# 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. Apr 5, 2017 at 16:13
• @MichaelChernick it relates to the main question in the question Apr 5, 2017 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 Apr 5, 2017 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. Apr 5, 2017 at 17:19
• The Y$_i$ are random variables. Their values are observations. Apr 5, 2017 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 Apr 5, 2017 at 17:35