Revision as of 07:38, 24 February 2021

Crash Course in Scientific Computing

VI. Random Numbers

Let us now play with things computers are good at (e.g., with statistics). We'll need another package first:

import Pkg; Pkg.add("Distributions")

Once this is done (once for ever on a given machine), you can then be:

using Distributions

Let us generate ten thousands random points following a squared-uniform probability distribution, $X^2$.

lst=[rand()^2 for i=1:10^5]

and after

using Plots
histogram(lst)

The theoretical result is obtained by differentiating the cumulative function, $f_{X^2}=dF_{X^2}/dx$, with

$$F_{X^2}(x)\equiv\mathbb{P}(X^2\le x)$$

but $\mathbb{P}(X^2\le x)=\mathbb{P}(X\le\sqrt{x})=\sqrt{x}$, since the probability is uniform. Therefore:

$$f_{X^2}(x)={1\over2\sqrt{x}}$$

Let us check:

f(x)=1/(2*sqrt(x))
histogram(lst,norm=true)
plot!(f,.01,1, linewidth = 4, linecolor = :red, linealpha=0.75)

The ! means to plot on the existing display. This seems to work indeed:

Let's have a look at the distribution of inverse random numbers, also sampled from a uniform distribution between 0 and 1. This time, the support is infinite (it is $[1,\infty[$). The theoretical distribution is $f_{1/X}=d\mathbb{P}_{1/X}/dx$ with $\mathbb{P}_{1/X}=\mathbb{P}(1/X\le x)=\mathbb{P}(X\ge 1/x)=1-1/x$ (being uniform again), so that $f_{1/X}(x)=1/x^2$.

This time we'll use a lot of points to make sure we get the exact result:

@time lst = [1/rand() for i=1:10^8]

The @time allows us to benchmark the time and memory resources involved. As you can see, on a 2020 machine, it's pretty fast (for a hundred million points!)

  0.427357 seconds (67.69 k allocations: 766.257 MiB, 0.65% gc time)

The result is also consequently very smooth:

@time histogram(lst,norm=true,bins=0:.01:10)

returning

4.644046 seconds (4.88 k allocations: 1.490 GiB, 1.74% gc time)

and, after also

f(x)=1/x^2
plot!(f,1,10,linewidth=4,linecolor=:red,linealpha=.75)

This is slightly off, clearly. The culprit is our distribution function, which density should be 1 at x=1 and is more than that. The problem is the norm option of histogram, that normalizes to what is plotted (or retained). And we have points beyond that. We can use this number to renormalize our theory plot. This is how many numbers we have:

sum(lst .< 10)

and this is the list of these numbers:

lst[lst .> 100]

When plotting pure functions, you might find that the density of points (PlotPoints in Mathematica) is not enough. This can be increased by specifying the step in the range, but in this case beware not to query your functions outside of its validity range, which would work otherwise. Compare:

plot(x->sqrt((15/x^2)-1),0,4,linewidth=5,linealpha=.5,linecolor=:red,ylims=(0,3))
plot!(x->sqrt((15/x^2)-1),0:.0001:3.8729,linewidth=2,linecolor=:blue,ylims=(0,3))

The first version can plot over the range 0–4 but does so by not plotting the full function, while the second (with x-steps of 0.0001 allowing for the nice curvature) shows everything but would return an error if going over the limit where the argument gets negative.

Let us now explore numerically a very interesting (and important) phenomenon. For that, we'll need the Cauchy function from the above-loaded "Distributions" package. This is a histogram of $10^5$ Cauchy-distributed points:

histogram(rand(Cauchy(),10^5))

Here the binning is mandatory:

histogram(rand(Cauchy(),10^5),bins=-10:.5:10,norm=true)

Let us now look at how to generate random numbers on our own. For instance, let us say we want to simulate the ground state of a particle in a box, with probability distribution over $[0,1]$ given by:

$$\psi^2(x)=2\sin^2(\pi x)$$

We can use unicode characters by entering \psi+TAB and call our density of probability ψ2, which in quantum mechanics is given by the modulus square of the probability amplitude (in 1D the wavefunction can always be taken real so we do not need worry about the modulus):

ψ2(x)=2*sin(pi*x)^2

That's our density of probability:

using LaTeXStrings
default(; w=3)
plot(ψ2,0,1, title=L"2\sin^2(x)", xlabel=L"x", ylabel=L"\psi^2(x)",dpi=150,legend=false)

where we decorated the plot with a lot of options, but the width which we gave as a default attribute, not to do it in the future (and since we like thick lines).

