# 4Probability Background

Monte Carlo methods involve computations done with the help of random numbers. To reason about Monte Carlo methods, we need a little background in probability theory. I assume that you have seen some probability theory before, and that this is just a reminder. If you need a more thorough refresher, the book by Ross (2014) is a popular introductory text that covers discrete and continuous problems, but not more general probability measures. Another good undergraduate text by Chung and AitSahlia (2003) includes a little bit of measure theory. Good graduate texts include the books by Billingsley (1995) and by Breiman (1992). If you want a reminder that is more thorough than the one we give here, but less than a full textbook, the treatment in is a good starting point.

## 4.1 Probability basics

When we do an experiment, there are a variety of possible outcomes that could result. These outcomes are described in terms of a sample space $$S$$. An event is a set $$A \subset S$$.1 A probability measure is a function $$P$$ mapping events to non-negative real numbers such that $$P(S) = 1$$ and $$P(\cup_{i=1}^\infty A_i) = \sum_{i=1}^\infty P(A_i)$$ for any countable collection of pairwise disjoint events $$A_i$$.

Rather than work directly with the sample space, we usually consider random variables. A random variable $$X$$ is a function on $$S$$.2 In the name of concise notation, we often suppress the argument to $$X$$, writing expressions like $$\{X \in A\}$$ to denote $$\{s \in S : X(s) \in A\}$$.

For a discrete random variable, we write $P\{X \in A\} = \sum_{x \in A} p_X(x)$ where $$p_X$$ is a probability mass function (pmf) which is everywhere between zero and one and which sums to one when the sum is taken over all possible outcomes. Similarly, for a continuous random variable3, we write $P\{X \in A\} = \int_{A} f_X(x) \, dx$ where $$f_X(x)$$ is a probability density function (pdf). When the outcomes are integers or real numbers, we sometimes also care about the cumulative distribution function (cdf) $F_X(x) = P\{X \leq x\}$ which we can get by summing the mass function or integrating the density function. The cdf is a monotonically increasing functions with limiting values $$\lim_{x \rightarrow -\infty} F_X(x) = 0$$ and $$\lim_{x \rightarrow \infty} F_X(x) = 1$$.

The expected value of a function $$g$$ of a random variable $$X$$ is $E[X] = \int_{\Omega} g(x) f_X(x) \, dx;$ in the discrete case, the integral is replaced by a sum. The variance of $$X$$ is $\operatorname{Var}[X] = E[(X-E[X])^2] = E[X^2]-E[X]^2.$ The standard deviation is the square root of the variance, and we can think of it as a measure of how far, on average, $$X$$ is from its expected value.

Random variables $$X$$ and $$Y$$ are independent if for general choices of events $$A$$ and $$B$$ we have $$P(\{ X \in A \} \cap \{ Y \in B \}) = P\{X \in A\} \cdot P\{Y \in B\}$$. In simple Monte Carlo calculations, we typically run repeated experiments that are independent and identically distributed (i.i.d.). If $$X_1, X_2, \ldots, X_N$$ are independently drawn from the same distribution (with finite mean and variance), then by the central limit theorem, the sample mean $\bar{X} = \frac{1}{N} \sum_{j=1}^N X_j$ is a random variable that is approximately normal (Gaussian) with mean $$E[X]$$ and variance $$\operatorname{Var}[X]/\sqrt{N}$$.

## 4.2 The Monte Carlo idea

Monte Carlo methods use random numbers to compute something that is not random. In the abstract, we write some quantity of interest $$A$$ as $A = E_f[V(X)],$ where $$X$$ is a collection of random variables whose joint distribution is $$f$$ (sometimes written $$X \sim f$$) and $$V(x)$$ is some quantity determined by $$X$$. A Monte Carlo code generates many samples $$X_k$$, $$k = 1, \ldots, N$$, from the distribution $$f$$, and then computes the approximate answer $A \approx \hat{A}_N = \frac{1}{N} \sum_{k=1}^N V(X_k).$ If the samples $$X_k$$ are independent, the error is roughly $$\sigma/\sqrt{N}$$, where $$\sigma^2 = \operatorname{var}_f(V(X))$$ is the variance of the random variable $$V(X)$$. If we don’t know the variance of $$V(X)$$ analytically (which is typically the case), we can use the estimate $\hat{\sigma}^2_N = \frac{1}{N-1} \sum_{k=1}^N (V(X_k)-A)^2.$ Sometimes we’re sloppy and divide by $$N$$; if $$N$$ is small enough that this makes a significant data, we ideally should run more experiments! When we approximate $$A$$ by $$\hat{A}_N$$, we call $$\hat{\sigma}_N$$ an “error bar”, since it describes a measure of the statistical error in our problem (the radius of a symmetric 67% confidence interval). The error bars are not the same as error bounds, of course, but they are useful for reasoning about the order of magnitude of the errors we expect to see.

