# Obtain marginal CDF from joint CDF by simulation

How can I evaluate the marginal cumulative distribution function of a set of random variables for which I do not have the CDF in closed form. I can, however, simulate from a joint distribution involving this set of variables.

To be more specific, assume I want to evaluate the CDF of $(X_1,X_2)$ but I only have a way to simulate from $(X_1,X_2,X_3)$.

Obviously I can approximate the CDF of $(X_1,X_2,X_3)$ by obtaining a large number of simulations and checking how many observations fall below the desired threshold. But how to get the CDF of $(X_1,X_2)$?. Can I just simply throw $X_3$ away and use the same procedure as for the joint PDF?

Obviously I cant get $F_X(x) = P(X \leq x) = \lim_{y \to \infty} P(X \leq x, Y \leq y) = \lim_{y \to \infty} F_{XY} (x, y)$ because no closed form CDF is available. I also do not want to involve the pdf.

• Yes, you can throw away $X_3$ and compute the empirical CDF of $(X_1, X_2)$. – Greenparker Apr 9 '16 at 21:32
• You are basically telling us you can freely simulate $(X_1,X_2)$ (simply by ignoring the values of $X_3$) and are asking how to "evaluate" the CDF of $(X_1,X_2)$. Could you please then tell us what you mean by "evaluate"? – whuber Apr 10 '16 at 15:47
• Well, sometimes you do not see the obvious... (that holds for me, at least). That's one reason why I am very greateful for this platform. I understand now that I can use the edf of $(X_1,X_2)$ to approximate the desired edf and it's also perfectly clear why. Thanks for your answers! – InfiniteVariance Apr 13 '16 at 11:19

## 1 Answer

As indicated in the earlier comments, once you get a sample from the joint distribution of $(X_1,X_2,X_3)$, $$(x_1^1,x_2^1,x_3^1),\ldots,(x_1^t,x_2^t,x_3^t)$$ the marginal sample $$(x_1^1,x_2^1),\ldots,(x_1^t,x_2^t)$$is indeed a sample from the marginal joint distribution of $(X_1,X_2)$ and you can ignore the simulated $x_3^j$'s. They can however be useful in Monte Carlo evaluations through a technique called Rao-Blackwellisation since the average$$\frac{1}{t}\sum_{i=1}^t h(x_1^i,x_2^i)$$is improved by the average$$\frac{1}{t}\sum_{i=1}^t \mathbb{E}[h(x_1^i,x_2^i)|x_3^i]$$as a conditional expectation shares the same expectation as the original but reduces the variance. See for instance this discussion on Cross Validated.