Consider a function \(g\) on the real line. Suppose we want to integrate \(g\) over \([a,b]\):

\[ \int_a^b g(x) \ dx. \]

Suppose that we have a density \(p\) for which \(p(x)>0\) for all \(x \in [a,b]\). Remember that the idea of Monte Carlo integration is to write the quantity of interest as an expectation, as follows:

\[ \begin{align*} \int_a^b g(x) \ dx & = \int_a^b \frac{g(x)}{p(x)} \ p(x) \ dx \\ & = \int \frac{g(x)}{p(x)} \ \text{I}\{ a \le x \le b \} \ p(x) \ dx \\ & = \text{E} \bigg( \frac{g(X)}{p(X)} \ \text{I}\{ a \le X \le b \} \bigg) \\ & = \text{E} ( h (X) ) \\ \end{align*} \]

Now do the usual steps in Monte Carlo integration (e.g., sampling many \(X\)’s, apply the function \(h\) to each \(X\), find the arithmetic mean, and assess Monte Carlo uncertainity).

Does it matter which density \(p\) we use? In theory, no. In practice, we want a density \(p\) having the following desirable properties:

- The density \(p\) is easy to evaluate,
- The distribution corresponding to the density \(p\) is easy to sample, and
- The variance of \(\frac{g(X)}{p(X)} \ \text{I}\{ a \le X \le b \}\) is small. That this, \(p\) matches the shape of \(g\) well on \(x \in [a,b]\).

Evaluate: \[ \int_1^3 \frac{1}{ x \ \Gamma(x) \ e^x } \ dx. \]

```
g <- function(x) 1 / ( x * gamma(x) * exp(x) )
curve(g(x),1,3)
```

Consider a density \(p\) of the following form: \(p(x) = \frac{c}{2x^3}\) for \(x \in [1,3]\).

```
curve(g(x),1,3)
curve(1/(2*x^3),1,3,add=TRUE,col="red")
```

Notice that \(p\) matches the shape of \(h\) well (“Desirable Properties 3”).

Since \(p\) is a density, we can find \(c\) by solving the following:

\[ 1 = \int_1^3 \frac{c}{2x^3} dx \ \implies c = 9/2. \]

So, \(p(x) = \frac{9}{4x^3}\) for \(x \in [1,3]\).

`p <- function(x) 9 / (4*x^3)`

Notice that \(p\) is easy to evaluate (“Desirable Properties 1”).

How can we sample from the distribution having density \(p\)? Use the probability intergral transform. First, find \(F\), the distribution function associated with the density \(p\):

\[ F(x) = \int_1^x \frac{9}{4t^3} \ dt = \frac{9}{8}\bigg( 1 - \frac{1}{x^2} \bigg) \ \ \text{for} \ x \in [1,3]. \]

Now find \(F^{-1}\), the inverse of the c.d.f. \(F\):

\[ F^{-1}(y) = \bigg( 1 - \frac{8}{9} y \bigg)^{-\frac{1}{2}} \ \ \text{for} \ y \in [0,1]. \]

Finally, get samples from the distribution having density \(p\) by apply the inverse c.d.f. to draws from the uniform distribution on the unit interval.

```
Finv <- function(y) 1/sqrt(1-8*y/9)
B <- 10000
x <- Finv(runif(B))
```

Notice that it is easy and quick to obtain samples from the distribution (“Desirable Properties 2”).

Check ourselves by noticing they seems to have a histogram with the right shape.

`hist(x,freq=FALSE)`

Now apply standard Monte Carlo integration techniques to \(h\).

```
h <- function(x) g(x)/p(x) * (x>=1) * (x<=3)
est <- mean(h(x))
cat(sprintf("Monte Carlo estimate is %7.5f.\n",est))
```

`## Monte Carlo estimate is 0.21303.`

```
ci <- est + c(-1,1)*qnorm(0.975)*sd(h(x))/sqrt(B)
cat(sprintf("The 95%% confidence interval is (%7.5f,%7.5f).\n",ci[1],ci[2]))
```

`## The 95% confidence interval is (0.21234,0.21372).`

Instead of this \(p(x) = \frac{9}{4x^3}\) for \(x \in [1,3]\), what if we had just used a uniform distribution \([1,3]\), i.e., \(p(x) = 1/2\) for \(x \in [1,3]\)?

```
x <- runif(B,1,3)
h <- function(x) g(x)/(1/2) * (x>=1) * (x<=3)
est <- mean(h(x))
cat(sprintf("Monte Carlo estimate is %7.5f.\n",est))
```

`## Monte Carlo estimate is 0.21336.`

```
ci <- est + c(-1,1)*qnorm(0.975)*sd(h(x))/sqrt(B)
cat(sprintf("The 95%% confidence interval is (%7.5f,%7.5f).\n",ci[1],ci[2]))
```

`## The 95% confidence interval is (0.20947,0.21725).`

What do you notice is difference about the two solutions?