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Monte Carlo Integration
Digital Image SynthesisYung-Yu Chuang
with slides by Pat Hanrahan and Torsten Moller
Introduction
• The integral equations generally don’t have analytic solutions, so we must turn to numerical methods.
• Standard methods like Trapezoidal integration or Gaussian quadrature are not effective for high-dimensional and discontinuous integrals.
)ωp,( ooL )ω,p( oeL
iiiio ωθcos)ωp,()ω,ωp,(2
dLf is
Numerical quadrature
• Suppose we want to calculate , but can’t solve it analytically. The approximations through quadrature rules have the form
which is essentially the weighted sum of samples of the function at various points
b
adxxfI )(
n
iii xfwI
1
)(ˆ
Midpoint rule
convergence
Trapezoid rule
convergence
Simpson’s rule
• Similar to trapezoid but using a quadratic polynomial approximation
convergence
assuming f has a continuous fourth derivative.
Curse of dimensionality and discontinuity• For an sd function f,• If the 1d rule has a convergence rate of
O(n-r), the sd rule would require a much larger number (ns) of samples to work as well as the 1d one. Thus, the convergence rate is only O(n-r/s).
• If f is discontinuous, convergence is O(n-1/s) for sd.
Randomized algorithms
• Las Vegas v.s. Monte Carlo• Las Vegas: always gives the right answer
by using randomness.• Monte Carlo: gives the right answer on the
average. Results depend on random numbers used, but statistically likely to be close to the right answer.
Monte Carlo integration
• Monte Carlo integration: uses sampling to estimate the values of integrals. It only requires to be able to evaluate the integrand at arbitrary points, making it easy to implement and applicable to many problems.
• If n samples are used, its converges at the rate of O(n-1/2). That is, to cut the error in half, it is necessary to evaluate four times as many samples.
• Images by Monte Carlo methods are often noisy. Most current methods try to reduce noise.
Monte Carlo methods
• Advantages– Easy to implement– Easy to think about (but be careful of statistical
bias)– Robust when used with complex integrands
and domains (shapes, lights, …)– Efficient for high dimensional integrals
• Disadvantages– Noisy– Slow (many samples needed for convergence)
Basic concepts
• X is a random variable• Applying a function to a random variable
gives another random variable, Y=f(X).
• CDF (cumulative distribution function)
• PDF (probability density function): nonnegative, sum to 1
• canonical uniform random variable ξ (provided by standard library and easy to transform to other distributions)
dx
xdPxp
)()(
}Pr{)( xXxP
Discrete probability distributions
• Discrete events Xi with probability pi
• Cumulative PDF (distribution)
• Construction of samples: To randomly select an event,
Select Xi if
0ip 1
1n
ii
p
1i iP U P
ip
1
j
j ii
P p
U
1
0Uniform random variable 3X
iP
Continuous probability distributions• PDF
• CDF1
0
( ) Pr( )P x X x
Pr( ) ( )
( ) ( )
X p x dx
P P
( )p x
10
Uniform
( ) 0p x
0
( ) ( )x
P x p x dx
(1) 1P ( )P x
Expected values
• Average value of a function f(x) over some distribution of values p(x) over its domain D
• Example: cos function over [0, π], p is uniform
Dp dxxpxfxfE )()()(
00
1cos)cos( dxxxE p
Variance
• Expected deviation from the expected value• Fundamental concept of quantifying the error
in Monte Carlo methods
2)()()]([ xfExfExfV
Properties
)()( xfaExafE
ii
ii XfEXfE )()(
)()( 2 xfVaxafV
22 )()()( xfExfExfV
Monte Carlo estimator
• Assume that we want to evaluate the integral of f(x) over [a,b]
• Given a uniform random variable Xi over [a,b], Monte Carlo estimator
says that the expected value E[FN] of the estimator FN equals the integral
N
iiN Xf
N
abF
1
)(
General Monte Carlo estimator
• Given a random variable X drawn from an arbitrary PDF p(x), then the estimator is
• Although the converge rate of MC estimator is O(N1/2), slower than other integral methods, its converge rate is independent of the dimension, making it the only practical method for high dimensional integral
N
i i
iN Xp
Xf
NF
1 )(
)(1
Convergence of Monte Carlo
• Chebyshev’s inequality: let X be a random variable with expected value μ and variance σ2. For any real number k>0,
• For example, for k= , it shows that at least half of the value lie in the interval
• Let , the MC estimate FN becomes
)(/)( iii XpXfY
N
iiN Y
NF
1
1
2
1}|Pr{|
kkX
2)2,2(
Convergence of Monte Carlo
• According to Chebyshev’s inequality,
• Plugging into Chebyshev’s inequality,
So, for a fixed threshold, the error decreases at the rate N-1/2.
