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 $$ f_{X_i}(x) = \int_{\mathbb{R}} f_{X_i|Z_i}(x|z)f_{Z_i}(z)dz $$


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 $\pmb{Y}$.

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

1 Answer 1


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 $\Phi\left(\frac{a}{\sqrt{1+b^2}}\right)$.


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.