Because statistical error is $$O(1/\sqrt{N})$$, it tends to be very expensive to get high accuracy with Monte Carlo methods. For some problems, though, particularly those in high dimensions, Monte Carlo methods are the most practical choice. The basic idea of Monte Carlo is simple, if expensive; much of the cleverness in Monte Carlo methods goes into variance reduction, which at least reduces the constant in the $$O(1/\sqrt{N})$$ expression. The good side of statistical error is that it is usually at least possible to estimate its order of magnitude (via error bars).

## 4.3 Examples

Monte Carlo methods have relatively low accuracy compared to deterministic methods, but they are particularly useful in a few cases:

1. Some problems are naturally probabilistic, and a Monte Carlo method may be an almost-direct translation of the problem statement. If we don’t mind low accuracy, this can be a very effective way to get a feel for the answer before diving into a more exact calculation (which we might have to spend more time debugging). The standard advice is to only use Monte Carlo for things that cannot be well managed by deterministic methods; this sort of exploratory computation might be an exception.
2. The cost of deterministic methods often grows exponentially with the dimension of the ambient space. This causes a problem when we’re interested in even moderately high dimensions. For computing integrals in high-dimensional spaces (including the sort of position-and-direction coordinates we need to describe particles in scattering problems like the one in HW 3), a Monte Carlo method is often appropriate.
3. Sometimes we are driven by data, and the data that we have is too huge to process all at once. Sampling the data by Monte Carlo methods can be a very effective approach in this case for the same reason it is effective for high-dimensional problems: the cost depends on the number of samples we draw, and not on the size or dimension of the underlying thing from which our samples are drawn.

## 4.4 Random number generation

In order to run Monte Carlo simulations, we need a source of pseudo-random numbers. One could teach an entire class on how to produce pseudo-random number generators, but we will simply state that it is a tricky business and you should use a well-designed library routine for your day-to-day draws of random bits or of numbers that are uniformly distributed in the interval $$[0,1]$$. In Julia, you can use rand to get such uniformly distributed random samples (and randn to get samples from a standard normal distribution). For our purposes, we simply need to know how to turn such uniform sampling procedures into methods to sample from other distributions. I know a handful of tricks for deriving new samplers; let’s investigate them by example.

### 4.4.1 Bernoulli random variables

A Bernoulli random variable generates 1 (success) with probability $$p$$ and 0 (failure) with probability $$1-p$$. In the problem du jour, we implicitly assumed that each question was a Bernoulli trial with $$p = 0.8$$. Generating a Bernoulli trial from a uniform sampler is relatively simple; in Julia, we might write

bernoulli(p) = if rand() < p 1 else 0 end
bernoulli (generic function with 1 method)

Note that $$P\{U < p\} = \int_0^p 1 \, du = p$$, so this sampler certainly has the right properties. Also notice that we can compute a vector or matrix of Bernoulli trials simultaneously with one call to rand, where the arguments give the output size.

### 4.4.2 Exponential random variables

An exponential random variable with rate parameter $$\lambda$$ has the density function $f(x; \lambda) = \lambda e^{-\lambda x}, \quad x \geq 0$ and the cumulative distribution function $F(x; \lambda) = 1-e^{-\lambda x}.$ Now, suppose that for a uniform sample $$U$$ we generate $$X$$ to satisfy $$F(X; \lambda) = U$$, i.e. $X = -\frac{1}{\lambda} \log(1-U).$ Then $P\{X \leq x\} = P\{F(X;\lambda) \leq F(x;\lambda)\} = P\{U \leq F(x;\lambda)\} = F(x; \lambda).$ This inverse transformation trick works whenever we have a simple way to compute a cumulative distribution function. In this particular case, we might also note that $$U$$ and $$1-U$$ have the same distribution, so we could also use $X' = -\frac{1}{\lambda} \log(U).$

rand_exp(λ) = -log(rand())/λ
rand_exp (generic function with 1 method)

### 4.4.3 Sampling from an empirical distribution

Suppose we have a histogram of results from some large number of real-world experiments. If the outcomes of the experiments are integers in the range from $$1$$ to $$m$$, we can define a probability mass function where $$p(j)$$ is the fraction of the experiments that had outcome $$j$$. There is a corresponding cumulative distribution function $$F(j) = \sum_{i=1}^j p(j)$$ that goes from $$F(0) = 0$$ to $$F(m) = 1$$. To draw a sample from this distribution, we would again use the inverse transformation trick: draw $$U$$ uniform between $$0$$ and $$1$$, then choose the smallest $$j$$ such that $$F(j) > U$$.