YVN
YVN
YVN
YN
VFVN
ii
N
ii
N
iiN
1111][
12
12
1
21
][|][|Pr N
NN
FVFEF
21
][1||Pr
YV
NIFN
Properties of estimators
• An estimator FN is called unbiased if for all N
That is, the expected value is independent of N.
• Otherwise, the bias of the estimator is defined as
• If the bias goes to zero as N increases, the estimator is called consistent
QFE N ][
QFEF NN ][][
0][lim N
NF
QFE NN
][lim
Example of a biased consistent estimator• Suppose we are doing antialiasing on a 1d
pixel, to determine the pixel value, we need to evaluate , where is the filter function with
• A common way to evaluate this is
• When N=1, we have
1
0)()( dxxfxwI )(xw
1
01)( dxxw
N
i i
N
i iiN
Xw
XfXwF
1
1
)(
)()(
IdxxfXfEXw
XfXwEFE
1
011
111 )()]([
)(
)()(][
Example of a biased consistent estimator• When N=2, we have
• However, when N is very large, the bias approaches to zero
Idxdxxwxw
xfxwxfxwFE
1
0
1
0 2121
22112 )()(
)()()()(][
N
i i
N
i ii
N
XwN
XfXwNF
1
1
)(1
)()(1
Idxxfxwdxxw
dxxfxw
XwN
XfXwNFE
N
i iN
N
i iiN
NN
1
01
0
1
0
1
1)()(
)(
)()(
)(1
lim
)()(1
lim][lim
Choosing samples
•
• Carefully choosing the PDF from which samples are drawn is an important technique to reduce variance. We want the f/p to have a low variance. Hence, it is necessary to be able to draw samples from the chosen PDF.
• How to sample an arbitrary distribution from a variable of uniform distribution?– Inversion– Rejection– Transform
N
i i
iN Xp
Xf
NF
1 )(
)(1
Inversion method
• Cumulative probability distribution function
• Construction of samplesSolve for X=P-1(U)
• Must know:1. The integral of p(x)
2. The inverse function P-1(x)
U
X
1
0
( ) Pr( )P x X x
Proof for the inversion method
• Let U be an uniform random variable and its CDF is Pu(x)=x. We will show that Y=P-1(U) has the CDF P(x).
Proof for the inversion method
• Let U be an uniform random variable and its CDF is Pu(x)=x. We will show that Y=P-1(U) has the CDF P(x).
because P is monotonic,
Thus, Y’s CDF is exactly P(x).
)())(()(Pr)(PrPr 1 xPxPPxPUxUPxY u
2121 if )()( xxxPxP
Inversion method
• Compute CDF P(x)
• Compute P-1(x)
• Obtain ξ
• Compute Xi=P-1(ξ)
Example: power function
It is used in sampling Blinn’s microfacet model.nxxp )(
Example: power function
• Assume( ) ( 1) np x n x
11 1
0 0
1
1 1
nn x
x dxn n
1( ) nP x x
1 1~ ( ) ( ) nX p x X P U U
1 2 1max( , , , , )n nY U U U U 1
1
1
Pr( ) Pr( )n
n
i
Y x U x x
Trick
It is used in sampling Blinn’s microfacet model.
(It only works for sampling power distribution)
• Compute CDF P(x)
• Compute P-1(x)
• Obtain ξ
• Compute Xi=P-1(ξ)
Example: exponential distribution
useful for rendering participating media.axcexp )(
• Compute CDF P(x)
• Compute P-1(x)
• Obtain ξ
• Compute Xi=P-1(ξ)
Example: exponential distribution
useful for rendering participating media.axcexp )(
10
dxce ax ac
axx as edsaexP 1)(0
)1ln(1
)(1 xa
xP
ln1
)1ln(1
aaX
Rejection method
• AlgorithmPick U1 and U2
Accept U1 if U2 < f(U1)
• Wasteful?
