Commit 42a62882 authored by Michael Wimmer's avatar Michael Wimmer
Browse files

add link to Monte Carlo intro

parent ebcd53e7
Pipeline #58174 passed with stages
in 1 minute and 49 seconds
......@@ -95,7 +95,7 @@ as evident from the growth procedure. However, in our model we said every polyme
$$P_\text{real}(\mathbf{r}_1, \dots, \mathbf{r}_L) = \frac{1}{T_L}\,$$
where $T_L$ is the total number of possible polymers with length $L$. These two probabilities differ: growing the polymer does not make them with equal probability, but biases them as adding a new subunit depends on the previous steps.
So we are sampling with a distribution that only *approximates* the real probability distribution. This an example of approximate importance sampling (**TODO add link to main lecture notes**), and accordingly we find that the average is given as:
So we are sampling with a distribution that only *approximates* the real probability distribution. This an example of [approximate importance sampling](proj2-monte-carlo.md#approximate-importance-sampling), and accordingly we find that the average is given as:
$$\langle r^2(L)\rangle \approx \frac{1}{N} \sum_{k=1}^N \frac{P_\text{real}}{P_\text{sample}} r_k^2(L) = \frac{1}{N}\sum_{k=1}^N \frac{w_k^{(L)}}{T_L} r_k^2(L)= \frac{1}{N} \frac{1}{T_L} \sum_{k=1}^N w_k^{(L)} r_k^2(L) \tag{3}$$
However, we still don't know $T_L$ exactly - we need to estimate it. To this end we can make use of a rather trivial exact identity:
$$T_L = \sum_\text{all SAW(L)} 1 = \sum_\text{all SAW(L)} P_\text{sample} \frac{1}{P_\text{sample}}$$
......
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment