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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 $f(y)=\int f(y|u)f(u)du$.

$x(t)$ is composite function of $u$. The problem is I need to generate random variable from $f(y)$, I am reading Monte Carlo method textbook, but there is no detail about it. 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. 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)?

Alex
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    The distribution on $Y$ is the marginal of the joint distribution on $(U,Y)$, which can be generated by (1) generating $U$, then (2) $Y$ conditional on $U$. – Xi'an Jun 10 '21 at 19:32
  • @Xi'an The thing you said is written on the book, but can you tell little detail? I can generate random variable from the distribution of u and the conditional distribution y|u, but how to link that with the distribution of y? – Alex Jun 10 '21 at 19:45
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    The answer to your question is given in the first two words and first two formulas of the first sentence: generate $u$ and then generate $Y$ given that value of $u$--presuming "$x(t)$" and "$\sigma_e^2$" are determined by $u$ (the notation doesn't suggest that, but maybe that's what you mean by "composite function"?). – whuber Jun 10 '21 at 19:58
  • @whuber So, Like we generate u from f(u) and y from f(y|u). Than what? Is it okay if we are not generating y from f(y) since we first generating u or I am getting more confused? – Alex Jun 10 '21 at 20:02
  • @whuber if we generate u from f(u) and then y from f(y|u), does it imply that we are generating y from f(y)? I mean can we use the random variable y that we are generating from f(y|u) for further procedure? – Alex Jun 10 '21 at 20:08
  • What would "$f(u)$" mean? You state $u$ is a standard Normal variable. – whuber Jun 10 '21 at 20:14
  • @whubur, yes that's what it means, f(u) is standard normal density – Alex Jun 10 '21 at 20:15
  • @whuber so am I thinking right? – Alex Jun 10 '21 at 20:25
  • If one generate $U\sim\mathcal N(0,1)$ and then $Y|U\sim\mathcal N(U,1)$ would you agree that one has generated $Y\sim\mathcal N(0,2)$, marginally? – Xi'an Jun 10 '21 at 20:39
  • @Xi'an, honestly I am not sure, I am very new about the generating random variable(I can can only do the very basic one). If your statement is true, how exactly, can you explain? or what can help me to understand this stuff clearly? – Alex Jun 10 '21 at 20:56

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