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Suppose that there is a dataset of 2D points $(x_i, y_i)$, consider the following statement:

"When the Pearson Correlation Coefficient(PCC) between $x$ and $y$ is equal to -1 or 1 (highest absolute values), data could be described by a straight line". That is, there are constants $a$ and $b$ such that:

$y_i = ax_i+b \;\;\: \forall i$

Now, I have two questions:

1) Is the above statement true?

2) More importantly, if the statement is true, please give a mathematical proof.

I have visually seen that a line can be fitted perfectly on such a dataset, but I am not sure this is always true, so I need a mathematical proof.

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    $\begingroup$ To answer 1): yes, a value of 1 implies that a linear equation describes the relationship between X and Y perfectly, with all data points lying on a line. $\endgroup$ – user2974951 Sep 5 '19 at 10:00
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    $\begingroup$ Part of the Cauchy-Schwarz Inequality asserts "Moreover, the two sides are equal if and only if $\mathbf {u}$ and $\mathbf {v}$ are linearly dependent," QED. $\endgroup$ – whuber Sep 5 '19 at 15:02
  • $\begingroup$ Just to be clear. Nothing guarantees that a newly chosen point will fall on the line. The function may only be linear in the region where the points were taken. It is also possible that a new point taken between two data points could fall off the line. Now if you have a strange error term distribution that has a probability less than 1 of being 0 & consequently a positive probability of being different from 0 a new point could fall off the line and the linear equation previously determined could still hold. $\endgroup$ – Michael R. Chernick Sep 14 '19 at 22:25
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For the answer 1), I just invite you to read the comment by @user2974951 which in a couple of words says exactly what is worth. I cite it "a value of 1 implies that a linear equation describes the relationship between X and Y perfectly, with all data points lying on a line". I stress the importance of linear, as the correlation is a measure for the linear interaction between variables.

Now for the point 2), I apologize if I only sketch a proof due to serious time constraints and very busy days on my side. But it will be enough. Consider that in a univariate regression with a single independent variable like $y_{i}=a+\beta x_{i}+ \epsilon_{i}$ with $i=1,...,n$, the $R^{2}$ of the regression can be retrieved as the square of the correlation coefficient between the two variable. Therefore setting the corr equal to 1 implies $R^{2}=1$. Take the formula for $R^{2}$ (here we take the one assuming there is an intercept as you hinted) given by:

$$R^2=1-\frac{\sum_{i=1}^{n}e_i^2}{\sum_{i=1}^{n}(y_i-\bar{y})^2}$$

Then set it to 1 and solve for the sum of squared errors at the numerator, then evaluate the implications for each observation residual. You get:

$$R^2=1 \rightarrow 1-\frac{\sum_{i=1}^{n}e_i^2}{\sum_{i=1}^{n}(y_i-\bar{y})^2} = 1 \rightarrow \frac{\sum_{i=1}^{n}e_i^2}{\sum_{i=1}^{n}(y_i-\bar{y})^2} = 0 \rightarrow \sum_{i=1}^{n}e_i^{2} = 0 \rightarrow e_i^{2}=0 \text{ for each i} \rightarrow e_i=0 \text{ for each i} \rightarrow y_{i}=a+\beta x_{i} \text{ for each i}$$

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  • $\begingroup$ For a much shorter demonstration, please see my comment to the question. $\endgroup$ – whuber Sep 5 '19 at 15:02
  • $\begingroup$ @whuber very nice! Thanks $\endgroup$ – Fr1 Sep 5 '19 at 15:40

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