Revision as of 12:09, 11 February 2020

Crash course in Julia (programming)

Julia is a powerful/efficient/high-level computer programming language. You can get into interacting mode right-away with:

julia

You may need to install packages, which can be done as follows:

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:

@@ Line 33: / Line 33: @@
 <center><wz tip="Distribution of $10^5$ squared random numbers.">[[File:julia-randX2.png|400px]]</wz></center>
+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:
+<code lang="julia">
+f(x)=1/(2*sqrt(x))
+histogram(lst,norm=true)
+plot!(f,.01,1, linewidth = 4, linecolor = :red, linealpha=0.75)
+</code>
+The ! means to plot on the existing display. This seems to work indeed:
+<center><wz tip="Comparison of the normalized histogram with the theoretical result.">[[File:julia-randX2th.png|400px]]</wz></center>