Chi-squared Distribution ($k=1$)

Theory

Given a standard random variable $X \sim \mathcal{N}(0,1)$ we would like to find the random variable $Y = X^2$. The analytic solution is known: $Y$ follows a chi-squared distribution with $k=1$ degree of freedom.

Using polynomial chaos expansion (PCE), the problem can be solved using Galerkin projection. Let $\{\phi_k \}_{k=0}^{n}$ be the monic orthogonal basis relative to the probability density of $X$, namely

\[f_X(x) = \frac{1}{\sqrt{2 \pi}} \exp \left( - \frac{x^2}{2} \right).\]

Then, the PCE of $X$ is given by

\[X = \sum_{k=0}^n x_k \phi_k,\]

with

\[x_0 = 0, \quad x_1 = 1, \quad x_i = 0 \quad \forall i =2,\dots,n.\]

To find the PCE coefficients $y_k$ for $Y = \sum_{k=}^n y_k \phi_k$, we apply Galerkin projection, which leads to

\[y_m \langle \phi_m, \phi_m \rangle = \sum_{i=0}^n \sum_{j=0}^n x_i x_j \langle \phi_i \phi_j, \phi_m \rangle \quad \forall m = 0, \dots, n.\]

Hence, knowing the scalars $\langle \phi_m, \phi_m \rangle$, and $\langle \phi_i \phi_j, \phi_m \rangle$, the PCE coefficients $y_k$ can be obtained immediately. From the PCE coefficients we can get the moments and compare them to the closed-form expressions.

Notice: A maximum degree of 2 suffices to get the exact solution with PCE. In other words, increasing the maximum degree to values greater than 2 introduces nothing but computational overhead (and numerical errors, possibly).

Practice

First, we create a orthogonal basis relative to $f_X(x)$ of degree at most $d=2$ (deg below).

Notice that we consider a total of Nrec recursion coefficients, and that we also add a quadrature rule by calling OrthoPolyQ().

using PolyChaos
k = 1
deg, Nrec = 2, 20
op = OrthoPoly("gaussian",deg;Nrec=Nrec);
opq = OrthoPolyQ(op) #OR: opq = OrthoPolyQ("gaussian",deg;Nrec=Nrec)

OrthoPolyQ(OrthoPoly("gaussian", 2, [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0], [1.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0], Measure("gaussian", PolyChaos.w_gaussian, (-Inf, Inf), true, Dict{Any,Any}())), Quad("golubwelsch", 19, [-7.38258, -6.26289, -5.32054, -4.46587, -3.66442, -2.89805, -2.1555, -1.42888, -0.712085, -2.10166e-16, 0.712085, 1.42888, 2.1555, 2.89805, 3.66442, 4.46587, 5.32054, 6.26289, 7.38258], [7.48283e-13, 1.22037e-9, 2.53222e-7, 1.53511e-5, 0.000378502, 0.00450724, 0.0286667, 0.103604, 0.220942, 0.283773, 0.220942, 0.103604, 0.0286667, 0.00450724, 0.000378502, 1.53511e-5, 2.53222e-7, 1.22037e-9, 7.48283e-13], Measure("gaussian", PolyChaos.w_gaussian, (-Inf, Inf), true, Dict{Any,Any}())))

What are the basis polynomials?

showbasis(opq; sym="ξ") # works for `op` too!

1
ξ
ξ^2 - 1.0

Note that the command showbasis is based on the more general showpoly:

showpoly(0:2:deg,opq)

1
x^2 - 1.0

Next, we define the PCE for $X$.

L = dim(opq)
mu, sig = 0., 1.
x = [ convert2affinePCE("gaussian",mu,sig); zeros(Float64,L-2) ]

3-element Array{Float64,1}:
 0.0
 1.0
 0.0

With the orthogonal basis and the quadrature at hand, we can compute the tensors t2 and t3 that store the entries $\langle \phi_m, \phi_m \rangle$, and $\langle \phi_i \phi_j, \phi_m \rangle$, respectively.

t2 = Tensor(2,opq);
t3 = Tensor(3,opq)

