# Simulation given conditional distribution

I have a question regarding conditional distributions and simulation.

Assume you have a random vector of dimension $$n$$ with distribution $$\pmb{Z}\sim N_n(0,\Sigma)$$ where $$\Sigma_{ii} = 1$$, so each component is such that $$Z_i \sim N(0,1)$$. Now, we have another variables $$X_i$$ such that $$X_i = f(Z_i) \sim Ber(p_i(Z_i))$$. So, to simulate each $$X_i$$ what I did was simulate the vector $$\pmb{Z}$$ and use the fact that I know the distribution of $$X_i | Z_i$$.

My question is, why can't I simulate directly from the marginal of each $$Z_i$$? This is, simulate $$Z_i = z$$ from the standard normal, then simulate $$X_i|Z_i$$. Is the conditioning actually on the whole vector?

Finally, if I want to calculate the marginal distribution of $$X_i$$, can I use the law of total probability conditioning only on $$Z_i$$ or do I need to condition on the whole vector? This is, is it correct to calculate $$f_{X_i}(x) = \int_{\mathbb{R}} f_{X_i|Z_i}(x|z)f_{Z_i}(z)dz$$

EDIT:

My question is more abstract than what I explained here. Basically, assume $$X_i = f(Y_i)$$, and also $$Y_i$$ is part of a vector $$\pmb{Y}$$. If I want to find the unconditional distribution of $$X_i$$, do I have to use $$X_i |\pmb{Y}$$ or can I use only $$X_i|Y_i$$ given that $$X_i$$ depends only on $$Y_i$$, even though $$Y_i$$ is related to the other components of $$\pmb{Y}$$.

• You did not tell us whether or not the matrix $\Sigma$ is diagonal. Oct 7, 2021 at 4:29
• It is not, just the diagonal is all 1's, but off diagonal can have positive or negative values. Oct 8, 2021 at 21:48

You do not have to consider all of the individual random variables in $$Z$$ and only need to consider $$Z_i$$ as you define the distribution of the conditional random variable $$X_i|Z_i$$ in terms of $$Z_i$$: $$X_i|Z_i=z_i \sim \text{Bernoulli}(p_i(z_i))$$.
I'm going to drop the subscripts as those are not necessary. The resulting marginal distribution of $$X$$ is $$\text{Bernoulli}(q)$$ where $$q$$ is the mean of $$p(Z)$$. In other words, however you define $$p(Z)$$, the marginal of $$X$$ will only depend on the mean of $$p(Z)$$.
In one of your other questions (https://math.stackexchange.com/questions/4513934/solve-integral-of-phiabxk1-phiabxn-k-phix) you define $$p(Z)=\Phi(a+b Z)$$. The mean of $$p(Z)$$ is given by
$$\int_{-\infty}^\infty \Phi(a+b z)\frac{e^{-z^2/2}}{\sqrt{2\pi}}dz=\Phi\left(\frac{a}{\sqrt{1+b^2}}\right)$$
So the marginal distribution of $$X_i$$ is a Bernoulli distribution with parameter $$\Phi\left(\frac{a}{\sqrt{1+b^2}}\right)$$.