How can a double integral be used to estimate pi using Monte Carlo integration?

[Sonden]

Homework Statement

I am supposed to calculate/estimate pi using a Monte Carlo method. This is of course trivial: create N uniformly distributed random pairs (x,y) in [0,1]^2 and check how many, M, that have x^2+y^2<1. Then M/N=pi/4.

Now, the strange thing is that I am not supposed to do it that way. Instead I'm supposed to use a double integral (Monte Carlo integration): "At a first glance, this may not seem like a integration problem, but it can be formulated as such. Write down the (double) integral (and thus the expected value) that correspond to this procedure, i.e. the integral I = 4E[g(X)] = pi", where E[g(X)] is the expected value for a function g and X a random variable. I've glanced at it for several hours now, I still haven't got a clue how to do it. Of course,
$$\int_0^1 \int_0^{2 \pi}rdrd\theta=\pi$$
but is this useful here?

 

[Dickfore]

This seems dumb, since the upper bound of the integral w.r.t. $\theta$ contains the mathematical constant you are trying to evaluate. I think your initial approach is correct.

 

[vela]

I think your original idea, the one you ruled out, is the method intended in the problem. It is essentially evaluating a double integral. Why do you think you're not supposed to do it that way?

 

[HallsofIvy]

Homework Statement

I am supposed to calculate/estimate pi using a Monte Carlo method. This is of course trivial: create N uniformly distributed random pairs (x,y) in [0,1]^2 and check how many, M, that have x^2+y^2<1. Then M/N=pi/4.

Now, the strange thing is that I am not supposed to do it that way. Instead I'm supposed to use a double integral (Monte Carlo integration): "At a first glance, this may not seem like a integration problem, but it can be formulated as such. Write down the (double) integral (and thus the expected value) that correspond to this procedure, i.e. the integral I = 4E[g(X)] = pi", where E[g(X)] is the expected value for a function g and X a random variable. I've glanced at it for several hours now, I still haven't got a clue how to do it. Of course,
$$\int_0^1 \int_0^{2 \pi}rdrd\theta=\pi$$
but is this useful here?

Useful? If that weren't true this method of approximating $\pi$ wouldn't work!

The whole point is that since that integral is equal to $\pi$, approximating that integral, with a Monte Carlo method, approximates $\pi$.

 

[Sonden]

vela: because the teacher wants MC *integration* (that the unit circle's area is pi perhaps follows from integration but...). :)

Is this true?:

4 times the expected value for a function g(X,Y), where X, Y are random variables with some probability distribution f(x,y), is

$$4E[g(X,Y)]=4\int\int_{\mathbb{R}^2}g(x,y)f(x,y)dxdy$$

Then if f(x,y)=1/4 in [-1,1]^2 and 0 elsewhere,

$$4\int\int_{\mathbb{R}^2}g(x,y)f(x,y)dxdy=\int\int_{[-1,1]^2}g(x,y)dxdy$$.

Now if g(x,y)=1 inside the unit circle and 0 elsewhere, we get

$$4E[g(X,Y)]=\int\int_{unit \; circle}dxdy=\pi$$

so

$$\pi/4=E[g(X,Y)] \approx 1/N\sum_{i=1}^N g(x_i,y_i)$$

according to the law of large numbers (I think?), where (x_i, y_i) are random pairs with f:s distribution, ie they are uniformly distributed on [-1,1]^2 and g(x_i,y_i)=1 if they are inside the unit circle, ie if x_i^2+y_i^2<1, and g(x_i,y_i)=0 otherwise (that is, between the unit circle and [-1,1]^2). Is this derivation correct?

 

[vela]

You do realize, I hope, that you essentially just rederived the "trivial" method you alluded to in your first post.

 

[Sonden]

But this time I used the law of large numbers, so I know that I can get as close to pi as I wish by taking N large enough. Which was obvious from the start, but still... ;)

 

[Dickfore]

If you denote:

$$I(A) = \left\{\begin{array}{ll} 1&, A \, \mathrm{occurs} \\ 0 &, \bar{A} \, \mathrm{occurs} \end{array}\right.$$

the indicator of a random event A ($P(A) = p, P(\bar{A}) = q, \, p + q = 1$), then:

$$E(I) = 1 \cdot p + 0 \cdot q = p$$

and

$$\sigma^{2}(I) = (1 - p)^2 \cdot p + (0 - p)^2 \cdot q = p q^{2} + p^{2} q = p q (p + q) = p q = p(1 - p) \le \frac{1}{4}$$

Thus, this random variable satisfies the provisions of the Law of Large numbers and the sample mean is a good estimate of the probability of the event.