Sampling from a continuous 2 dimensional probability distribution function for importance sampling

I just want to clarify a few points with regarding to sampling from a continuous 2-dimensional probability density function. If I want to sample from this pdf, I could sample from a 1D pdf, $$P(x)$$, along $$x$$ and one along $$y$$ using the inverse transform sampling method. This method starts with cumulatively integrating a 1D pdf to calculate the cumulative density function,

$$C(x) = \int^{x}_{-\infty} \frac{1}{A}P(x') \ dx'$$

Once the cdf has been calculated, it can be inverted to produce the ppf function. Once that has been done, the last step is to sample uniformly in [0,1), then pass those values through the ppf function which redistributes the initially uniformed sampled points in accordance to the original pdf.

In the case of the 2D pdf, I can repeat this process along $$x$$ and along $$y$$ to get samples for the 2D pdf. Now, if I want to calculate an integral via importance sampling MC. When I sample in accordance with the 2D pdf for $$N$$ samples, do I sample along $$x$$ and $$y$$ with $$N$$ samples each so I would have an array of (N,2) or do I sample along $$x$$ and $$y$$ with $$N$$ samples and calculate a $$N$$ by $$N$$ mesh of all combinations of $$x$$ and $$y$$?

• What should the definition of the cdf be then? Commented Jun 13, 2020 at 13:55

Actually, the inverse cdf method only works in dimension one as $$(F_X(X),F_Y(Y))$$ is not a Uniform $$\mathcal U(0,1)^2$$ variate, unless $$X$$ and $$Y$$ are independent. To simulate from an arbitrary bivariate distribution with joint pdf $$p_{X,Y}(x,y)$$, one (among many) possible solution is to
1. Generate $$X$$ from the marginal $$p_X(x)=\int_{\mathcal Y} p_{X,Y}(x,y)\,\text{d}y$$, possibly by inverse cdf means
2. Generate $$Y$$ given $$X=x$$ from the conditional $$p_{Y|X}(y|x)=p_{X,Y}(x,y)/p_X(x)$$, again possibly by inverse cdf means
As an example, consider a joint distribution defined by the density $$p(x,y)=\frac{x^y}{y!}\exp\{-2x\}\mathbb I_{x>0}\mathbb I_{\mathbb N}(y)$$Then $$p_{Y|X}(y|x)\propto \frac{x^y}{y!} \mathbb I_{\mathbb N}(y)$$ meaning that $$Y$$ conditional on $$X=x$$ is distributed as a Poisson $$\mathcal P(x)$$ variate. And $$p_X(x)=\dfrac{p(x,y)}{p_{Y|X}(y|x)}=\dfrac{\frac{x^y}{y!}\exp\{-2x\}}{\frac{x^y}{y!}\exp\{-x\}}=\exp\{-x\}$$ implies that the marginal distribution of $$X$$ is an Exponential $$\mathcal E(1)$$ distribution. Hence simulating from the joint distribution can be done by
1. generate an Exponential $$\mathcal E(1)$$ variate, $$X=-\log(U_1)$$
2. generate a Poisson $$\mathcal P(x)$$ variate, $$Y=\sum_{i=1}^\infty \mathbb I_{U_2\le\sum_{j=0}^i \frac{x^i}{i!}\exp\{-x\}}$$
• So, if I calculate the marginal $P_x(x)$ and sample from that, from perhaps an inverse cdf method. How exactly is $y$ calculated then? Because in order to calculate the numerator in the conditional probability definition I need a value of $y$. Are both values of $x$ and $y$ drawn from their own respective marginals? Then passed through the joint distribution? Sorry for the silly question, I'm just a bit confused by step 2 in your explanation. Commented Jun 13, 2020 at 14:00
• $p_{Y|X}(y|x)=p_{X,Y}(x,y)/p_X(x)$ is the density of the distribution one need simulate from in Step 2. This distribution also has a cdf, which can be inverted to produce a simulation of $Y$, at least ideally. Commented Jun 13, 2020 at 14:28
• Ah, so in a way, I need to do the cumulative integral of $p_{Y \vert X}(y|x)$ w.r.t $y$. That would then give me a conditional cdf with arguments of $x$. I generate $x$ from the marginal $p_x(x)$ then generate $y$ based on this conditional cdf? If I have assumed this procedure correctly? Commented Jun 13, 2020 at 23:36