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

# Conjugate Gradient Monte Carlo (CGMC)

where MTM refers to
and its simplified version orientational bias Monte Carlo (OBMC) when
$w(\mathbf x, \mathbf y)=\pi(\mathbf x)$
:

## Bimodal Example

whose contour plot is
For the general MH procedure:
function propose(x; a = 4)
θ = rand() * 2π
r = rand() * 4
return [x + r*cos(θ), x + 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 / 2π * exp(-0.5*sum(x.^2))
term2 = w / (2π * 0.19) * exp( -0.5 * transpose(x.+6) * invΣ2 * (x.+6) )
term3 = w / (2π * 0.19) * exp( -0.5 * transpose(x.-4) * invΣ3 * (x.-4) )
term = term1 .+ term2 * invΣ2 + term3 * invΣ3
dx1 = term[1,1]*x + term[1,2]*x
dx2 = term[2,1]*x + term[2,2]*x
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,