Gibbs抽样详解：入门级MCMC在自然语言处理中的应用

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Gibbs Sampling是一种基于概率的迭代抽样方法，它在统计学和计算机科学中尤其在马尔可夫链蒙特卡罗(Markov Chain Monte Carlo, MCMC)技术中占据重要地位。MCMC是一类通过模拟随机过程来估计难以直接计算的复杂概率分布的方法，常用于解决高维贝叶斯模型的参数估计问题，如在文本处理中的应用。在这篇名为《Gibbs Sampling for the Uninitiated》的论文中，作者Philip Resnik和Eric Hardisty针对对计算机科学家而言，特别是那些希望尝试使用MCMC技术进行文本处理中贝叶斯模型推理的人群进行了详尽的讲解。他们强调了理论知识的最小化，但又提供了比常规入门教程更明确的细节和示例。论文首先介绍了Gibbs抽样的基本原理，它利用了条件独立性假设，即在高维空间中，通过只考虑与当前状态相关的其他状态来逐步更新变量的值，从而达到探索整个分布的目的。这在贝叶斯网络中尤其有效，因为它允许我们有效地处理变量间的复杂依赖关系。接着，论文深入探讨了Gibbs抽样在朴素贝叶斯模型（Naive Bayes）中的具体应用。朴素贝叶斯模型是一种简单而有效的分类算法，它假设特征之间相互独立，尽管在现实世界中这种假设可能不成立，但在许多场景下仍具有良好的性能。通过Gibbs抽样，可以有效地估计每个特征的后验概率，这对于文档分类、情感分析等任务中的特征推断至关重要。在论文中，作者详细展示了如何为朴素贝叶斯模型构建一个Gibbs sampler，包括步骤、公式和具体的实现细节。他们还讨论了这个过程中的注意事项，以及如何处理可能出现的收敛问题和采样效率。这篇论文为计算机科学家提供了一个易于理解的Gibbs抽样入门指南，尤其是在文本挖掘和机器学习领域使用贝叶斯方法时。通过实际的Naive Bayes模型案例，读者能够掌握如何将Gibbs抽样技巧应用于实际问题，并进一步了解如何将其与其他MCMC方法结合，以提高模型的准确性和效率。

1.2 Why sampling?

The trouble with integrals, of course, is that they can be very diﬃcult to calculate. The methods we learned

in calculus class are ﬁne for classroom exercises, but often cannot be applied to interesting problems in the

real world. Indeed, analytical solutions to (8) and the denominator of (9) might be impossible to obtain, so

we might not be able to determine the exact form of P(π|X ). Gibbs sampling allows us to sample from a

distribution that asymptotically follows P (π|X ) without having to explicitly calculate the integrals.

1.2.1 Monte Carlo: a circle, a square, and a bag of rice

Gibbs Sampling is an instance of a Markov Chain Monte Carlo technique. Let’s start with the “Monte

Carlo” part. You can think of Monte Carlo methods as algorithms that help you obtain a desired value by

performing simulations involving probabilistic choices. As a simple example, here’s a cute, low-tech Monte

Carlo technique for estimating the value of π (the ratio of a circle’s circumference to its diameter).

Draw a perfect square on the ground. Inscribe a circle in it — i.e. the circle and the square are centered

in exactly the same place, and the circle’s diameter has length identical to the side of the square. Now take

a bag of rice, and scatter the grains uniformly at random inside the square. Finally, count the total number

of grains of rice inside the circle (call that C), and inside the square (call that S).

You scattered rice at random. Assuming you managed to do this pretty uniformly, the ratio between the

circle’s grains and the square’s grains (which include the circle’s) should approximate the ratio between the

area of the circle and the area of the square, so

≈

π(

)

. (10)

Solving for π, we get π ≈

You may not have realized it, but we just solved a problem by approximating the values of integrals. The

true area of the circle, π(

)

, is the result of summing up an inﬁnite number of inﬁnitessimally small points;

similarly for the the true area d

of the square. The more grains of rice we use, the better our approximation

will be.

1.2.2 Markov Chains: walking the right walk

In the circle-and-square example, we saw the value of sampling involving a uniform distribution, since the

grains of rice were distributed uniformly within the square. Returning to the problem of computing expected

values, recall that we’re interested in E

p(x)

[f(x)] (equation 7), where we’ll assume that the distribution p(x)

is not uniform and, in fact, not easy to work with analytically.

Figure 2 provides an example f(z) and p(z) for illustration. Conceptually, the integral in equation (7)

sums up f (z)p(z) over inﬁnitely many values of z. But rather than touching each point in the sum exactly

once, here’s another way to think about it: if you sample N points z

(0)

, z

(1)

, z

(2)

, . . . , z

(N)

at random from

the probability density p(z), then

p(z)

[f(z)] = lim

N→∞

t=1

f(z

(t)

). (11)

That looks a lot like a kind of averaged value for f, which makes a lot of sense since in the discrete case

(equation 6) the expected value is nothing but a weighted average, where the weight for each value of z is

its probability.

Notice, though, that the value in the sum is just f(z

(t)

), not f (z

(t)

)p(z

(t)

) as in the integral in equation (7).

Where did the p(z) part go? Intuitively, if we’re sampling according to p(z), and count(z) is the number of

We’re elaborating on the introductory example at http://en.wikipedia.org/wiki/Monte Carlo method.

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