# PBMCMC

This note is mainly based on Chapter 12 of Liu (2008).

where MTM refers to

and its simplified version

**orientational bias Monte Carlo (OBMC)**when$w(\mathbf x, \mathbf y)=\pi(\mathbf x)$

:whose contour plot is

For the general MH procedure:

function propose(x; a = 4)

θ = rand() * 2π

r = rand() * 4

return [x[1] + r*cos(θ), x[2] + r*sin(θ)]

end

function iMH(M = 200000)

# start points

x = [rand() - 0.5, rand() - 0.5]

X = ones(M, 2)

num = 0

for m = 1:M

xs = propose(x)

ρ = targetpdf(xs) / targetpdf(x)

if rand() < ρ

x = xs

num += 1

end

X[m,:] = x

end

return X, num/M

end

For CGMC, there is a long way to go. Firstly, find out the derivative of target pdf and find the anchor point:

# derivative of targetpdf

function dm(x; step = 0.05, nmaxstep = 100)

term1 = w[1] / 2π * exp(-0.5*sum(x.^2))

term2 = w[2] / (2π * 0.19) * exp( -0.5 * transpose(x.+6) * invΣ2 * (x.+6) )

term3 = w[3] / (2π * 0.19) * exp( -0.5 * transpose(x.-4) * invΣ3 * (x.-4) )

term = term1 .+ term2 * invΣ2 + term3 * invΣ3

# negative of gradient direction (remove the minus symbol already)

dx1 = term[1,1]*x[1] + term[1,2]*x[2]

dx2 = term[2,1]*x[1] + term[2,2]*x[2]

dx = [dx1, dx2]

# find local maximum

bst = targetpdf(x)

i = 0

while i < nmaxstep

i += 1

cur = targetpdf(x .- step * dx)

if cur > bst

bst = cur

x = x .- step * dx

else

break

end

end

return x

end

Then implement MTM algorithm for sampling radius

$r$

:function MTM(anchor, direction;M = 100, k = 5, σ = 10)

# use the simplified version: OBMC algorithm

# initial x

x = 2*rand()-1

num = 0

for m = 1:M

y = ones(k)

wy = ones(k)

for i = 1:k

y[i] = rand(Normal(x, σ))

wy[i] = f(y[i], anchor, direction)

end

#wy = wy ./ sum(wy)

yl = sample(y, pweights(wy))

# draw reference points

xprime = ones(k)

xprime[k] = x

wxprime = ones(k)

wxprime[k] = f(x, anchor, direction)

for i = 1:k-1

xprime[i] = rand(Normal(yl, σ))

wxprime[i] = f(xprime[i], anchor, direction)

end

# accept or not

ρ = sum(wy) / sum(wxprime)

if rand() < ρ

x = yl

num += 1

end

end

return x, num/M

end

Finally, we can combine these sub-procedures:

function cgmc(M = 200000)

# start points (each col is a stream)

x = rand(2, 2) .- 0.5

X = ones(M, 4)

for m = 1:M

# randomly choose xa

xaIdx = sample([1,2])

xa = x[:,xaIdx]

# find gradient of pdf at xa (or conjugate gradient of log pdf at xa)

y = dm(xa)

xcIdx = 3 - xaIdx

xc = x[:,xcIdx]

e = y - xc

e = e / norm(e)

# sample r

r, = MTM(y, e)

# update xc

x[:,xcIdx] = y .+ r*e

X[m, :] .= vec(x)

end

return X

end

Refer to bimodal-example.jl for the complete source code. Run the program, I can reproduce Fig 11.1 successfully,

Last modified 4yr ago