We will compute $\int_0^1\psi^2(x)\,dx$ by computing Riemann sums, i.e., as a sum of parallelograms defined in a partition of the support of the function (here $[0,1]$) cut into $n$ intervals. Calling $f$ the function to integrate, this is defined as:

$$R_n=\sum_{i=1}^{n}f(x_i)\delta(i)$$

a, b = 0, 1; # boundaries
lx=range(a, stop=b, length=11) # list of x intervals
δ=diff(lx) # widths of intervals
lxi=lx[1:end-1] # x_i points of parallelograms

We can now check the normalization:

sum([ψ2(lxi[i])*δ[i] for i=1:length(lxi)])

1.0

The cumulative is obtained as:

Ψ=[sum([ψ2(lxi[i])*δ[i] for i=1:j]) for j=1:length(lxi)]

This is the result for 100 intervals (length=101 above):

plot(lxi, Ψ, legend=false, title="Cumulative")

Now the generation of a random number following the initial distribution works as follows. We select randomly (uniformly) a number on the $y$-axis of the cumulative and find the corresponding $x$ such that $F(x)=y$. Those $x$ are $f$&nbps;distributed.

This can be simply achieved as follows (later we'll see a more sophisticated way):

yr=rand()
fx=lx[[findmin(abs.(Ψ.-rand()))[2] for i=1:10]]

Here are some positions where our ground state collapsed:

10-element Array{Float64,1}:
 0.55
 0.96
 0.24
 0.5 
 0.58
 0.16
 0.67
 0.74
 0.08
 0.46

histogram(lx[[findmin(abs.(Ψ.-rand()))[2] for i=1:10^7]], bins=0:.01:1,norm=true)

Note that the binning cannot be less than the $\delta$ step, otherwise we will get "holes" (and also the wrong normalization):

histogram(lx[[findmin(abs.(Ψ.-rand()))[2] for i=1:10^7]], bins=0:δ[1]/2:1,norm=true)

We would typically want to embed our results together in a function of our own:

function myRiemann(f, n)
    a, b = 0, 1; # boundaries
    lx=range(a, stop=b, length=n) # list of x intervals
    δ=diff(lx); # widths of intervals
    lxi=lx[1:end-1]; # x_i points of parallelograms
    sum([f(lxi[i])*δ[i] for i=1:length(lxi)])
end

Then we can try:

myRiemann(g, 10^3)

The result is fairly close to the exact $e-1\approx1.718281828459045$:

1.717421971020861

However we'll find that our method is fairly limited:

scatter([abs(myRiemann(g, i)-(exp(1)-1)) for i=2:100:1002], 
  ylims=(10^-4,1), yscale=:log10, xticks=(1:11, 2:100:1002), 
  xlabel="Num steps", ylabel="error")

It is the purpose of numerical methods to learn how to use algorithms that are efficient, in the sense that they are accurate, fast and resource-effective.

It is easy to be brutal and terribly inefficient with a computer. In fact a fairly trivial enhancement of our method leads to considerable improvement:

function myRiemann(f, n, method="mid")
    a, b = 0, 1; # boundaries
    lx=range(a, stop=b, length=n) # list of x intervals
    δ=diff(lx); # widths of intervals
    if method == "left"
        lxi=lx[1:end-1]; # x_i points of parallelograms
    elseif method == "right"
        lxi=lx[2:end];
    else
        lxi=(lx[1:end-1]+lx[2:end])/2;
    end
    sum([f(lxi[i])*δ[i] for i=1:length(lxi)])
end

Of course, other people already wrote such functions, that are available through packages. For numerical integration of 1D functions, one can use QuadGK which is based on the so-called Gauss–Kronrod quadrature formula, which basically figures out how to best choose the points $x_i$ and how to weight them:

using QuadGK
@time quadgk(g, 0, 1)

Unlike our naive implementation, the result is pretty accurate:

  0.000034 seconds (9 allocations: 304 bytes)
(1.718281828459045, 0.0)

That's the best we can achieve (with mid-points and a hundred-million intervals!)

@time myRiemann(g, 10^8)

    9.762264 seconds (500.00 M allocations: 8.941 GiB, 11.01% gc time)
1.7182818284590453

It differs from the exact result only in the last digit, but took about 10s & 9GiB of memory.