1
0
( )
( )
y f x
I f x dx
dx dy
( )y f x
Efficiency = Area / Area of rectangle
• Sometimes, we can’t integrate into CDF or invert CDF
Rejection method
• Rejection method is a dart-throwing method without performing integration and inversion.
1. Find q(x) so that p(x)<Mq(x)
2. Dart throwinga. Choose a pair (X, ξ), where X is sampled from
q(x)
b. If (ξ<p(X)/Mq(X)) return X
• Equivalently, we pick a point (X, ξMq(X)). If it lies beneath p(X) then we are fine.
Why it works
• For each iteration, we generate Xi from q. The sample is returned if ξ<p(X)/Mq(X), which happens with probability p(X)/Mq(X).
• So, the probability to return x is
whose integral is 1/M• Thus, when a sample is returned (with
probability 1/M), Xi is distributed according to p(x).
M
xp
xMq
xpxq
)(
)(
)()(
Example: sampling a unit disk
void RejectionSampleDisk(float *x, float *y) {
float sx, sy;
do {
sx = 1.f -2.f * RandomFloat();
sy = 1.f -2.f * RandomFloat();
} while (sx*sx + sy*sy > 1.f)
*x = sx; *y = sy;
}
π/4 ～ 78.5% good samples, gets worse in higher dimensions, for example, for sphere, π/6 ～ 52.4%
Transformation of variables
• Given a random variable X from distribution px(x) to a random variable Y=y(X), where y is one-to-one, i.e. monotonic. We want to derive the distribution of Y, py(y).
• • PDF:
)(}Pr{)}(Pr{))(( xPxXxyYxyP xy
dx
xdP
dx
xydPxy )())((
)(xpx
dx
dy
dy
ydP
dx
dyyp y
y
)()(
)()(1
xpdx
dyyp xy
x
y
Px(x)
Py(y)
Example
XY
xxpx
sin
2)(
2
11
1
sin2
cos
2)()(cos)(
y
y
x
xxpxyp xy
Transformation method
• A problem to apply the above method is that we usually have some PDF to sample from, not a given transformation.
• Given a source random variable X with px(x) and a target distribution py(y), try transform X into to another random variable Y so that Y has the distribution py(y).
• We first have to find a transformation y(x) so that Px(x)=Py(y(x)). Thus,
))(()( 1 xPPxy xy
Transformation method
• Let’s prove that the above transform works. We first prove that the random variable Z=
Px(x) has a uniform distribution. If so, then
should have distribution Px(x) from the inversion method.
Thus, Z is uniform and the transformation works.
• It is an obvious generalization of the inversion method, in which X is uniform and Px(x)=x.
)(1 ZPy
xxPPxPXxXPxZ xxxx ))(()(Pr)(PrPr 11
Example
xxpx )( yy eyp )(y
Thus, if X has the distribution , then the random variable has the distribution
Example
xxpx )( yy eyp )(
2)(
2xxPx y
y eyP )(
2lnln2)2
ln())(()(2
1 xx
xPPxy xy
yyPy ln)(1
2lnln2 XYxxpx )(
yy eyp )(
y
Multiple dimensions
We often need the other way around,
Spherical coordinates
• The spherical coordinate representation of directions is
cos
sinsin
cossin
rz
ry
rx
sin|| 2rJT
),,(sin),,( 2 zyxprrp
Spherical coordinates
• Now, look at relation between spherical directions and a solid angles
• Hence, the density in terms of
ddd sin , dpddp )(),(
)(sin),( pp
Multidimensional sampling
• Separable case: independently sample X from px and Y from py. p(x,y)=px(x)py(y)
• Often, this is not possible. Compute the marginal density function p(x) first.
• Then, compute the conditional density function
• Use 1D sampling with p(x) and p(y|x).
dyyxpxp ),()(
)(
),()|(
xp
yxpxyp
Sampling a hemisphere
• Sample a hemisphere uniformly, i.e.
