# Generate R.V.

An important Lemma.

Note: Unless otherwise stated, the algorithms and the corresponding screenshots are adopted from Robert and Casella (2005).

# Box-Muller Algorithm

function boxmuller()    x = ones(2)    u = rand(2)    logu = sqrt(-2*log(u))    x = logu * cos(2*pi*u)    x = logu * sin(2*pi*u)    return(x)end​## exampleboxmuller()

# Accept-Reject Method

Example:

$f(x)\propto \exp(-x^2/2)(\sin(6x)^2+3\cos(x)^2\sin(4x)^2+1)$

The Julia code is as follows:

function AccRej(f::Function, M)    ## use normal distribution N(0, 1) as sampling function g    while true        x = randn()        u = rand()        cutpoint = f(x)/(M*g(x))        if u <= cutpoint            return([x, f(x)])        end    endend​## density function of N(0, 1)function g(x)    return(exp(-0.5*x^2)/sqrt(2*pi))end​## example function and ignore the normalized constantfunction f(x)    return(exp(-x^2/2)*(sin(6*x)^2 + 3*cos(x)^2*sin(4*x)^2 + 1))end​## exampleN = 500;data = ones(N, 2);for i = 1:500    data[i,:] = AccRej(f, sqrt(2*pi)*5)end

# Envelope Accept-Reject

It is easy to write the Julia code:

function EnvAccRej(f::function, M, gl::function)    while true        x = randn() # still assume gm is N(0,1)        u = rand()        cutpoint1 = gl(x)/(M*g(x))        cutpoint2 = f(x)/(M*g(x))        if u <= cutpoint1            return(x)        elseif u <= cutpoint2            return(x)        end    endend

# Atkinson's Poisson Simulation

It is necessary to note that the parameters in the algorithm are not same with those in the density function. In other words, the corresponding density function of the algorithm should be

$f(x) = \beta \frac{\exp(\alpha-\beta x)}{[1+\exp(\alpha-\beta x)]^2}.$

The following Julia code can generate the poisson random variable with respect to $\lambda$.

function AtkinsonPois(lambda)    # parameters    beta = pi/sqrt(3*lambda)    alpha = lambda*beta    c = 0.767 - 3.36/lambda    k = log(c) - lambda - log(beta)    # step 1: propose new x    u1 = rand()    while true        global x        x = (alpha - log((1-u1)/u1))/beta        x > -0.5 && break    end    while true        # step 2: transform to N        N = floor(Int, x)        u2 = rand()        # step 3: accept or not        lhs = alpha - beta*x + log(u2/(1+exp(alpha-beta*x))^2)        rhs = k + N*log(lambda) - log(factorial(lambda))        if lhs <= rhs            return(N)        end    endend​## exampleN = 100;res = ones(Int, N);for i = 1:N    res[i] = AtkinsonPois(10)end# ans: 8 9 13 10 12 .......

As mentioned in the above exercise, another poisson generation method can be derived from the following exercise.

We can write the following Julia code to implement this generation procedure.

function SimplePois(lambda)    s = 0    k = 0    while true        u = rand()        x = -log(u)/lambda        s = s + x        if s > 1            return(k)        end        k = k + 1    endend​## example for simple poissonres2 = ones(Int, N);for i = 1:N    res2[i] = SimplePois(10)end# ans: 5 7 16 7 10 .......

# ARS Algorithm

ARS is based on the construction of an envelope and the derivation of a corresponding Accept-Reject algorithm. It provides a sequential evaluation of lower and upper envelopes of the density $f$ when $h=\log f$ is concave.

Let be a set of points $x_i,i=0,1,\ldots,n+1$, in the support of $f$ such that $h(x_i)=\log f(x_i)$ is known up to the same constant. Given the concavity of $h$, the line $L_{i,i+1}$ through $(x_i,h(x_i))$ and $(x_{i+1},h(x_{i+1}))$ is below the graph of $h$ in $[x_i,x_{i+1}]$ and is above this graph outside this interval.

