# Integrals and Monte Carlo Methods

I am teaching surface integrals right now. Searching for some interesting applications away from physics, I came up with mean values along surfaces. After all, mean or „expected“ values are a center part of probability theory. It is tempting to try to verify the analytic results by Monte Carlo methods. This can be done nicely in Euler Math Toolbox.

E.g., an integral on a finite interval [a,b] can be simulated this way. We generate random numbers in [a,b] uniformly, and compute the mean of the values of the function. The expected mean value is the integral of the function divided by the length of the interval.

>&integrate(log(x),x,0,2), %()

2 log(2) - 2

-0.61370563888
>N = 10 000 000;
>x = random(N)*2;
>mean(log(x))*2
-0.61394326247


Syntax notes for EMT: The blanks in integer numbers can be generated by pressing Ctrl-Space. This generates a hard space which is skipped when scanning the number. The random() function generates uniform numbers in [0,1] which we need to be scaled properly to fall into [0,2].

The accuracy of these Monte Carlo methods are like 1/√N. So, we cannot expect more than 4 digits, even with ten million samples.

As another example, the area of the circle can be simulated by throwing random numbers into the unit cube and counting the number of hits inside the circle. We expect a proportion of pi/4 there.

>N = 10 000 000;
>x = 2*random(N)-1; y=2*random(N)-1;
>sum(x^2+y^2<=1)/N
0.7854814
>pi/4
0.785398163397


Syntax notes for EMT: Applying the boolean „≤“ to a vector generates a vector of zeros and ones. Summing this vector up counts the „true“ cases.

This rejection method is also a simple way to create random numbers inside the unit circle. It is not as bad as it sounds, since more than every second pair of numbers hits inside the circle.

For a further test, let us compute the integral of r^2 on the unit circle (which is proportional to the energy in a rotating disk). The integral is the mean value of r^2 on the unit circle times the area of the unit circle.

>N = 10 000 000;
>x = 2*random(N)-1; y=2*random(N)-1;
>r = sqrt(x^2+y^2);
>i = nonzeros(r<=1);
>mean(r[i]^2)*pi
1.57088571913
>&2*pi*integrate(r^3,r,0,1), %()

pi
--
2

1.57079632679


Syntax notes for EMT: The nonzeros() function generates a vector of indices of nonzero elements. We can use that vector to select the hits from the vector r.

For the correct integral, we used the formula for rotational invariant functions on circles.

As a side remark, there is a direct method to generate random numbers inside the unit circle. For this, we need to select a random angle between 0 and 2pi. The radius must be selected in proportion to the length of the circle with this radius, i.e. with density 0≤r≤1. The following does the trick. We explain it after the code.

>function circrandom (N) ...
$t = random(N)*2*pi;$  r = sqrt(1-random(N));
$return {r*cos(t),r*sin(t)}$endfunction
>{x,y} = circrandom(10000);
>plot2d(x,y,>points,style=".",r=1.2):


This will generate the following points, which look quite nicely distributed in the unit circle.

Why does this work? The reason is that any probability distribution can be simulated by inverting its accumulated distribution, in this case

$$F(R) = p(r \ge R) = \int\limits_R^1 2r \, dr = 1-r^2$$

Here, we take 2r because this generates a probability distribution with density proportionally to r. Then we create a random number between 0 and 1 and apply the inverted function.

Using this trick, we can do another Monte Carlo simulation of the integral.

>{x,y} = circrandom(N);
>r = sqrt(x^2+y^2);
>i = nonzeros(r<=1);
>mean(r[i]^2)*pi
1.57153749805


Syntax notes on EMT: Note that circrandom() returns two values. It is possible to assign these two values to two variables.

Finally, we try an integral on a surface. We want to know the mean geographical width of a point in the Northern Hemisphere.

We try a simple rejection method. For this, we generate random numbers in the unit cube and reject those with z<0, r>1 or r<1/2. The remaining points, projected to the surface, are randomly distributed on the Northern Hemisphere.

>N = 10 000 000;
>x = random(N)*2-1; y = random(N)*2-1; z=random(N)*2-1;
>r = sqrt(x^2+y^2+z^2);
>i = nonzeros(z>0 && r<1 && r>1/2);
>mean(asin(z[i]/r[i]))
0.570688989539
>&2*pi*integrate(asin(z),z,0,1) / (2*pi) | factor, ...
>   %(), %->"°"

pi - 2
------
2

0.570796326795
32.7042204869°


The analytic surface integral is done by parameterizing the surface using

$$g(z,t) = [\sqrt{1-z^2}\cos(t),\sqrt{1-z^2} \sin(t),z]$$

We can compute the Gram determinant for this parameterization, which is actually 1 (or use a geometric reasoning with the same result). Thus, for any function on the hemisphere H which depends on the height z only, we get

$$\int_H f(x,y,z) \,dS = 2 \pi \int\limits_0^1h(z) \,dz$$

The rejection trick works on symmetric surfaces only, where we can easily do an orthogonal projection and know the distance to the surface. Otherwise, we have to distribute the random points on the parameters in accordance with the distribution of the surface measure. This can be very tricky.

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