### 4.4.4 Sampling from the unit disk

Suppose we want to draw $$(X,Y)$$ uniformly at random from the interior of the unit circle. One way to do this is with polar coordinates: if $$U_1$$ and $$U_2$$ are uniform on $$(0,1)$$, we can generate $$\Theta = 2 \pi U_1$$ and $$R = \sqrt{U_2}$$ (the cdf for $$R$$ should be $$F_R(r) = r^2$$ on $$[0,1)$$, so we can use the inverse transformation trick from above). Then we could compute $$(X,Y) = R (\sin \Theta, \cos \Theta)$$. But suppose we didn’t know this, or suppose that we’re thinking of the disk as a proxy for some more complicated set sitting inside the unit square. What other tactics could we use?

One simple idea is rejection sampling. The basic idea is

1. Draw a sample from an easy distribution $$g$$. In this case, we might use the uniform distribution on $$[-1,1]^2$$ (i.e. $$g(x,y) = 1/4$$ on $$[-1,1]^2$$ and zero elsewhere).
2. Accept the sample with probability that is a function of the sample values. In this case, we have $p(x,y) = \begin{cases} 1, & x^2+y^2 < 1 \\ 0, & \mbox{otherwise}. \end{cases}$ In this case, we accept with probability one if $$X^2+Y^2 < 1$$ and with probability zero otherwise.

We then keep repeating until acceptance. The probability density associated with the accepted values is then $f(x,y) = \frac{1}{Z} g(x,y) p(x,y)$ where $$Z$$ is some normalization constant chosen so that the acceptance probability is one. In our case, this gives us a density that is a nonzero constant on the circle and zero elsewhere, which is what we wanted.

A more geometric way of seeing rejection sampling is that we fit some shape that completely surrounds the graph of our density function (in this case, that shape is a three-dimensional box). We then draw uniformly at random from within that shape, and discard the samples that do not fall under the graph of the density function. The probability that we succeed in any given trial is equal to the fraction of the area inside the shape that lies underneath the graph of the density function.

### 4.4.5 Distribution with an exponential tail

Let’s look at another example of rejection sampling. Suppose I wanted to sample from $$f(x) = C^{-1} g(x) e^{-x}$$ on $$[0,\infty)$$, where $$C$$ is some (possibly unknown) normalization constant and $$0 < g(x) < G$$. Then I could compute samples from $$f$$ using the following procedure:

function sample_exp_tail(g, gmax)
p = 0.0
X = 0.0
while rand() > p
X = -log(rand())
p = g(x)/G
end
X
end
sample_exp_tail (generic function with 1 method)

The probability of success in this problem is the ratio of the area under the histogram for $$f$$ to the area under $$G e^{-x}$$, or $$1/G$$. The expected number of rounds until success is therefore $$G$$.

## 4.5 Variance reduction

The problem above is a simple example of Monte Carlo integration. We now want to see how to make this simple example more efficient by reducing the variance of the estimator. We will approach this in a few different ways.

### 4.5.1 Importance sampling

Let us consider the computation $\sqrt{2\pi} = 2 \int_{0}^\infty \exp(-x^2/2) \, dx.$ Using the idea of the problem du jour, we could estimate $$\sqrt{2\pi}$$ by drawing uniform samples on $$[0,L]$$ for $$L$$ large enough. But this estimator has rather high variance, and the variance gets larger the larger $$L$$ is. This is intuitive in that most of the sample points don’t really matter to the computation, since $$\exp(-x^2/2)$$ decays very quickly away from zero.

The integrand $$\exp(-x^2/2)$$ is largest near the origin, so we get the most contribution to our integral when we have samples near zero. Therefore, it makes sense to use a method that samples more frequently near the origin, rather than sampling uniformly over some large range of $$x$$ values $\sqrt{2\pi} = 2\int_{0}^\infty \frac{\exp(-x^2/2)}{\exp(-x)} \exp(-x) \, dx = 2E[\exp(-X^2/2)/\exp(-X)]$ where $$X$$ is an exponential random variable.

function zmean2(N=1000)
Y = -log.(rand(N))
fY = exp.(-Y.^2/2)
gY = exp.(-Y)/2
lY = fY./gY
mean(lY), std(lY)/sqrt(N)
end

μ, σ = zmean2()
μ-sqrt(2π), σ
(-0.0429320967078084, 0.026675651732767005)

