Rachel Villa

2022-05-23

A random sample of size $n$ from a bivariate distribution is denoted by $({x}_{r},{y}_{r}),r=1,2,3,...,n$. Show that if the regression line of $y$ on $x$ passes through the origin of its scatter diagram then

$$\overline{y}\sum _{r=1}^{n}{x}_{r}^{2}=\overline{x}\sum _{r=1}^{n}{x}_{r}{y}_{r}$$

where $(\overline{x},\overline{y})$ is the mean point of the sample.

$$\overline{y}\sum _{r=1}^{n}{x}_{r}^{2}=\overline{x}\sum _{r=1}^{n}{x}_{r}{y}_{r}$$

where $(\overline{x},\overline{y})$ is the mean point of the sample.

Harper Heath

Beginner2022-05-24Added 9 answers

Recall that the OLS estimators are

$${\hat{\beta}}_{0}={\overline{Y}}_{n}-{\hat{\beta}}_{1}{\overline{X}}_{n},\phantom{\rule{1em}{0ex}}{\hat{\beta}}_{1}=\frac{\sum {X}_{i}{Y}_{i}-n\overline{X}\overline{Y}}{\sum {X}_{i}^{2}-n{\overline{X}}^{2}},$$

because the line passes through the origin, you have that ${\hat{\beta}}_{0}=0=\overline{Y}-{\hat{\beta}}_{1}\overline{X}$, thus

$$\frac{\overline{Y}}{\overline{X}}=\frac{\sum {X}_{i}{Y}_{i}-n\overline{X}\overline{Y}}{\sum {X}_{i}^{2}-n{\overline{X}}^{2}},$$

or

$$\overline{Y}\sum {X}_{i}^{2}-\overline{Y}n{\overline{X}}^{2}=\overline{X}\sum {X}_{i}{Y}_{i}-n\overline{Y}{\overline{X}}^{2},$$

hence

$$\overline{Y}\sum {X}_{i}^{2}=\overline{X}\sum {X}_{i}{Y}_{i}.$$

$${\hat{\beta}}_{0}={\overline{Y}}_{n}-{\hat{\beta}}_{1}{\overline{X}}_{n},\phantom{\rule{1em}{0ex}}{\hat{\beta}}_{1}=\frac{\sum {X}_{i}{Y}_{i}-n\overline{X}\overline{Y}}{\sum {X}_{i}^{2}-n{\overline{X}}^{2}},$$

because the line passes through the origin, you have that ${\hat{\beta}}_{0}=0=\overline{Y}-{\hat{\beta}}_{1}\overline{X}$, thus

$$\frac{\overline{Y}}{\overline{X}}=\frac{\sum {X}_{i}{Y}_{i}-n\overline{X}\overline{Y}}{\sum {X}_{i}^{2}-n{\overline{X}}^{2}},$$

or

$$\overline{Y}\sum {X}_{i}^{2}-\overline{Y}n{\overline{X}}^{2}=\overline{X}\sum {X}_{i}{Y}_{i}-n\overline{Y}{\overline{X}}^{2},$$

hence

$$\overline{Y}\sum {X}_{i}^{2}=\overline{X}\sum {X}_{i}{Y}_{i}.$$

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