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I am not quite familiar with the concept of correlation. The Pearson's correlation coefficient is defined as:

$$\rho_{X,Y}=\mathrm{corr}(X,Y)={\mathrm{cov}(X,Y) \over \sigma_X \sigma_Y} ={E[(X-\mu_X)(Y-\mu_Y)] \over \sigma_X\sigma_Y}$$

which makes use of Mean and Standard deviation. But, is it strict to the normal distributed data ? Since Gaussian distribution is configured by mean and variance.

I currently have some which is apparently not following normal distribution. When assessing the correlation between them, is correlation appropriate here ?

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    $\begingroup$ The correlation coefficient exists whenever the relevant expectations exist. Very much not confined to the normal. $\endgroup$ Jun 4, 2014 at 20:14

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Correlation makes sense in any case where the two standard deviations are finite and not zero

As a non-normal distribution example, if you have two independent Poisson distributed random variables, $X$ with mean $\lambda$ and $Z$ with mean $\mu$, and you let $Y=X+Z$ then you can make the following useful statements:

  • $Y$ is a Poisson distributed random variable with mean $\lambda + \mu$
  • $X$ has standard deviation $\sigma_X^{\,}=\sqrt{\lambda}$
  • $Y$ has standard deviation $\sigma_Y^{\,}=\sqrt{\lambda+\mu}$
  • the covariance of $X$ and $Y$ is $\mathrm{cov}(X,Y)=\lambda$
  • the correlation of $X$ and $Y$ is $\rho_{X,Y}^{\,}=\mathrm{corr}(X,Y)=\dfrac{1}{\sqrt{1+\frac{\mu}{\lambda}}}$
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The Pearson's correlation is often used for normal distributions. It doesn't need to assume normality, although it assumes finite variances and finite covariance. When the variables are bivariate normal, it represents a reliable and exhaustive measure of association.

However, if you have some non-normal distribution, you can consider other measures, such as the Spearman's rho (particularly if you are interested in monotonic rather than linear association), or the Kendall's tau.

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  • $\begingroup$ i dont understand why the downvotes; in fact, if there is a heavily skewed distribution under consideration, the mean (which makes a significant part of Pearson's correlation coefficient) is not a relaible "functional" of the underlying distribution. $\endgroup$
    – lmaosome
    Apr 1, 2023 at 1:20

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