Commit f839d649 authored by Michael Wimmer's avatar Michael Wimmer
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fix math errors and formatting

parent ff558aa4
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......@@ -21,7 +21,7 @@ where $x_i = a + \left(i-\frac{1}{2}\right) \frac{b-a}{N}$. This the simplest ap
A different approach to evaluating the integral can be taken if we use concepts from the theory of random variables. To this end, consider a probability distribution
$p(x)$ of a random variable $x$ (then $\int dx p(x) = 1$). We can compute the expectation value of a function $f(x)$ of this random variable as
$$\int p(x)f(x)dx \underbrace{\approx}_\text{$x_i$ sampled from $p(x)$ \frac{1}{N}\sum_i f(x_i) + \mathcal{O}(\frac{1}{\sqrt{N}}\,. \tag{2}$$
$$\int p(x)f(x)dx \underbrace{\approx}_\text{$x_i$ sampled from $p(x)$}\frac{1}{N}\sum_i f(x_i) + \mathcal{O}(\frac{1}{\sqrt{N}}\,. \tag{2}$$
Here we are approximating the expectation value by taking a finite sample of the random variable $x$, we "sample" $x_i$ from the probability
distribution $p(x)$.
......@@ -69,8 +69,8 @@ probability distribution that is concentrated in the physically relevant space.
### Sampling *almost* the right probability distribution
Let us consider the general case of computing the expectation value of a function $A(R)$ for a random variable distributed according to $p_\text{real}(R)$
(in the previous example, $p_\text{real} = e^{-\beta H(R)}/Z$. We consider the general case here).
Let us consider the general case of computing the expectation value of a function $A(R)$ for a random variable distributed according to $p_\text{real}(R)$.
(In the previous example, $p_\text{real}(R) = e^{-\beta H(R)}/Z$. We consider the general case here.)
We then have to calculate the integral
$$\int p_\text{real}(R) A(R) dR\,.$$
Ideally, we would now generate random variables $R_i$ that are distributed according to $p_\text{real}(R)$. However, this may be impractical. In this
......@@ -81,12 +81,12 @@ In this way we can make sure to approximately focus on the physically relevant c
Doing this requires us to rewrite the integral as
$$\begin{split}
\int p_\text{real}(R) A(R) dR = &
\int p_\text{sampling}(R) \underbrace{\frac{p_\text{real}(R)}{p_\text{sampling}(R)}}_{=w(R)} A(R) dR\\
\int p_\text{sampling}(R) \underbrace{\frac{p_\text{real}(R)}{p_\text{sampling}(R)}}}_{=w(R)} A(R) dR\\
= & \int p_\text{sampling}(R) w(R) A(R) dR\\
\approx \frac{1}{N} \sum_{i=1}^N w(R_i) A(R_i) \tag{3}
\end{split}$$
where the configurations $R_i$ are now sampled from $p_\text{sampling}(R)$. When using this approximate probability distribution
we thus have to introduce *weights$* $w(R)$ into the average.
we thus have to introduce *weights* $w(R)$ into the average.
### Why approximate importance sampling eventually fails
......@@ -109,12 +109,12 @@ This is also a problem when we sample with only a nearby probability distributio
dimensionality of the configuration space increases, the overlap between the actual and the nearby
probability distributions vanishes, and we keep on mostly sampling uninteresting space.
This effect directly shows in the weights. Let us demonstrate this using a simpel example. Consider the case
This effect directly shows in the weights. Let us demonstrate this using a simple example. Consider the case
where
$$\p_\text{real}(x_1, \dots x_d) = (2 \pi \sigma_\text{real})^{d/2} e^{-\frac{\sum_k=1^d x_k^2}{2\sigma_\text{real}}$$
$$\p_\text{real}(x_1, \dots, x_d) = (2 \pi \sigma_\text{real})^{d/2} e^{-\frac{\sum_k=1^d x_k^2}{2\sigma_\text{real}}}$$
is a normal distribution with standard deviation $\sigma_\text{real}$. For the sampling distribution we use
also a normal distribution, but with a slightly differen standard deviation $\sigma_\text{sampling}$:
$$\p_\text{sampling}(x_1, \dots x_d) = (2 \pi \sigma_\text{sampling})^{d/2} e^{-\frac{\sum_k=1^d x_k^2}{2\sigma_\text{sampling}}\,.$$
$$\p_\text{sampling}(x_1, \dots, x_d) = (2 \pi \sigma_\text{sampling})^{d/2} e^{-\frac{\sum_k=1^d x_k^2}{2\sigma_\text{sampling}}}\,.$$
We will now compute how the weights $p_\text{real}/p_\text{sampling}$ are distributed for different dimensionality
$d$. In the example below we have chosen $\sigma_\text{real} = 1$ and $\sigma_\text{sampling} = 0.9$ and sampling over $N=10000$
configurations:
......@@ -151,9 +151,9 @@ We can thus generate a desired probability distribution $p(R)$ by choosing the r
probability distribution? The Metropolis algorithm solves this problem!
The Metropolis algorithm uses the following ansatz: $T(R \rightarrow R^\prime) = \omega_{RR^\prime} \cdot A_{RR^\prime}$. Here, the generation of the
new state in the Markov chain is split into two phases. First, starting from the previous state $R = R_i$, we generate a candidate state $R^\rprime$
new state in the Markov chain is split into two phases. First, starting from the previous state $R = R_i$, we generate a candidate state $R^\prime$
with probability $\omega_{XX^\prime}$. This is the so-called "trial move". We then accept this trial move with probability $A_{RR^\prime}$,
i.e. set $R_{i+1} = R'. If we don't accept it, we take the old state again, $R_{i+1} = R$. Altogether, the probability of going to a state new state ($T(X^\prime \rightarrow X)$)
i.e. set $R_{i+1} = R'$. If we don't accept it, we take the old state again, $R_{i+1} = R$. Altogether, the probability of going to a state new state ($T(X^\prime \rightarrow X)$)
is the product of proposing it ($\omega_{RR^\prime}$) and accepting it ($A_{RR^\prime})$.
The problem can we further simplified by demanding that $\omega_{RR^\prime}=\omega_{R^\prime R} - the trial move should have a symmetric probability of going from $R$ to $R^\prime$
......
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