If I have a set of random variables $X_1\dots X_n$ with joint density $f(x_1,\dots,x_n)$, if I wanted the joint density of any (say) two random variables $X_i$ and $X_j$, can I find this using the following:

$$f_{X_i, X_j}(x_i,x_j)=\int\dots\int f(x_1,\dots,x_n)dx_1\dots dx_{i-1}dx_{i+1}\dots dx_{j-1}dx_{j+1}\dots dx_n$$

Essentially, just integrate $f(x_1,\dots,x_n)$ with respect to the variables that are not in the subset that you're trying to find the joint density of?


1 Answer 1


Yes, you can find via that integral. And, it's the most general way of doing it. The procedure is called marginalization. Depending on the distributions, you might even get rid of the integration, e.g. for jointly normal RVs, you can write the joint distribution of any subset or RVs by just getting relevant entries in the covariance matrix and the mean vector.

  • $\begingroup$ So if I have $n$ RVs that are jointly normal, is any subset of this is also jointly normal because by definition, the $n$ RVs are only jointly normal if any linear combination the RVs is normal. As a result, this means the linear combination of any subset of those $n$ RVs is normal, so any subset satisfies the definition of a jointly normal distribution? $\endgroup$
    – Yandle
    Commented Mar 9, 2020 at 17:17
  • $\begingroup$ Yes, any subset of jointly normals is also jointly normal. $\endgroup$
    – gunes
    Commented Mar 9, 2020 at 19:38

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