2
$\begingroup$

I am using a Monte Carlo method to estimate the expected value of the results of certain simulations.

Consider this simplified case: $X, Y$ are independent random variables and $g(X,Y)$ is a nonlinear function of $X,Y$. I would like to use Monte Carlo to estimate $\mathbb{E}_{X,Y} g(X,Y)$.

In my code I do the following:

  • Draw $N_Y$ random samples ($y_1, \dots, y_{N_y}$) of $Y$

  • For each $y_j$, draw $N_X$ fresh new random samples ($x_1^{(j)}, \dots, x_{N_x}^{(j)}$) of $X$ and compute $$ g_j = \frac{1}{N_x} \sum_{i=1}^{N_x} g(x_i^{(j)}, y_j)$$

  • I then estimate $$\mathbb{E}g(X,Y) \approx \frac{1}{N_y} \sum_{j=1}^{N_y} g_j$$

The reason why I am using this method (rather than just take $N_x \times N_y$ random samples $(x_i, y_i)$) is because it is computationally heavy to draw random samples of $Y$ (large Erdos-Renyi graphs), and in this way, I can take $N_y << N_x$.

I know that because of the linearity of the expected value, the above estimate is indeed $\mathbb{E}g(X,Y)$. My question is regarding its uncertainties.

My question is:

  • How do you compute the uncertainties for it, e.g. std? How do they compare with the standard Monte Carlo method?

Edit after @Xi'an's answer

For easiness let me introduce some notation. There are three possible estimators:

  1. Draw $N_y$ random samples of $Y$ and $N_x$ random samples of $X$, then estimate $\mathbb{E}g(X,Y)$ by $S_1$ $$ S_1 = \frac{1}{N_y N_x} \sum_{j=1}^{N_y} \sum_{i=1}^{N_x} g(x_i, y_j)$$

  2. Draw $N_y \times N_x$ random pairs of $Y$ and $X$, then estimate $\mathbb{E}g(X,Y)$ by $S_2$ $$ S_2 = \frac{1}{N_y N_x} \sum_{j=1}^{N_y N_x} g(x_j, y_j)$$

  3. Draw $N_y$ random samples of $Y$ and for each $y_j$, draw $N_X$ fresh new random samples ($x_1^{(j)}, \dots, x_{N_x}^{(j)}$) of $X$ , then estimate $\mathbb{E}g(X,Y)$ by $S_3$ $$ S_3 = \frac{1}{N_y N_x} \sum_{j=1}^{N_y} \sum_{i=1}^{N_x} g(x_i^{(j)}, y_j)$$

My question is: Can we say anything, a priori, about the variance of these estimators or when one outperforms the others?

Notice that these estimators have very different amount of "randomness".

$\endgroup$
9
  • $\begingroup$ I feel it is impossible to derive a closed form adjustment coefficient for an arbitrary function and arbitrary X and Y random variables' distributions. I would avoid such biased Monte Carlo by all costs. $\endgroup$
    – Alex
    Oct 5, 2022 at 17:25
  • 1
    $\begingroup$ Would it be fair to say you are estimating $E[g(X,Y)] = E[E[g(X,Y)\mid Y]]$? Why not, then, propagate the estimation errors through the two univariate steps? $\endgroup$
    – whuber
    Oct 5, 2022 at 17:46
  • 1
    $\begingroup$ @Alex why should this Monte Carlo be biased? The expected value of the estimator is the actual expected value $\endgroup$
    – 123prior
    Oct 5, 2022 at 18:25
  • $\begingroup$ The issue with the three estimators is that they require very different computing times, hence comparing their variances is not necessarily relevant. $\endgroup$
    – Xi'an
    Oct 5, 2022 at 20:09
  • 1
    $\begingroup$ Yes, and if one still prefers $S_3$, one may want to admit that its uncertainty is comparatively uncertain. $\endgroup$
    – Alex
    Oct 5, 2022 at 21:44

1 Answer 1

1
$\begingroup$

Since $X$ and $Y$ are independent, there is not need to simulate new samples $x^{(j)}$ for different $j$'s. That is, $$\frac{1}{N_xN_y}\sum_{i=1}^{N_x}\sum_{j=1}^{N_y}g(x_i,y_j)\tag{1}$$ is an unbiased estimator of $\mathbb E_{X,Y}[g(X,Y)]$. Indeed, for all pairs $(i,j)$, $$\mathbb E_{X,Y}[g(X_i,Y_j)]=\int_\mathfrak{X}\int_\mathfrak{Y}g(x,y)f_X(x)f_Y(y)\,\text dx\text dy$$ The variance of (1) may be larger than the variance of $$\frac{1}{N_xN_y}\sum_{i=1}^{N_x}\sum_{j=1}^{N_y}g(x_i^{(j)},y_j)\tag{2}$$ since $g(X_i,Y_j)$ is likely to be positively correlated with $g(X_i,Y_k)$ for $j\ne k$, but the cost of producing the $N_x\times N_y$ $X_i^{(j)}$'s may offset the gain in variance provided by (2).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.