# A Quick Introduction to Importance Sampling

You might have encountered the term **importance sampling** in a machine learning paper. This article provides a quick hands-on introduction to this topic. Hopefully after reading it you will have know understand how to use this technique in practice.

## A quick review of sample averages

Let $x \sim f(x)$ be a continuous random variable. The expected value of $x$ is

\[E_{x\sim f}\left[x\right] = \int x f(x) dx\]More generally the expected value of any function $h(x)$ is

\[E_{x\sim f}\left[h(x)\right] = \int h(x) f(x) dx\]Sometimes such as when $f(x)$ is the Gaussian distribution this integral can be evaluated exactly. But when this is not possible you can approximate using **sample averages**. Take $N$ samples $x_i \sim f(x)$ and approximate the integral by a the mean of these samples

By the **law of large numbers** as $N \rightarrow \infty$ this approaches the exact expected value.

As an example consider $h(x) = x^2$ and $f(x) = \mathcal{N}(0, 1)$. Here we know for a fact that $E[x^2] = 1$.

```
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm
```

```
plt.figure(figsize=(8, 4))
N = np.logspace(1, 4, 21).astype('int')
E_xsq = []
for Ni in N:
samples = np.random.normal(size=[Ni])
E_xsq.append((samples**2).mean())
plt.semilogx(N, np.abs(np.subtract(1, E_xsq)), marker='x')
plt.title('$|1 - (1/N)\sum_{x_i \sim f}{x_i^2}|$ for different values of $N$');
```

We can see that as $N$ increases the error $\left\vert1 - (1/N)\sum_{x_i \sim f}{x_i^2}\right\vert$ decreases. Now we shall plot a histogram of 1000 estimates for $N = 100, 1000, 10000$.

```
plt.figure(figsize=(8, 4))
for Ni in [100, 1000, 10000]:
samples = np.random.normal(size=[Ni, 1000])
plt.hist((samples**2).mean(0), alpha=0.5, bins=100, label=f'N={Ni}');
plt.legend();
```

Note that as $N$ increases the histogram becomes narrower indicating that there is is less variance in the estimates.

## The need for importance sampling

The sample average estimate is simple and frequently used. The loss function in training neural networks is often a sample average, for example the softmax cross-entropy loss, where $y$ is a onehot vector

\[L_\text{BCE} = E_{(x, y) \sim p_\text{data}}\left[-y^T\log h_\theta\left(x\right)\right] \approx -\frac{1}{N}\sum_{(x_i, y_i) \sim p_\text{data}}y_i^T\log h_\theta\left(x_i\right)\]However there are some potential issues

- It might be hard to sample from $f(x)$
- The samples might not be informative - for example the overlap between the support of $f(x)$ and the domain of $h(x)$ is small.

In this post we will focus on the second issue. Previously we saw that as the number of samples $N$ increased the variance of the estimate decreased. We might ask if we by having more informative samples we could reduce the variance without increasing the number of samples.

In **importance sampling** we find another distribution or a “proposal” distribution $q(x)$ and to turn $E_{x\sim f}\left[h(x)\right]$ into an expectation over $q$

Now we can approximate $E_{x\sim q}\left[f(x) \frac{p(x)}{q(x)} \right]$ as before

\[E_{x\sim p}\left[f(x)\right] = E_{x\sim q}\left[f(x) \frac{p(x)}{q(x)} \right] \approx \frac{1}{N}\sum_{i=1}^N f(x_i)\frac{p(x_i)}{q(x_i)} = \frac{1}{N}\sum_{i=1}^N f(x_i)w(x_i)\]where the samples are weighted by $w(x_i)$. If $x_i$ is equally likely according to $q$ as well as $p$ then the sample is treated the same way as in the earlier approximation. If it is less likely according to $p$ relative to $q$ it is scaled down but this will be offset by the fact that you will have more samples from $q$ with a high probability.

