# Generating random variable from no closed-form marginal density [closed]

Suppose $$u\sim N(0,I_p)$$ and $$Y|U\sim N(x(t),\sigma_e^2I_m)$$, and the marginal distribution of $$y$$ is $$f(y)=\int_u f(y|u)f(u)du$$.

$$x(t)$$ is composite function of $$u$$, basically $$x(t)$$ is a function of $$z(t)$$ and $$z(t)$$ is a function of $$u$$. I can provide the expression of $$x(t)$$ in terms of $$u$$, it's too complicated(involves exponent terms, integrand in dinominator), but I feel it's probably unnecessary. Another info is that the marginal pdf of $$y$$ has no-closed form expression.

I think what I need to do is generate random variable from $$f(y)$$, I am reading Monte Carlo method textbook, but there is no detail or probably I am newbie about generating random variable. The textbook is giving an example of normal mixture and the example is given Generating random variables from a mixture of Normal distributions here too, but it's not similar what I am dealing with.

I have also seen this how to generate data from cdf which is not in closed form? and Multivariate and Marginal simulations, but still so confused. The fact is how do I assume the distribution(marginal distribution) of $$y$$. Or do I need to go for indirect method (eg. accept-reject method)?

Like in the last link they are telling to generate $$y_1^{(1)}$$ from $$f(y_1)$$ first. But how do I do that if I don't know the $$f(y_1)$$

I want to implement the algorithm in r, but I am stuck at the very beginning of making the algorithm.

• You asked the same question yesterday and it was answered through the comments, before getting closed as similar questions had been posted and answered on X validated. The issue is not one with simulation but with probability theory: a random variable $Y$ can have simultaneously both a marginal and a conditional distribution, and remain the same random variable. – Xi'an Jun 11 at 6:44
• This question was reposted after being closed and without significant changes. – Tim Jun 11 at 9:35

A marginal distribution on $$Y$$ associated with a joint distribution on $$(U,Y)$$ is by definition the distribution of $$Y$$ on its own, that is, ignoring the realisation of $$U$$.

As an example, consider the joint density $$f(u,y)=\varphi(u;0,1)\times\varphi(y;u,1)$$ for which

1. $$Y|U=u$$ is conditionally distributed as a $$\mathcal N(u,1)$$
2. $$Y$$ is marginally distributed as a $$\mathcal N(0,2)$$ variate.

and both statements are simultaneously correct. This implies that if one jointly simulates $$(U,Y)$$ from the joint Normal distribution, the resulting $$Y$$ will be distributed from $$\mathcal N(0,2)$$. As an illustration, here are 100 $$\mathcal N(u_i,1)$$ densities when $$U_1,\ldots,U_{100}\stackrel{\text{iid}}{\sim} \mathcal N(0,1)$$ and their average density (in red): which is very close to a $$\mathcal N(0,2)$$ density.

A mistake in considering that since $$Y\sim \mathcal N(U,1)$$ it cannot be $$\mathcal N(0,2)$$ at the same time is to forget that $$U$$ is a random variable. (There is some similarity with accept-reject in that, while all simulations are from the proposal distribution $$g$$, those that are accepted are also simulations from the target distribution $$f$$.)

Another illustration is provided by finite mixtures of distributions: if $$Y\sim f(y)=\sum_{i=1}^k\omega_i f_i(y)\qquad\sum_{i=1}^k\omega_i=1$$ this distribution can be represented as the marginal distribution (in $$Y$$) of a joint distribution on $$(U,Y)$$ such that $$U$$ is marginally distributed as a Multinomial variable: $$\mathbb P(U=i)=\omega_i \qquad i=1,\ldots,k$$ and the conditional distribution of $$Y$$ given $$U$$ is $$Y|U=i \sim f_i(y)$$ Generating an index $$i$$ with probability $$\omega_i$$ and then a realisation from $$f_i(\cdot)$$ is equivalent to generating from $$f(\cdot)$$.