Tensor(3,   [1 ]  =  1.0
  [4 ]  =  1.0
  [7 ]  =  2.0
  [9 ]  =  2.0
  [15]  =  8.0, getfield(PolyChaos, Symbol("#g#27")){Int64,OrthoPolyQ,SparseArrays.SparseVector{Float64,Int64}}(3, OrthoPolyQ(OrthoPoly("gaussian", 2, [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0], [1.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0], Measure("gaussian", PolyChaos.w_gaussian, (-Inf, Inf), true, Dict{Any,Any}())), Quad("golubwelsch", 19, [-7.38258, -6.26289, -5.32054, -4.46587, -3.66442, -2.89805, -2.1555, -1.42888, -0.712085, -2.10166e-16, 0.712085, 1.42888, 2.1555, 2.89805, 3.66442, 4.46587, 5.32054, 6.26289, 7.38258], [7.48283e-13, 1.22037e-9, 2.53222e-7, 1.53511e-5, 0.000378502, 0.00450724, 0.0286667, 0.103604, 0.220942, 0.283773, 0.220942, 0.103604, 0.0286667, 0.00450724, 0.000378502, 1.53511e-5, 2.53222e-7, 1.22037e-9, 7.48283e-13], Measure("gaussian", PolyChaos.w_gaussian, (-Inf, Inf), true, Dict{Any,Any}()))),   [1 ]  =  1.0
  [4 ]  =  1.0
  [7 ]  =  2.0
  [9 ]  =  2.0
  [15]  =  8.0), OrthoPolyQ(OrthoPoly("gaussian", 2, [0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0], [1.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0], Measure("gaussian", PolyChaos.w_gaussian, (-Inf, Inf), true, Dict{Any,Any}())), Quad("golubwelsch", 19, [-7.38258, -6.26289, -5.32054, -4.46587, -3.66442, -2.89805, -2.1555, -1.42888, -0.712085, -2.10166e-16, 0.712085, 1.42888, 2.1555, 2.89805, 3.66442, 4.46587, 5.32054, 6.26289, 7.38258], [7.48283e-13, 1.22037e-9, 2.53222e-7, 1.53511e-5, 0.000378502, 0.00450724, 0.0286667, 0.103604, 0.220942, 0.283773, 0.220942, 0.103604, 0.0286667, 0.00450724, 0.000378502, 1.53511e-5, 2.53222e-7, 1.22037e-9, 7.48283e-13], Measure("gaussian", PolyChaos.w_gaussian, (-Inf, Inf), true, Dict{Any,Any}()))))

With the tensors at hand, we can compute the Galerkin projection.

y = [ sum( x[i]*x[j]*t3.get([i-1,j-1,m-1])/t2.get([m-1,m-1])  for i=1:L, j=1:L ) for m=1:L ]

3-element Array{Float64,1}:
 1.0               
 0.0               
 1.0000000000000033

Let's compare the moments via PCE to the closed-form expressions.

moms_analytic(k) = [k, sqrt(2k), sqrt(8/k)]
function myskew(y)
   e3 = sum( y[i]*y[j]*y[k]*t3.get([i-1,j-1,k-1]) for i=1:L,j=1:L,k=1:L )
   μ = y[1]
   σ = std(y,opq)
   (e3-3*μ*σ^2-μ^3)/(σ^3)
end

print("Expected value:\t\t$(moms_analytic(k)[1]) = $(mean(y,opq))\n")
print("\t\t\terror = $(abs(mean(y,opq)-moms_analytic(k)[1]))\n")
print("Standard deviation:\t$(moms_analytic(k)[2]) = $(std(y,opq))\n")
print("\t\t\terror = $(moms_analytic(k)[2]-std(y,opq))\n")
print("Skewness:\t\t$(moms_analytic(k)[3]) = $(myskew(y))\n")
print("\t\t\terror = $(moms_analytic(k)[3]-myskew(y))\n")

Expected value:		1.0 = 1.0
			error = 0.0
Standard deviation:	1.4142135623730951 = 1.4142135623730998
			error = -4.6629367034256575e-15
Skewness:		2.8284271247461903 = 2.8284271247462023
			error = -1.199040866595169e-14

Let's plot the probability density function to compare results. We first draw samples from the measure with the help of sampleMeasure(), and then evaluate the basis at these samples and multiply times the PCE coefficients. The latter stop is done using evaluatePCE(). Finally, we compare the result agains the analytical PDF $\rho(t) = \frac{\mathrm{e}^{-0.5t}}{\sqrt{2 t} \, \Gamma(0.5)}$ of the chi-squared distribution with one degree of freedom.

using Plots
gr()
Nsmpl = 10000
#ξ = sampleMeasure(Nsmpl,opq)
#ysmpl = evaluatePCE(y,ξ,opq)
ysmpl = samplePCE(Nsmpl,y,opq)
histogram(ysmpl;normalize=true,xlabel="t",ylabel="\rho(t)")


import SpecialFunctions: gamma
ρ(t) = 1/(sqrt(2)*gamma(0.5))*1/sqrt(t)*exp(-0.5*t)
t = range(0.1; stop=maximum(ysmpl), length=100)
plot!(t,ρ.(t),w=4)