g2esebyy7

2022-04-25

Simple Linear Regression - Difference between predicting and estimating?

Here is what my notes say about estimation and prediction:

Estimating the conditional mean

"We need to estimate the conditional mean $\beta}_{0}+{\beta}_{1}{x}_{0$ at a value $x}_{0$, so we use $\hat{{Y}_{0}}=\hat{{\beta}_{0}}+\hat{{\beta}_{1}}{x}_{0}$ as a natural estimator." here we get

$\hat{{Y}_{0}}~N({\beta}_{0}+{\beta}_{1}{x}_{0},{\sigma}^{2}{h}_{00})\text{}\text{where}{h}_{00}=\frac{1}{n}+\frac{{({x}_{0}-\stackrel{\u2015}{x})}^{2}}{(n-1){s}_{x}^{2}}$

with a confidence interval for $E\left({Y}_{0}\right)={\beta}_{0}+{\beta}_{1}{x}_{0}$ is

$(\hat{{b}_{0}}+\hat{{b}_{1}}{x}_{0}-cs\sqrt{{h}_{00}},\hat{{b}_{0}}+\hat{{b}_{1}}{x}_{0}+cs\sqrt{{h}_{00}})$

where $c={t}_{n-2,1-\frac{\alpha}{2}}$ Where these results are found by looking at the shape of the distribution and at $E\left(\hat{{Y}_{0}}\right)$ and $var\left(\hat{{Y}_{0}}\right)$

Predicting observations

"We want to predict the observation $Y}_{0}={\beta}_{0}+{\beta}_{1}{x}_{0}+{\u03f5}_{0$ at a value $x}_{0$"

$E(\hat{{Y}_{0}}-{Y}_{0})=0\text{}\text{and}var(\hat{{Y}_{0}}-{Y}_{0})={\sigma}^{2}(1+{h}_{00})$

Hence a prediction interval is of the form

$(\hat{{b}_{0}}+\hat{{b}_{1}}{x}_{0}-cs\sqrt{{h}_{00}+1},\hat{{b}_{0}}+\hat{{b}_{1}}{x}_{0}+cs\sqrt{{h}_{00}+1})$

Colonninisxi

Beginner2022-04-26Added 16 answers

You have to distinguish between estimating $E[Y\mid X={x}_{0}]$ and $\left\{Y\right\}\left({x}_{0}\right)$

For the former you have that$\hat{E[Y\mid X=x]}=\hat{{\beta}_{0}}+{\hat{\beta}}_{1}{x}_{0},$

such that$Var\left(\hat{E[Y\mid X=x]}\right)={\sigma}^{2}[\frac{1}{n}+\frac{{({x}_{0}-\stackrel{\u2015}{x})}^{2}}{(n-1){s}_{x}^{2}}]={\sigma}^{2}{h}_{00},$

where for the latter you have that

$Var\left(\hat{Y}\left({x}_{0}\right)\right)=Var\left(\hat{E[Y\mid X=x]}\right)+Var\left({\u03f5}_{0}\right)={\sigma}^{2}(1+{h}_{00}).$

Namely, for the former case you are estimating the conditional mean of Y at$x}_{0$ , while for the latter you are estimating (predicting) the value itself. The conditional mean "smooths out" the variance of the error term as $E[Y\mid X=x]={\beta}_{0}+{\beta}_{1}x$ , however the prediction of a new value $Y\left({x}_{0}\right)$ should consider also the (estimated) variance of the error term (to account for the fluctuation around the conditional mean).

For the former you have that

such that

where for the latter you have that

Namely, for the former case you are estimating the conditional mean of Y at

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