## Importance sampling example

Consider the integral

\[\int_0^\pi x \sin x \cdot dx\]We know how to integrate this and the answer is in fact

\[\left.x\sin x \right\vert_0^\pi = \pi\]This can be written in terms of an expectation over $x \sim U(0, \pi)$

\[\int_0^\pi x \sin x \cdot dx = \pi\int_0^\pi \frac{1}{\pi} x \sin x \cdot dx = \pi E_{x \sim U(0, \pi)}\left[x \sin x\right]\]Let us take a look at the function

```
plt.figure(figsize=(8, 4))
hx = lambda x: x*np.sin(x)
xmin = 0
xmax = np.pi
x = np.linspace(xmin, xmax, 1000)
plt.plot(x, hx(x), label='$x\sin x$');
plt.hlines(xmin=xmin, xmax=xmax, y=1/xmax, label='$1/\pi$', color='k');
```

Here you can see that the peak of the function around $x=2$ will contribute most to the integral and the values close to the endpoints much less. However if you sample from $f(x) = U(0, \pi)$ you would sample equally from the whole interval $[0, \pi]$. We would like to sample more points from round the peak.

Since $x\sin x$ looks roughly like a Gaussian with a peak round $x=2$ let us try using as a proposal $N(2, 0.7)$ after scaling it so that its support is the same as the domain of $h(x)$ i.e. $\left[0, \pi\right]$.

```
plt.figure(figsize=(8, 8))
mu = 2
sigma = .7
n_samples = 1000
n_estimates = 1000
normal = norm(loc=mu, scale=sigma)
# Scale so that it covers the same domain as h(x)
denom = normal.cdf(xmax) - normal.cdf(xmin)
qx = lambda x: normal.pdf(x) / denom
wx = lambda x: (1/xmax) / qx(x)
xmin = 0
xmax = np.pi
x = np.linspace(xmin, xmax, n_samples)
plt.plot(x, hx(x), label='$x\sin x$');
plt.hlines(xmin=xmin, xmax=xmax, y=1/xmax, label='$1/\pi$', color='k');
plt.plot(x, qx(x), label='$q(x)$');
samples = np.random.normal(loc=mu, scale=sigma, size=n_samples);
plt.plot(samples, wx(samples), '.', label='$w(x)$', alpha=0.1);
plt.ylim(0, 2);
plt.xlim(xmin, xmax);
plt.legend();
```

First take samples from $f(x)$ and find estimates.

```
uniform_estimates = np.stack([
hx(np.random.uniform(xmin, xmax, size=[n_samples])).mean() * np.pi
for i in range(n_estimates)
])
```

Now do importance sampling. Here we need to be a bit careful since we want to restrict samples to the domain of $h(x)$ so we retain only those samples are in this domain.

```
is_estimates = []
num_rejected = []
for i in range(n_estimates):
n_samples_left = n_samples
samples = []
n = 0
while n_samples_left > 0:
s = np.random.normal(loc=mu, scale=sigma, size=n_samples_left)
s = s[(s<xmax) & (s>xmin)]
n += (n_samples_left - len(s))
samples = np.concatenate([samples, s])
n_samples_left -= len(s)
num_rejected.append(n)
is_estimates.append(
(hx(samples) * wx(samples)).mean() * np.pi
)
```

Out of interest the number of rejected samples is small relative to total number of samples

```
np.mean(num_rejected / np.add(n_samples, num_rejected))
```

```
0.053700062413939764
```

Comparing the histograms for the two sets of estimates we can see that the variance is much smaller when we using importance sampling.

```
plt.figure(figsize=(8, 8))
plt.hist(uniform_estimates, bins=30, alpha=0.5, label='Samples from $f(x)$');
plt.hist(is_estimates, bins=30, alpha=0.5, label='Importance sampling');
plt.legend();
```

```
print("Exact solution: ", np.pi)
print("Mean of estimates using x ~ f(x): ", np.mean(uniform_estimates))
print("Standard deviation of estimates using x ~ f(x) ", np.std(uniform_estimates))
print("Mean of estimates using importance sampling: ", np.mean(is_estimates))
print("Standard deviation of estimates using importance sampling: ", np.std(is_estimates))
```

```
Exact solution: 3.141592653589793
Mean of estimates using x ~ f(x): 3.1405385491946087
Standard deviation of estimates using x ~ f(x) 0.06185643911655511
Mean of estimates using importance sampling: 3.1414545418966644
Standard deviation of estimates using importance sampling: 0.015353005384556133
```

## References

- The importance sampling example is borrowed from the lecture 4 of Harvard’s AM207: Stochastic Optimisation which was run in 2017. The codebase and resources for this course can be found here.