Let $Y_{i} = \beta_{0} + \beta_{1}X_{i} + \epsilon_{i}$ be a simple linear regression model with independent errors and iid normal distribution. If $X_{i}$ are fixed what is the distribution of $Y_{i}$ given $X_{i} = 10$?

I am preparing for a test with questions like these but I am realizing I am less up to date on these things than I thought. Could anyone explain the thought process used to approach this kind of question?


Let $\epsilon_i \sim N(0,\sigma^2)$. Then, we have:

$$Y_i \sim N(\beta_0 + \beta_1 X_i,\sigma^2)$$

Further clarification:

The above uses the following facts:

(a) Expectation is a linear operator,

(b) Variance of a constant is $0$,

(c) Covariance of a random variable with a constant is $0$ and finally,

(d) A linear combination of normals is also a normal.

Does that help?

  • $\begingroup$ I think I see why that's true since the mean of the line is shifted from the mean of the error BECAUSE they're independent, right?. But I think I am more lost than that, I can't put together how $X_{i}$ being fixed, independent errors or any of that affects the question. Can we in some way split that term, the line you wrote, up since the errors are independent? $\endgroup$ – user78504 May 23 '13 at 14:08
  • $\begingroup$ That does help a lot, I really appreciate it as I would not have seen all those moving parts. How is expectation being a linear operator used here? I mean I understand at the basic level that (redacted - incorrect) but you're applying that somewhere in the background I think. In fact the one part I do see the use of is that a linear combination of normals is also normal. $\endgroup$ – user78504 May 23 '13 at 14:16
  • $\begingroup$ $E(Y_i) = E(\beta_0 + \beta_1 X_i + \epsilon_i) = E(\beta_0) + E(\beta_1 X_i) + E(\epsilon_i) = \beta_0 + \beta_1 X_i $ $\endgroup$ – response May 23 '13 at 14:18
  • $\begingroup$ I had that backwards as well! Clearly expectation of a constant is constant, and $E(e)$ = 0 seems familiar too but I can't pinpoint why it's true. $\endgroup$ – user78504 May 23 '13 at 14:19
  • $\begingroup$ That is by assumption. Look up the assumptions of the linear regression model. $\endgroup$ – response May 23 '13 at 14:21

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