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This is a follow-up question to 1 and 2. So we have the normal linear model \begin{align*} \textbf{Y} = \textbf{X}\beta + \epsilon \end{align*}

where $\epsilon\sim\mathcal{N}(\textbf{0},\sigma^{2}\textbf{I})$, $\mu_{i} = \beta_{0} + \beta_{1}x_{i1} + \ldots + \beta_{p}x_{ip}$ and $\mu = \textbf{X}\beta$. As far as I have understood, we take $n$ observations \begin{align*} Y_{1} & = \beta_{0} + \beta_{1}x_{11} + \ldots + \beta_{p}x_{1p} + \epsilon_{1}\\ Y_{2} & = \beta_{0} + \beta_{1}x_{21} + \ldots + \beta_{p}x_{2p} + \epsilon_{2}\\ &\vdots\\ Y_{n} & = \beta_{0} + \beta_{1}x_{n1} + \ldots + \beta_{p}x_{np} + \epsilon_{n}\\ \end{align*}

and apply the least square method, for instance, to obtain $\hat{\beta} = (\textbf{X}^{T}\textbf{X})^{-1}\textbf{X}^{T}\textbf{Y}$.

The problem which concerns me is the interpretation of such process. Let us suppose, for example, that $Y$ represents the income, $x_{1}$ indicates the gender and $x_{2}$ stands for the age. Thus we draw someone from target population and obtain the first observation

\begin{align*} Y_{1} = \beta_{0} + \beta_{1}x_{11} + \beta_{2}x_{12} + \epsilon_{1} \end{align*}

After so, we draw another person (with replacement) from the target population and obtain the second observation \begin{align*} Y_{2} = \beta_{0} + \beta_{1}x_{21} + \beta_{2}x_{22} + \epsilon_{2} \end{align*}

We repeat such process until $n$ observations are made. Once we have $\hat{\beta}$ at hand, we can estimate $\mu$ according to $\hat{\mu} = \textbf{X}\hat{\beta}$. Moreover, we can also estimate the variance $\sigma^{2}$ through the estimator \begin{align*} S^{2} = \frac{(\textbf{Y} - \textbf{X}\hat{\beta})^{T}(\textbf{Y} - \textbf{X}\hat{\beta})}{n-p-1} \end{align*}

where it is assumed that $\operatorname{Rank}(\textbf{X}) = p+1$ and $\textbf{X}$ has full rank.

My first question is: am I describing the observation process rightly?

My second question is: how should we interpret the distribution $\textbf{Y} = \mathcal{N}(\textbf{X}\beta,\sigma^{2}\textbf{I})$?

The last question may be confusing me because, in the context of inference, we normally assume the sample consists in independent identically distributed random variables and in the multiple linear regression problem we just assume independence. In other words, the distribution of $\textbf{Y}$ corresponds to the distribution of the sample $(Y_{1},Y_{2},\ldots,Y_{n})$ and the means $\mu_{i}$ do not need to be the same. Is it correct?

Any help is appreciated. Thanks in advance!

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  • $\begingroup$ The first two-thirds of your post explicitly answers the second question. The first question is almost tautological, but maybe what one could add is that this "observation process" is one of many ways where this multiple regression formulation is applicable. For the last question, you have implicitly assumed more than you state at the point where you first refer to "the variance." What is this the variance of? Thinking about this ought to help you understand the answer. $\endgroup$ – whuber May 17 at 15:18
  • $\begingroup$ The variance $\sigma^{2}$ refers to the distribution of $\textbf{Y}$, right? As to the hypothesis concerning the distribution of $\textbf{Y}$, each marginal is independent from the others, but they are not identically distributed. Am I reasoning correctly? $\endgroup$ – user1337 May 17 at 16:12
  • $\begingroup$ The variance is an assumed, common variance among all the $\epsilon_i.$ That's the only way you can justify the estimator $S^2.$ $\endgroup$ – whuber May 17 at 16:16
  • $\begingroup$ Thanks for the response but I am still concerned about my second question. What we are modeling is the sample distribution, where each $Y_{i}$ is independent from the others, but they are not equally distributed. At least that's the conclusion I've got. $\endgroup$ – user1337 May 17 at 16:22
  • $\begingroup$ That's correct: the $Y_i$ are not equally distributed. Your account of that is very clear. Their expectations vary, as given by $X\beta.$ $\endgroup$ – whuber May 17 at 16:59
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Let us be explicit about what it is that you observe and what it is that you do not observe.

You observe $Y_i, x_{i1}, x_{i2}$ for $i=1,\ldots,n.$

You do not observe $\beta_0,$ $\beta_1,$ $\beta_2,$ or $\varepsilon_i.$

What you are calling $\widehat\mu_i$ is usually called $\widehat Y_i$ and is called the $i$th fitted value. It is the estimated expected value of a $Y$ value corresponding to the observed values of $x_{i1},x_{i2}.$ And as you say, the values of $\widehat Y_i$ for $i=1,\ldots,n$ in general differ from each other.

I don't understand what you mean by interpreting the distribution of $\textbf{Y} = \mathcal{N}(\textbf{X}\beta,\sigma^{2}\textbf{I}),$ but normally one writes $$ \textbf{Y} \sim \mathcal{N}(\textbf{X}\beta,\sigma^{2}\textbf{I}) $$ rather than $$ \textbf{Y} = \mathcal{N}(\textbf{X}\beta,\sigma^{2}\textbf{I}). $$ One could say that for any measurable set $A\subseteq\mathbb R^n$ one has \begin{align} & \Pr(\textbf{Y}\in A) \\[10pt] = {} & \frac 1 {\sqrt{(2\pi)^n\det(\sigma^2 \textbf{I})}} \\[8pt] & {} \times \int\limits_A \exp\left( \frac{-1}2 (\textbf{y} - \textbf{X}\beta)^\top \left( \sigma^2 \textbf{I}\right)^{-1} ( \textbf{y} - \textbf{X}\beta) \right) \, d\textbf{y} \end{align}

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  • $\begingroup$ That's exactly my doubt: how do we interpret such integral? $\endgroup$ – user1337 May 20 at 22:18
  • $\begingroup$ @user1337 : I'm still not sure how to understand your last question. Maybe this will shed some light. One writes $\textbf{Y} \sim \operatorname N(\textbf{X}\beta, \sigma^2 I_n),$ but one could just as well write $\textbf{Y} = \textbf{X}\beta + \varepsilon$ and $\varepsilon \sim\operatorname N(0,\sigma^2 I_n).$ To say that $\varepsilon \sim\operatorname N(0,\sigma^2 I_n)$ means that the scalar components $\varepsilon_1,\ldots,\varepsilon_n$ are jointly normally distributed and each has expectation $0$ and variance $\sigma^2$ and the covariance between any two of them is $0. \qquad$ $\endgroup$ – Michael Hardy May 20 at 23:25
  • $\begingroup$ In the first place, thanks for the response. My question is: how should we interpret each marginal distribution? Does it give the distribution corresponding to each sampling? $\endgroup$ – user1337 May 20 at 23:49

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