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Consider a random variable $x_{ijk}$, where $ijk$ is a subscript that indicates that the variable varies with 3 dimensions (e.g., firm, product, and country). $x_{ijk}$ is known to be i.i.d. log-normally distributed.

How to find the distribution of $x_{ijk}$ where $k$ is fixed? Is $x_{ijk}$ still an i.i.d. drawn sample and is $x_{ijk}$ still lognormally distributed?

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  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$ Commented Mar 8, 2023 at 16:48

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Due to the fact that your multivariate-normal distribution is independent (the covariance matrix is diagonal), it will be more intuitive to treat each dimension as being its 'own' random variable; i.e., let $(I, J, K) \sim (\log(X), \log(Y), \log(Z))$, where $X, Y, Z$ are all normal distributions with their own mean and variance parameters.

Now, due to the independence, the joint distribution is given by $f(i, j, k) = f_I(i)f_J(j)f_K(k)$, where $f_A(a)$ is the marginal density function of RV $A$. If you want to 'fix', aka condition on knowing $k$, then you want the distribution $I, J|K$. The density for this can be found with Bayes rule: $$f(i, j|k) = \frac{f(i, j, k)}{f_K(k)}\;;$$ but given we can factorise $f(i, j, k)$ using the above independence assumption we now have $$f(i, j|k) = \frac{f(i, j, k)}{f_K(k)} = \frac{f_I(i)f_J(j)f_K(k)}{f_K(k)} = f_I(i)f_J(j)\;.$$ So, you can see that the distribution of $I, J|K$ has the same distribution as the product of the two random variables $I$ and $J$. Putting this together, we can deduce that the distribution is still log-normally distributed (indeed, it is the product of two independent log-normal distributions) and they will still be independent.

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  • $\begingroup$ Hi David: thanks for the answer. It all makes total sense to me... just one thing. x_{ijk} does not have a multivariate distribution. (Think of it as a measurement error term in a regression). So, it has a single variate log-normal distribution. What allows you to think of x_{ijk} as a vector (X,Y,Z) of independent variables? Do you treat them as projections on the XYZ space? Thank you! $\endgroup$ Commented Mar 10, 2023 at 11:13
  • $\begingroup$ Maybe I misunderstood, but if your random variable $x$ depends on three things ($i,j,k$) then I took this to mean it is 3-dimensional? if it is a scalar, then you would have to be more clear on what it is you're doing, is it a scalar conditioned on these three values? $\endgroup$
    – David
    Commented Mar 10, 2023 at 14:34
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    $\begingroup$ @David I like your answer. I voted for you! $\endgroup$ Commented Dec 5, 2023 at 22:49

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