# Prediction Interval, linear regression - why is a future response a random variable but other responses are not random variables?

We have a new observation $x_0$, whose response will be $Y_0 = \beta_0+\beta_1x_0+\epsilon_0$. We want to predict $Y_0$.

The estimator that we use is $\hat{Y}_0 = \hat{\beta}_0+\hat{\beta}_1x_0$.

The books goes on finding $E[\hat{Y}_0-Y_0]=0$ and $var[\hat{Y}_0-Y_0]=\sigma^2(1+h_{00})$. Also it says that since $Y_0$ is a random variable and it is normally distributed, we know that we can write $$0.95 = P\left(-c_1 < \frac{\hat{Y}_0 - Y_0}{\hat{\sigma} (1+h_{00})} < c_2 \right)$$ and then goes on finding the prediction interval.

So a couple of my questions are:

• How do we know $Y_0$ is normally distributed? But more importantly, how do we know it is a random variable?
• if $Y_0$ is a random variable, then why the responses $Y_i = \beta_0+\beta_1x_i+\epsilon_i$ are not treated as random variables? (or are they?)
• Finally, if $Y_0$ and $Y_i$ are treated in the same way, i.e. they have the same distribution, how come their estimators $\hat{Y}_0 \equiv \hat{\beta}_0+\hat{\beta_1}x_0$ and $\hat{Y}_i \equiv \hat{\beta}_0+\hat{\beta_1}x_i$ are not the same thing?
• The actual questions in the text do not relate to the question in the title. – Michael R. Chernick Apr 5 '17 at 16:13
• @MichaelChernick it relates to the main question in the question – Euler_Salter Apr 5 '17 at 16:17

I will assume that $x_i$ are a set of observations with known values of $Y_i$ to which you are fitting a model. Once the model has been fitted, you are left with known values of $\beta_0$ and $\beta_1$.

Q2

As you correctly state in your question, $Y_i = \beta_0+\beta_1x_i+\epsilon_i$. From this, you can calculate all the values of $\epsilon_i$ which are the model residuals. They are known because you know both $x_i$ and $Y_i$.

When you introduce a new observation, $x_0$, you can use the equation, $Y_0 = \beta_0+\beta_1x_0+\epsilon_0$, to calculate $Y_0$. Except, you do not know $\epsilon_0$. It is a random variable that leads to $Y_0$ being a random variable.

$\epsilon_i$, $\beta_0$, $\beta_1$, and $x_i$ are fixed values, so $Y_i$ is a fixed value. While $\epsilon_0$ is unknown, we do know that it is normally distributed with a mean of zero ($\hat{\epsilon_0} = 0$). For any given value of $x_0$ we expect $\epsilon_0$ to be a random number sampled from this distribution.

• Although for example I often see $Y_i \sim N(\beta_0+\beta_1x_i, \sigma^2)$, which would make me think that $Y_i$ are actually random variables – Euler_Salter Apr 5 '17 at 17:07
• The tilde means "has the probability distribution of" so the formula $Y_i \tilde N(\beta_0 + \beta_1 x_i, \sigma^2)$ is stating that the values of $Y_i$ are normally distributed around the mean of the model, $\hat{Y_i}$. The difference is that $Y_i$ is a set of many values which are normally distributed whereas $\epsilon_0$ has only a single value but the probability of which value it will take is normally distributed. – Eumenedies Apr 5 '17 at 17:19
• The Y$_i$ are random variables. Their values are observations. – Michael R. Chernick Apr 5 '17 at 17:30
• okay so now here it comes the super question: if here we have $var[\hat{Y}_0-Y_0]=\sigma^2(1+h_{00})$, how come that $var(R_i)=var(residuals)=var(Y_i-\hat{Y}_i) \neq \sigma^2(1+h_{ii})$ but rather $var(R_i)=\sigma^2(1-h_{ii})$? In the book explaination they say that the covariance here doesn't vanish. But how come it doesn't? We just said that it does and it produces the above result – Euler_Salter Apr 5 '17 at 17:35