For $x\in [x_i,x_{i+1}]$, define

$\bar h_n(x)=\min\{L_{i-1,i}(x),L_{i+1,i+2}(x)\}\quad\text{and}\quad \underline h_n(x)=L_{i,i+1}(x)\,,$

the envelopes are

$\underline h_n(x)\le h(x)\le \bar h_n(x)$

uniformly on the support of $f$. Thus,

$\exp \underline h_n(x) = \underline f_n(x) \le f(x)\le \bar f_n(x)=\exp\bar h_n(x) = \bar w_ng_n(x)\,.$

Davison (2008) provides another version of ARS

and gives an illustration example.

Let's consider the slightly simple verison in which we do not need to consider $h_-(y)$. It is obvious that

Consider the CDF of $g_+=\exp(h_+)$:

\begin{aligned} G_+(y) & = \int_{-\infty}^y \exp(h_+(x)) dx \\ & = \int_{-\infty}^{\min\{z_1,y\}} \exp(h_+(x)) dx \\ & \qquad + \int_{z_1}^{\min\{z_2,\max\{y, z_1\}\}} \exp(h_+(x)) dx \\ & \qquad + \cdots \\ & \qquad + \int_{z_{k-1}}^{\min\{z_k,\max\{y, z_{k-1}\}\}} \exp(h_+(x)) dx \\ & \qquad + \int_{z_k}^{\max\{z_k,y\}} \exp(h_+(x)) dx \end{aligned}

and

$\int_{z_{j}}^{z_{j+1}}\exp(h_+(x))dx = \exp({h(y_{j+1})})\cdot \frac{1}{h'(y_{j+1})}\exp((y-y_{j+1})h'(y_{j+1}))\mid_{z_j}^{z_{j+1}},\; j=1,\ldots,k-1$

Then use the inverse of $G_+(y)$ to sample random variable whose density function is $g_+(x)$. So we can use the following Julia program to sample from $g_+(x)$.

