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Let $X=(1,2,3,...,20)$. Suppose that $Y=(y_1,y_2,...,y_{20})$ with $y_i=x_i^2$ and $Z=(z_1,z_2,...,z_{20})$ with $z_i=e^{x_i}$. Pearson correlation coefficient is defined by formula \begin{equation} \rho(X,Y)=\frac{\sum_{i=1}^{20} (x_i-\bar{x})(y_i-\bar{y})}{\sqrt{(\sum_{i=1}^{20}(x_i-\bar{x})^{2})(\sum_{i=1}^{20}(y_i-\bar{y})^{2})}} \end{equation}

If $\rho(X,Y)=1$, we can say that $X$ and $Y$ have a linear correlation. If $0.7\leq\rho(X,Y)<1$ then $X$ and $Y$ has a strong linear correlation, if $0.5\leq\rho(X,Y)<0.7$ then $X$ and $Y$ has a modest linear correlation, and if $0\leq\rho(X,Y)<0.5$ then $X$ and $Y$ has a weak linear correlation. Using this formula, we get $\rho(X,Y)=0.9$ and $\rho(X,Z)=0.5$. However, the relationship between $X$ and $Y$ is actually quadratic but they have the high correlation coefficient that indicate linear correlation.

So, my question is what is "linear correlation" actually between $X$ and $Y$ ? Since $\rho(X,Z)=0.5$ indicate the modest correlation coefficient, what is another intepretation of this value? What is the difference between $\rho(X,Y)$ and $ \rho(X,Z)$, noting that $Y$ and $Z$ is not a linear function of $X$.

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can you correct the outlook of the question? – Seyhmus Güngören Mar 17 '13 at 23:50
In other word, correlation betwen X and Y tend to 1 but $y_i=x_i^2$ is not linear? How come? – beginner Mar 18 '13 at 0:06
It is okay. Because when you identify a squared function by a linear function, your loss is not too much, especially compared to other non-linear functions. – Seyhmus Güngören Mar 18 '13 at 0:08
Could you give any reference about this material? – beginner Mar 18 '13 at 0:10 this could help at first hand. – Seyhmus Güngören Mar 18 '13 at 0:12

A linear function is given as $ax+b$ for some coefficients $a$ and $b$. If you have $y=ax+b$, then you can obtain $y$ from $x$ via some mean shifting and scaling, which are quite simple operations. If you have two such functions, then whenever you compute $a$ and $b$, you are done. They are completely correlated. If there is another function and you do linear regression and you found some $a$ and $b$ but the variance of the error of this regression is quite large then your correlation will degrade, it will seem that you need quite much than just mean shifting and scaling.

In your example $\exp(x)$ is a quite non-linear function especially compared to $x^2$. You can see this via Tailor expansion; $\exp(x)$ has infintely many terms, indicating the strength of the non-linearity compared to $x^2$ which is a single term. Therefore the correlation is much lesss.

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I also try $X=(1, 1.1 ,1.2,...,5)$ and $Z=(z_i)$ with $z_i=e^{x_i}$ and we get $\rho(X,Z)=0.89$. How come? – beginner Mar 18 '13 at 0:17
do you have a programming software? plot them. You will see immediately why. – Seyhmus Güngören Mar 18 '13 at 0:25
SO your argument using taylor(MacLaurin) expansion is not true in general right? – beginner Mar 18 '13 at 0:30
it is true in general for the same input data. Try to plot everything we've talked about. You will understand what i am talking about. – Seyhmus Güngören Mar 18 '13 at 0:36
@ Seyhmus Güngören : WHat is the mathematical ecplaination about this phenomena? – beginner Mar 18 '13 at 0:37

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