• Sample first
• Now sampling
2
1)( p
2
sin),( p
)(1 p2
1c
sin2
sin),()(
2
0
2
0
ddpp
2
1
)(
),()|(
p
pp
cp )(
Sampling a hemisphere
• Now, we use inversion technique in order to sample the PDF’s
• Inverting these:
cos1''sin)(0
dP
2'
2
1)|(
0
dP
2
11
2
cos
Sampling a hemisphere
• Convert these to Cartesian coordinate
• Similar derivation for a full sphere
2
11
2
cos
1
211
211
cos
1)2sin(sinsin
1)2cos(cossin
z
y
x
Sampling a disk
1
2
2 U
r U
1
2
2 U
r U
RIGHT Equi-ArealWRONG Equi-Areal
Sampling a disk
RIGHT Equi-ArealWRONG Equi-Areal
1
2
2 U
r U
1
2
2 U
r U
Sampling a disk
• Uniform
• Sample r first.
• Then, sample .
• Invert the CDF.
1
),( yxp
ryxrprp ),(),(
rdrprp 2),()(2
0
2
1
)(
),()|(
rp
rprp
2)( rrP 2
)|( rP
1r 22
Shirley’s mapping
1
2
14
r U
U
U
Sampling a triangle
121 1 1
0 0 00
(1 ) 1(1 )
2 2
u uA dv du u du
( , ) 2p u v
0
0
1
u
v
u v
u
v
1u v
( ,1 )u u
)(
),()|(
vp
vupvup
Sampling a triangle
• Here u and v are not independent!• Conditional probability
( , ) 2p u v
00 0 00 0
0 0
1( | ) ( | )
(1 ) (1 )o ov v v
P v u p v u dv dvu u
1( | )
(1 )p v u
u
0 20 00
( ) 2(1 ) (1 )u
P u u du u
1
0
( ) 2 2(1 )u
p u dv u
0 11u U
0 1 2v U U
dvvupup ),()(
Cosine weighted hemisphere
cos)( p
dp )(1
2
0 0
2 sincos1 ddc
ddd sin
2
0sincos21
dc
1c
sincos1
),( p
Cosine weighted hemisphere
sincos1
),( p
2sinsincos2sincos1
)(2
0 dp
2
1
)(
),()|(
p
pp
12
12cos
2
1)( P
22)|(
P
)21(cos2
11
1
22
Cosine weighted hemisphere
• Malley’s method: uniformly generates points on the unit disk and then generates directions by projecting them up to the hemisphere above it.
Vector CosineSampleHemisphere(float u1,float u2){
Vector ret;
ConcentricSampleDisk(u1, u2, &ret.x, &ret.y);
ret.z = sqrtf(max(0.f,1.f - ret.x*ret.x -
ret.y*ret.y));
return ret;
}
r
Cosine weighted hemisphere
• Why does Malley’s method work?• Unit disk sampling• Map to hemisphere
r
rp ),(
),(sin),( r
sinr
),( rY ),( XT
)()())((1
xpxJxTp xTy
cos10
0cos)(
xJT
Cosine weighted hemisphere
sinr
),( rY ),( XT
)()())((1
xpxJxTp xTy
cos10
0cos)(
xJT
sincos
),(),( rpJp T
Sampling Phong lobe
np cos)(
ncp cos)(
2
0
2/
0
1sincos ddc n
0
1cos
1coscos2
dc n 11
2
n
c
2
1
nc
sincos2
1),( nn
p
Sampling Phong lobe
'cos1
1
cos)1(coscos)1(
sincos)1()'(
sincos)1(sincos2
1)(
1
'cos
1cos
1'
0
'
0
2
0
n
nn
n
nn
nndn
dnP
ndn
p
11
1cos n
sincos2
1),( nn
p
Sampling Phong lobe
2
'
2
1)|'(
2
1
sincos)1(
sincos
)(
),()|(
'
0
21
dP
np
pp
n
nn
22
sincos2
1),( nn
p
Sampling Phong lobe
)2,(cos),( 1,n 211
211 2),21(cos
2
1),(
cosine-weighted hemisphere
2
12cos
2
1)( P 21 cos1cos1)( nP
222 cos1sin2
1)sin21(
2
1
2
12cos
2
1
When n=1, it is actually equivalent to
