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edit: I completely thought I was answering a different topic. Please disregard the quoted, but I will leave it for completeness. > I'm not quite sure what you

edit: I completely thought I was answering a different topic. Please disregard the quoted, but I will leave it for completeness. > I'm not quite sure what you mean by this question. > What do you mean by "not sampling at each step"? The idea isn't to characterize the entirety of the sampling function along its support, but to draw a sequence of random variables from it.

murbard2 12y ago
Yes but that is not what you do. You calculate the expected proportion of each atom at each step, but: 1) this does not tell you what the distribution is. Decay is a random process... what is the probability that you have more than 90% of a left at step 1000? 2) this only works because you're taking a limit over an infinite number of atoms, which takes the variance to 0. If you have a discrete system, things become hairier. Or take the following system for instance: There are 100 cells, each of which has the same size. At every time step, a little bit of food falls on the cells and is eaten by a single cell. A cell gets the food with a probability proportional to its weight, it then grows by one unit. If you use expectations, everyone grows at the same rate. If you simulate, you'll see that the results follow a Pareto distribution.
- JHonaker 12y ago
  Whoops, I thought I was answering a question about my rejection sampling article. My mistake. I'll address this though. 1) No it doesn't. I don't make any distributional claims for these calculations. You're right. I haven't addressed questions like P(>90%|t=1000). 2) Actually no, it doesn't take the variance to 0. For any given t, the Markov approximation is equal to the expected value of the given atom at time t. There is a variance for every atom count at every time point. I didn't address calculating variance at all. We're speaking in expected values here, the interpretation of 54.8 atoms at time t is the same as the average american household having 2.5 children. Something I addressed in a comment when I posted this on reddit was that the Markov chain approach is the theoretical mean. If you were to actually observe a decay process like this in the real world, it would essentially be one possible realization of the simulated example. As far as the Pareto distribution is concerned, I didn't know that actually. That's really interesting.
  - murbard2 12y ago
    Ah got it, you simulate individually per atom, I thought you were describing the entire population. Indeed, you can always use that vector as the parameter to a multinomial distribution and recover the real distribution. Regarding the Pareto distribution, random growth model are probably the reason why power laws are everywhere.
- omalleyt 12y ago
  The cell example is very cool. It's not a Markov chain though. The probability that a cell grows from size 2 to size 3 (in probabilistic notation, P(2->3|2) ), for instance, would be 2/(sum of all cell sizes), and the sum of all cell sizes is growing by 1 each period (assuming the unit of food given per period is 1, etc. etc.)
  - murbard2 12y ago
    It's a markov chain if your state is the vector representing the size of each cell

murbard2 12y ago

Yes but that is not what you do. You calculate the expected proportion of each atom at each step, but: 1) this does not tell you what the distribution is. Decay is a random process... what is the probability that you have more than 90% of a left at step 1000? 2) this only works because you're taking a limit over an infinite number of atoms, which takes the variance to 0. If you have a discrete system, things become hairier. Or take the following system for instance: There are 100 cells, each of which has the same size. At every time step, a little bit of food falls on the cells and is eaten by a single cell. A cell gets the food with a probability proportional to its weight, it then grows by one unit. If you use expectations, everyone grows at the same rate. If you simulate, you'll see that the results follow a Pareto distribution.

JHonaker 12y ago
Whoops, I thought I was answering a question about my rejection sampling article. My mistake. I'll address this though. 1) No it doesn't. I don't make any distributional claims for these calculations. You're right. I haven't addressed questions like P(>90%|t=1000). 2) Actually no, it doesn't take the variance to 0. For any given t, the Markov approximation is equal to the expected value of the given atom at time t. There is a variance for every atom count at every time point. I didn't address calculating variance at all. We're speaking in expected values here, the interpretation of 54.8 atoms at time t is the same as the average american household having 2.5 children. Something I addressed in a comment when I posted this on reddit was that the Markov chain approach is the theoretical mean. If you were to actually observe a decay process like this in the real world, it would essentially be one possible realization of the simulated example. As far as the Pareto distribution is concerned, I didn't know that actually. That's really interesting.
- murbard2 12y ago
  Ah got it, you simulate individually per atom, I thought you were describing the entire population. Indeed, you can always use that vector as the parameter to a multinomial distribution and recover the real distribution. Regarding the Pareto distribution, random growth model are probably the reason why power laws are everywhere.
omalleyt 12y ago
The cell example is very cool. It's not a Markov chain though. The probability that a cell grows from size 2 to size 3 (in probabilistic notation, P(2->3|2) ), for instance, would be 2/(sum of all cell sizes), and the sum of all cell sizes is growing by 1 each period (assuming the unit of food given per period is 1, etc. etc.)
- murbard2 12y ago
  It's a markov chain if your state is the vector representing the size of each cell