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Suppose $\mathbf X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$. After sampling $n$ samples, we repeat the sampling process $m$ times and the sampling data is stored in an $m\times n$ matrix.

Then, is the data on a specific column (not row) also normally distributed?

My simulation in R shows that it's also normally distributed but I'm not sure if there's a proof for that. Also what about other distributions say chi-Square, exponential, etc.?

More context:

My question actually came from the assumption for the Simple Linear Regression model:

\begin{align} Y=\beta_0 + \beta_1\cdot X + \epsilon \end{align}

where $\epsilon$ is normal random variable (random error). But I also see the same model is written as:

\begin{align} Y_i = \beta_0 + \beta_1\cdot X_i + \epsilon_i \end{align}

where $\epsilon_i$ also normally distributed, and $X_i$ is the specific pair $(X_i, Y_i)$ available in the data set. For which, I suspect that the two models' assumptions are equivalent.

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  • $\begingroup$ what do you mean by what about other distributions say chi-Square, exponential, etc.? Which property are you asking about them? Please try to be specific. $\endgroup$
    – utobi
    May 17, 2023 at 10:55
  • $\begingroup$ Columns $2$ through $n$ are irrelevant in this question, so why mention them? Maybe this comes down to what precisely you mean by "normally distributed with mean $\mu$ and standard deviation $\sigma.$" Could you explain? $\endgroup$
    – whuber
    May 17, 2023 at 15:55
  • $\begingroup$ @utobi I updated the question to add more context. $\endgroup$
    – Tran Khanh
    May 18, 2023 at 2:50
  • $\begingroup$ @whuber the question has been updated, two the current answers seem plausible to me at the moment, but I'm not sure that I understand the problem correctly. $\endgroup$
    – Tran Khanh
    May 18, 2023 at 2:51
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    $\begingroup$ If I understand you right, you simulate: $m$ times [a sample of $n$ independent random variables] and put each of these $m$ samples as a row of a matrix. After that you look at the columns. Regardless of the distribution of your random variables (norma, chi-Square...) you will always find that the columns consist each of $m$ independent random variables from the distribution that you chose. This is because all single entries in the matrix are independent simulations from the same distribution :-) $\endgroup$
    – Ute
    May 22, 2023 at 17:22

3 Answers 3

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For the time being, let $X = (X_1,\ldots, X_p) \sim N_p(\mu_p, \Sigma)$, $\mu = (\mu_1,\ldots,\mu_p)$ and $a$ a $p\times1$ vector. Then, for $Y = a^\top X$ it holds

$$E(Y) = E(a^\top X) = a^\top \mu$$ $$\text{var}(Y) = a^\top \Sigma a$$

and

\begin{align*} Y \sim N(a^\top \mu, a^\top \Sigma a).\tag{*} \end{align*}

Now take $a = (1,0,\ldots,0)$ and you are done.

Property $(*)$ can be proved by using the characteristic function of $X$. I'll give a proof of this later on.

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    $\begingroup$ The question is vague as to whether these components are jointly normal, but if they're jointly normal, then normality of the scalar-valued random variable that you've called $Y$ follows from any reasonable definition of joint normality without going into characteristic functions. (In particular, the theorem that the normal distribution is completely determined by its characteristic function is not needed.) And the assertions about the mean and variance of $Y$ follow from the definitions without anything about characteristic functions even if you don't have normality. $\endgroup$ May 17, 2023 at 15:54
  • $\begingroup$ @MichaelHardy I've updated the question! $\endgroup$
    – Tran Khanh
    May 18, 2023 at 7:45
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Suppose $\mathbf X = (X_1, X_2, \ldots, X_n)$ is a random vector with $n$ components is normally distributed with mean $\mu$ and standard deviation $\sigma$.

If you mean that $X_1,\ldots,X_n$ are jointly normal then the definition of joint normality answer your question. And in that case the expected value is an $n$-component vector and the variance is an $n\times n$ matrix.

Joint normality means every linear combination $a_1X_1+\cdots+a_nX_n$ is normally distributed, where $a_1,\ldots,a_n$ are not random.

In particular $X_1$ is the special case where $a_1=1$ and $a_2=\cdots=a_n=0.$

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  • $\begingroup$ To me it seems the question is more about notation. It can be confusing when you start learning statistics, that sometimes random variables come with an index, like $X_i$, and sometimes without, that is $X$, and still on other occasions, $\textbf{X}$ means a vector $\endgroup$
    – Ute
    May 22, 2023 at 17:18
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When you make $m$ times a sample of $n$ i.i.d. random variables, then you have generated in all $m\cdot n$ i.i.d. random variables. All these $m\cdot n$ variables are still independent of each other, the computer takes care of that. We could just number them from $X_1$ to $X_{mn}$.

Now you have stored them row by row in a matrix, so row 1 consists of $X_1,..., X_n$, row 2 consists of $X_{n+1},...,X_{n+n}$, row 3 of $X_{2n+1},...,X_{2n+n}$ and so on, and row $m$ of $X_{(m-1)n+1},...,X_{mn}$. If you look at any column, for example the first column $(X_1, X_{n+1},\dots, X_{(m-1)n +1})$, or the $j$-th, $(X_j, X_{n+j},\dots, X_{(m-1)n +j})$, the variables are still independent of each other.

This also holds when the simulated $X_1, \dots, X_{mn}$ have a distribution that is not normal. You only need that they are independent :-)

In many texts about linear regression, one writes $$ \mathbf{Y} = \beta_0 + \beta_1 \mathbf{X} + \epsilon $$ and in reality means $$ \begin{array}{1} Y_1\\ \vdots\\Y_n \end{array} \begin{array}{1} =\\ \ \\= \end{array} \begin{array}{1} \beta_0 +\beta_1 X_1 +\epsilon_1\\\qquad \vdots\\ \beta_0 +\beta_1 X_n +\epsilon_n \end{array}. $$

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