# example 3.22 in Davison(2008)r = 2; m = 10; mu = 0; sig2 = 1;# range of yyl = -2; yu = 2;# function of hfunction h(y)    return(r * y - m * log(1 + exp(y)) - (y - mu)^2 / (2 * sig2))end​function h(y::Array)    return(r * y - m * log.(1 .+ exp.(y)) .- (y .- mu) .^2 ./ (2 * sig2))end​# derivative of hfunction dh(y)    return(r - m * exp(y) / (1 + exp(y)) - (y - mu) / sig2)end​function dh(y::Array)    return(r .- m * exp.(y) ./ (1 .+ exp.(y)) .- (y .- mu)./sig2)end​# intersection pointfunction zfix(yfixed::Array)    yf0 = yfixed[1:end-1]    yf1 = yfixed[2:end]    zfixed = yf0 .+ (h(yf0) .- h(yf1) .+ (yf1 .- yf0) .* dh(yf1)) ./ (dh(yf1) .- dh(yf0))    return(zfixed)end​# evaluate log-density (not necessary)function hplus(y::Float64, yfixed::Array)    zfixed = zfix(yfixed)    n = size(zfixed, 1)    for i = 1:n        if i == 1 && y < zfixed[i]            return(h(yfixed[i]) + (y - yfixed[i]) * dh(yfixed[i]))        elseif i < n && y >= zfixed[i] && y < zfixed[i+1]            return(h(yfixed[i+1]) + (y - yfixed[i+1]) * dh(yfixed[i+1]))        elseif i == n && y >= zfixed[n]            return(h(yfixed[i]) + (y - yfixed[i]) * dh(yfixed[i]))        end    endend​# calculate G_+(z_i)function gplus_cdf(yfixed::Array, zfixed::Array)    n = size(zfixed, 1)    s = zeros(n+1)    pr = zeros(n+1)    for i = 1:(n+1)        ## integral from -infty to zi        if i == 1        #    s[i] = exp(h(yfixed[i])) / dh(yfixed[i]) * (exp((zfixed[i]-yfixed[i]) * dh(yfixed[i])) - )            s[i] = exp(h(yfixed[i])) / dh(yfixed[i]) * (exp((zfixed[i]-yfixed[i]) * dh(yfixed[i])) - exp((yl-yfixed[i]) * dh(yfixed[i])))        elseif i == n+1            s[i] = exp(h(yfixed[i])) / dh(yfixed[i]) * (exp((yu-yfixed[i]) * dh(yfixed[i])) - exp((zfixed[n]-yfixed[i]) * dh(yfixed[i])))        else            s[i] = exp(h(yfixed[i])) / dh(yfixed[i]) * (exp((zfixed[i]-yfixed[i]) * dh(yfixed[i])) - exp((zfixed[i]-yfixed[i]) * dh(yfixed[i])))        end    end    pr = s / sum(s)    return cumsum(pr), sum(s)end​# sample from gplus densityfunction gplus_sample(yfixed)    zfixed = zfix(yfixed)    gp = gplus_cdf(yfixed, zfixed)    zpr = gp    norm_const = gp    n = size(zfixed, 1)    u = rand()    # Invert the gplus pdf    for i = 1:n        if i == 1 && u < zpr[i]            ey = u * dh(yfixed[i]) * norm_const / exp(h(yfixed[i])) + exp((yl-yfixed[i])*dh(yfixed[i]))            return(yfixed[i] + log(ey)/dh(yfixed[i]))            elseif i == n && u >= zpr[i]            ey = (u - zpr[i]) * dh(yfixed[i+1]) * norm_const / exp(h(yfixed[i+1])) + exp((zfixed[i]-yfixed[i+1])*dh(yfixed[i+1]))            return(yfixed[i+1] + log(ey)/dh(yfixed[i+1]))        elseif i < n && u >= zpr[i] && u < zpr[i+1]            ey = (u - zpr[i]) * dh(yfixed[i+1]) * norm_const / exp(h(yfixed[i+1])) + exp((zfixed[i]-yfixed[i+1])*dh(yfixed[i+1]))            return(yfixed[i+1] + log(ey)/dh(yfixed[i+1]))        end    endend

Back to the main sampling algorithm, we can implement the procedure as follows:

## adaptive rejection samplingfunction ars(yfixed::Array)    x = gplus_sample(yfixed)    u = rand()    if u <= exp(h(x)-hplus(x, yfixed))        return(x)    else        return(ars(append!(yfixed, x)))    endend​## exampleN = 100res = ones(N);for i = 1:N    res[i] = ars([-1.8,-1.1,-0.5,-0.2])end

Based on the ARS algorithm, we can also get the Supplemental ARS algorithm:

# Exponential Random Variable

The inverse transform sampling from a uniform distribution can be easily used to sample an exponential random variable. The CDF is

$F(x) = 1-\exp(-\lambda x)\,,$

whose inverse function is

$F^{-1}(y) = -\frac{\log(1-y)}{\lambda}\,.$

Thus, we can sample $U\sim U(0, 1)$, and then perform the transfomation

$-\frac{\log U}{\lambda}\,.$

Today, I came across another method from Xi'an's blog, which points to the question on StackExchange.

Still confused about the details, as commented in the code.

The theoretical part is as follows,

I rewrite the provided C code in Julia, and compare the distribution with samples from inverse CDF and the package Distributions

a = [exp_rand() for i=1:1000]b = [-log(rand()) for i=1:1000]c = rand(Exponential(1), 1000)histogram(a, bins=40, label = "sexp")histogram!(b, bins=40, alpha=0.5, label = "invF")histogram!(c, bins=40, alpha=0.5, label = "rand")