the bills too... :)
Mar
17
comment Computing the derivative of a quadratic form and matrix chain rule
Yes, I am also working with wage data to predict yogurt brand purchase decisions based on wage data. The wage data comes from a well-known paper, Abowd and Card (1989) "On the Covariance Structure of Earnings and Hours Changes," Econometrica, 57 (2), 441-445. This GMM part is about estimating a parameter for an autoregressive model on the earnings.
Mar
17
awarded  Student