Suppose I want to simulate the value of the integral $\int_{0}^{1} \int_{2}^{3} 2xy \ dx dy$ using Monte Carlo methods.

So, now, I draw a random sample from $U_1,U_2,...,U_n$ from $U(0,1)$ and for each $i \in [1,n]$, I take $X_i =U_i+2$. So, now, I get a random sample of $X_i$'s from $U(2,3)$.

Again I take another sample $Y_1,Y_2,...,Y_n$ from $U(0,1)$.

So, by WLLN, I calculate the mean of $2X_iY_i$ defined over the space $[0,1] \times [2,3] $ to get the estimate.

My question is do we always need to draw another sample for $Y_1,Y_2,..,Y_n$?

That is, if I had taken $Y_i=U_i$ , i.e. the initial drawn sample what would go wrong?

I get it that if I don't draw a fresh sample, then $X_i, Y_i$'s are not independent. But do we at all require independence of $X_i,Y_i$ here?


1 Answer 1


First, let me point out [somewhat pedantly!] that one does not simulate the value of the integral since this is a fixed number.

Second, if using the same $U_i$'s for simulating $X_i$ and $Y_i$, as e.g. $Y_i=U_i$ and $X_i=U_i+1$, the functional dependence between these two rvs modifies the value of the integral $$\int_0^1\int_2^3 2xy\text{d}x\text{d}y=x^2|_0^1 y^2/2|_2^3=5/2$$ versus $$\int_0^1 2x(x+2)\text{d}x=2x^2/3+2x^2|^1_0=5/2=8/3$$ as can be checked by simulation

  > y=runif(1e7)
  > x=y+2
  > print(mean(2*x*y))
  [1] 2.667504 #8/3
  > x=runif(1e7)
  > print(mean(2*x*y))
  [1] 2.500612 #5/2

The simulation of the points $(X_i,Y_i)$'s in $(0,1)\times(2,3)$ need mimick a Uniform generation over the unit square. They may be dependent in $i$ but not between $X_i$ and $Y_i$ since this modifies the joint distribution. An alternative is to simulate directly $2X_iY_i$.


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