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If our population regression function is ($Y = B_0 + B_1 \times X + U$) where $U$ is a disturbance term with $Var[U]$ = $\sigma^2$...

How would one classify the following (random variable or not):

  1. $X_i$
  2. $\hat{B}_0$
  3. $\hat{B}_1$
  4. $U$
  5. $\sigma^2$
  6. $Y_i$

It seems clear to me that $Y_i$ is a random variable because it has a probability distribution for what values it could take on. What is least clear is $\hat{B}_1$ and $\hat{B}_0$? Those have some definite value for our sample, but would change if we took a new sample. The same is true for the $X_i$th value of $X_i$.

Samuel
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Jamal Mantburg
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1 Answers1

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Now that this is several months old, I am going to assume it is safe to give a full answer, without unfairly advantaging you in any academic assessment. So, have a look again at the model form:

$$Y_i = B_0 + B_1 X_i + U_i \quad \quad \quad \mathbb{V}(U_i) = \sigma^2.$$

In a classical linear regression model, the explanatory variables $X_i$ are considered to be fixed (since the problem is to find the conditional behaviour of the response) and the parameter values $B_0$, $B_1$ and $\sigma$ are considered to be "unknown constants", meaning they are also treated as fixed.$^\dagger$ So the only thing on the right-hand-side of the equation that is random is the error term $U_i$.

Since $U_i$ is a random variable, this means that anything that depends on $U_i$ is also a random variable. Hence, the response $Y_i$ is also a random variable (when not conditioning on it having been observed) and the estimators $\hat{B}_0$ and $\hat{B}_0$, which depend on these response values, are also random variables.

Hence, in this question, items 2, 3, 4 and 6 would all be considered random variables.

$^\dagger$ This assumes you are doing your analysis in the classical frequentist statistical framework. Using Bayesian analysis the parameters would be treated as random variables.

Ben
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