User:TheLarry/Notebook/Larrys Notebook/2009/09/22

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Finding average time to absorb

Yesterday i said how the expectation value of t was not working out well. It turns out what i was comparing it to was incorrect. I wasn't taking enough intervals in the histogram from the simulation. But now i am and that probability density curve i get from the simulation matches exp(Q t). Taking the expectation value from there gives me the hitting times. So it all looks good, and i am feeling pleased about myself.

With this process I can put in any complicated matrix I need for Q which contains the rate constants, and figure out the expectation value of t.

In this picture you can see the histogram generated from the simulation and the predicted P(t) from exp(Q t) from theory.
In this picture you can see the histogram generated from the simulation and the predicted P(t) from exp(Q t) from theory.

The expected time for completion is 1.26 seconds. This makes k effective .79 1/s. We still don't know what to do with that k since it isn't actually k since it isn't a one step chemical reaction, but it is something like a k effective since it is inverse the average time it takes to complete the process.

This is the same thing but with all the k's 1 instead of a mixture. This produced an expected time of 6 and a k effective of .166
This is the same thing but with all the k's 1 instead of a mixture. This produced an expected time of 6 and a k effective of .166


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