Answers.Body: <p>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 <span class="math-container">$(I, J, K) \sim (\log(X), \log(Y), \log(Z))$</span>, where <span class="math-container">$X, Y, Z$</span> are all normal distributions with their own mean and variance parameters.</p>
  <p>Now, due to the independence, the joint distribution is given by <span class="math-container">$f(i, j, k) = f_I(i)f_J(j)f_K(k)$</span>, where <span class="math-container">$f_A(a)$</span> is the marginal density function of RV <span class="math-container">$A$</span>. If you want to 'fix', aka <em>condition</em> on knowing <span class="math-container">$k$</span>, then you want the distribution <span class="math-container">$I, J|K$</span>. The density for this can be found with Bayes rule:
  <span class="math-container">$$f(i, j|k) = \frac{f(i, j, k)}{f_K(k)}\;;$$</span>
  but given we can factorise <span class="math-container">$f(i, j, k)$</span> using the above independence assumption we now have
  <span class="math-container">$$f(i, j|k) = \frac{f(i, j, k)}{f_K(k)} = \frac{f_I(i)f_J(j)f_K(k)}{f_K(k)} = f_I(i)f_J(j)\;.$$</span>
  So, you can see that the distribution of <span class="math-container">$I, J|K$</span> has the same distribution as the product of the two random variables <span class="math-container">$I$</span> and <span class="math-container">$J$</span>. 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.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2023-03-10T11:13:26.287Z
Answers.Text: 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!
Answers.UserDisplayName: user2518768
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2023-03-10T14:34:37.743Z
Answers.Text: 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?
Answers.UserDisplayName: David
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2023-12-05T22:49:39.797Z
Answers.Text: @David I like your answer. I voted for you!
Answers.UserDisplayName: Zollikofen4
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2023-03-08T11:45:16.027Z
Answers.LastActivityDate: 2023-03-08T11:45:16.027Z
Answers.UserDisplayName: David
Body: <p>Consider a random variable <span class="math-container">$x_{ijk}$</span>, where <span class="math-container">$ijk$</span> is a subscript that indicates that the variable varies with 3 dimensions (e.g., firm, product, and country). <span class="math-container">$x_{ijk}$</span> is known to be i.i.d. log-normally distributed.</p>
  <p>How to find the distribution of <span class="math-container">$x_{ijk}$</span> where <span class="math-container">$k$</span> is fixed? Is <span class="math-container">$x_{ijk}$</span> still an i.i.d. drawn sample and is <span class="math-container">$x_{ijk}$</span> still lognormally distributed?</p>
  
Comments.ContentLicense: CC BY-SA 4.0
Comments.CreationDate: 2023-03-08T16:48:10.967Z
Comments.Text: Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
Comments.UserDisplayName: Anshuman Kumar
Tags: <statistics>
Title: Distribution of a log-normal random variable with fixed dimension
UserDisplayName: user2518768
ThreadText: # Distribution of a log-normal random variable with fixed dimension
  
  (Asked by: username_0 on 2023-03-08)
  
  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?
  
  **Tags:** statistics
  
  > **Comments:**
  > **username_1**: Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_2 on 2023-03-08. Score: 1
  
  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.
  
  > **Comments:**
  > **username_0**: Hi username_2: 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!  
  > **username_2**: 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?  
  > **username_3**: @username_2 I like your answer. I voted for you!  
  
  ---
  
  