### 4.5.2 Control variates

I want to compute the expectation of $$l(x) = \exp(x-x^2/2)$$, but perhaps I’ve decided its too hard. But I know that most of the interesting behavior is near the origin, so perhaps I can approximate $$l(x)$$ by a polynomial over some interval close to zero. Let’s try just interpolating by a quadratic at $$x = 0$$, $$x = 1$$, and $$x = 2$$, and discarding everything past $$x = 2$$: $h(x) = \begin{cases} \sqrt{e}-(\sqrt{e}-1)(x-1)^2, & x \in [0,2] \\ 0, & \mbox{otherwise} \end{cases}.$ While $$h(X)$$ is not identical to $$l(X)$$, the two random variables surely are correlated. Furthermore, we can compute $$E[h(X)]$$ analytically; a somewhat tedious calculus exercise yields $E[h(X)] = \sqrt{e}(1-e^{-2}) - (\sqrt{e}-1)(1-5e^{-2}).$ The fact that $$h(X)$$ and $$l(X)$$ should be correlated, together with the fact that we can compute $$E[h(X)]$$ in closed form, makes $$h(X)$$ an ideal candidate to serve as a control variate with which we can construct a better estimator, as we shall now see.

First, note that $E[l(X)] = E[l(X)-ch(X)]+cE[h(X)].$ So $$\hat{l}_c(X) = l(X)-ch(X)+cE[h(X)]$$ has the same expected value that $$l(X)$$ does; but $\operatorname{Var}[\hat{l}_c(X)] = \operatorname{Var}[l(X)] -2c \operatorname{Cov}[l(X),h(X)] + c^2 \operatorname{Var}[h(X)].$ If we choose $$c_* = \operatorname{Cov}[l(X),h(X)]/\operatorname{Var}[h(X)]$$, we have \begin{align*} \operatorname{Var}[\hat{l}_{c_*}(X)] = \operatorname{Var}[l(X)] \left( 1-\operatorname{corr}[l(X),h(X)]^2 \right). \end{align*} If $$l(X)$$ and $$h(X)$$ are highly correlated, then $$\hat{l}_{c_*}(X)$$ may have a much lower variance than $$l(X)$$. Of course, computing the covariance analytically is hard, but we can always do it numerically.

function zmean3(N=1000)
e   = exp(1.0)
Y   = -log.(rand(N))
fY  = exp.(-Y.^2/2)
gY  = exp.(-Y)/2
lY  = fY./gY
hY  = (sqrt(e).-(sqrt(e)-1)*(Y.-1.0).^2) .* (Y.<2)
EhY = (sqrt(e)*(1-e^-2) - (sqrt(e)-1)*(1-5*e^-2))
cs  = -sum((lY.-mean(lY)) .* (hY.-EhY))/sum((hY.-EhY).^2)
W   = lY + cs*(hY.-EhY)
mean(W), std(W)/sqrt(N)
end

μ, σ = zmean3()
μ-sqrt(2π), σ
(-0.0016202242656349064, 0.008885346339627058)

### 4.5.3 Antithetic variates

Now let’s turn to the problem of computing $$\pi/4$$ by throwing darts at $$[0,1]^2$$ and seeing what fraction lie inside the unit circle. Note that if $$(X_i,Y_i)$$ is a uniform random sample from the square, then $$(1-X_i,1-Y_i)$$ is a correlated sample. It turns out that if $$\phi$$ is the indicator for the unit circle, then $$\phi(X_i,Y_i)$$ and $$\phi(1-X_i,1-Y_i)$$ have negative covariance; this makes sense, since only one of them can be outside the unit circle (though both could be the same). Therefore, the estimator $$\phi(X,Y)/2+\phi(1-X,1-Y)/2$$ actually has lower variance than $$\phi(X,Y)$$. This is the method of antithetic variables.

function pi_mc(N=1000)
XY  = rand(2,N)
XY2 = 1.0 .- XY
trials1 = [xyj[1]^2+xyj[2]^2 < 1 for xyj in eachcol(XY) ]
trials2 = [xyj[1]^2+xyj[2]^2 < 1 for xyj in eachcol(XY2)]
trials = (trials1+trials2)/2
mean(trials1), std(trials1)/sqrt(N), mean(trials), std(trials)/sqrt(N)
end

pi_mc()
(0.768, 0.0133549374522816, 0.7865, 0.007824894822231095)

1. When the sample space is uncountable, we cannot generally define probabilities for arbitrary subsets of $$S$$. We therefore require events belong to a sigma algebra (also called a Borel field) $$\mathcal{B}$$; this class of sets must contain the empty set and be closed under complement and countable union. A set in the sigma algebra is called a measurable set.↩︎

2. A random variable $$X : S \rightarrow T$$ must be measurable; that is, if $$A$$ is a measurable set in $$T$$, then $$\{X \in A\} = \{s \in S : X(s) \in A\}$$ must also be measurable.↩︎

3. In order to have a probability density function, we technically want an absolutely continuous random variable.↩︎