Add material on monte carlo

_toc.yml

@@ -28,6 +28,8 @@ parts:
         - file: random
         - file: lcg
         - file: generating_random
+        - file: monte_carlo_integration
+        - file: metropolis
         - file: random_solutions
           sections:
             - file: prob_distributions_solutions

metropolis.md

@@ -0,0 +1,37 @@
+# Metropolis-Hastings algorithm
+
+Another way to sample a probability distribution is to use a **Markov Chain Monte Carlo** method. In these methods, we generate a sequence (or chain) of $x$ values that sample a given probability distribution $f(x)$. Each sample $x_i$ is generated from the previous one $x_{i-1}$ in some probabilistic way, so $x_i$ depends only on the state before it $x_{i-1}$ (this is the Markov chain part).
+
+The **Metropolis-Hastings algorithm** is a simple method to do this. To generate the next value in  the chain $x_{i+1}$ given the current value $x_i$, the procedure is as follows:
+
+- Propose a jump from the current value $x_i=x$ to a new candidate value $x^\prime$ according to a proposal distribution $p(x^\prime|x)$. The proposal distribution should be symmetric, ie. the probability of choosing a new  location $x^\prime$ given that we are now at $x$ should be the same as the probability of choosing $x$ if we were located at $x^\prime$. For example, in 1D we could choose $x^\prime = x_i + \Delta x$ with $\Delta x$ drawn from a Gaussian distribution.
+
+- Evaluate the ratio $\alpha = f(x^\prime)/f(x)$. If $\alpha>1$, ie. the new point is located in a region with larger $f(x)$, then we accept the move and set $x_{i+1} = x^\prime$. If $\alpha<1$, we accept the move with probability $\alpha$. If the move if rejected, then we set $x_{i+1} = x_i$.  (Note that this entire step can be achieved by choosing a uniform deviate $u$ between 0 and 1, and accepting the move if $u<\alpha$.)
+
+After an initial *burn-in* phase, the chain will reach equilibrium and the samples $\{x_i\}$ will be distributed according to the probability distribution $f(x)$. The reason this works is that in equilibrium, the rate at which we move from point $x$ to $x^\prime$ should be equal to the rate at which we move in the reverse direction, from $x^\prime$ to $x$. This is the principle of detailed balance. Therefore
+
+$$n(x) \times (\mathrm{rate\ of}\ x\rightarrow x^\prime)
+=
+n(x^\prime) \times (\mathrm{rate\ of}\ x^\prime\rightarrow x)
+$$
+
+If $x^\prime$ has a larger value of the probability $f(x)$, then the transition from $x$ to $x^\prime$ will always happen (we always accept), i.e. the rate is 1. The rate of the reverse transition from $x^\prime$ to $x$ is then $f(x)/f(x^\prime)<1$. Therefore
+
+$$n(x) = n(x^\prime) f(x)/f(x^\prime)$$
+
+$${n(x)\over n(x^\prime)}= {f(x)\over f(x^\prime)}$$
+
+which shows that $n(x)$ will be distributed as $f(x)$.
+
+```{admonition} Exercise
+
+Code up this algorithm and generate $10^4$ samples from the distribution $f(x) = \exp(-\left|x^3\right|)$. Confirm that you are getting the correct distribution by comparing the histogram of $x$-values with the analytic expression.
+
+Plot the values $\{x_i\}$ against iteration number $i$, and plot $x_i$ against $x_{i+1}$ for the values in the chain. What do you notice?
+How do your results change when you take different values for the width of the proposal distribution?
+
+Compare your results with the same plots for $10^4$ values from $f(x)$ chosen by the rejection method.
+```
+
+
+

monte_carlo_integration.md

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+# Monte Carlo integration
+
+In our discussion of the rejection method, you might have wondered whether we could use the samples of $f(x)$ to calculate the integral under the curve. Indeed, for a uniform sampling of $x$ values $\left\{x_i\right\}$ and the corresponding function values $f(x_i)$, 
+we can estimate the value of the integral
+
+$$\int_a^b dx f(x) \approx {(b-a)\over N} \sum_i f(x_i).$$
+
+Depending on the shape of the integral, this estimate can be improved by using a different distribution for the $x$-values which concentrates the samples of $x$ in the regions where the integrand is largest (the ideal case would be to sample from $f(x)$ itself). In this case, we have to rescale the weight of a given $x$-value according to the ratio $f(x)/p(x)$. The integral then becomes
+
+$$ {1\over N} \sum_i {f(x_i)\over p(x_i)}.$$
+
+In the case of a uniform distribution, $p(x) = 1/(b-a)$, we recover the expression above. 
+
+```{admonition} Exercise
+
+Calculate the Monte Carlo estimate for the integral of $\exp(-\left|x\right|^3)$ between $x=0$ and $\infty$. 
+
+Use both uniform sampling of $x$ (in which case you will have to put in an upper limit on the $x$-value), and Gaussian sampling of $x$. 
+
+In each case, take the number of samples to be $N=10^4$ and repeat the integration 1000 times. Plot a histogram of the values of the integral that you obtain. Calculate the mean and standard deviation and compare the uniform and Gaussian sampling, and also with the exact value of the integral (you can use `scipy.integrate.quad`).
+
+Repeat for different $N$ values. How does the error in your value for the integral scale with $N$?
+
+```