4
$\begingroup$

I'd need to know if it's possible to reach [-1,1] bounds with Pearson's correlation with a generic pair of Gamma random variables. The problem as you may imagine is there's no known closed form for the quantile function (at least to my knowledge). Is there a solution? Even a research paper could be enough, actually I haven't been finding anything.

$\endgroup$
4
  • 1
    $\begingroup$ Are both variables drawn from the same Gamma distributions, or are you asking about (say) a Gamma and a Gaussian? $\endgroup$ Commented May 13, 2015 at 18:58
  • $\begingroup$ How would the quantile function be related to bounds on correlation coefficients? $\endgroup$
    – whuber
    Commented May 13, 2015 at 19:33
  • 1
    $\begingroup$ I was trying to replicate this reasoning... stats.stackexchange.com/questions/66775/… ps a generic pair of Gamma variables $\endgroup$ Commented May 14, 2015 at 9:17
  • $\begingroup$ An exponential variate is a $\Gamma(1)$ variable, so the answer you reference applies to any two Gamma variables with the same shape parameters both equal to $1$. It generalizes easily to other common shape parameter values. Although formulae may be difficult to obtain, two things can immediately be deduced from basic properties of Gamma variables (there is a one-to-one relationship between shape and skewness) and of correlation: the lower bound of $-1$ cannot be attained for different variables and the upper bound can be attained only when they have the same shape parameter. $\endgroup$
    – whuber
    Commented May 14, 2015 at 18:58

1 Answer 1

5
$\begingroup$

If $X$ and $Y$ are perfectly correlated random variables (the Pearson correlation coefficient $\rho$ has value $+1$ or $-1$), then it must be that $Y = \rho aX + b = \pm aX+b$ where $a >0$.

A Gamma random variable $X$ with shape parameter $t>0$ and scale parameter $\theta>0$ has density function $$f_X(x) = \frac{1}{\theta \cdot \Gamma(t)}\left(\frac{x}{\theta}\right)^{t-1}\exp\left(-\frac{x}{\theta}\right)\mathbf 1_{\{x\colon x>0\}}, ~~t > 0, ~~\theta > 0. \tag{1}$$ We write $X \sim \Gamma(t,\theta)$.

With this characterization, if $X$ and $Y$ are perfectly positively correlated random variables and $X \sim \Gamma(t,\theta)$, then $Y = aX+b \sim \Gamma(t, a\theta)$ provided that $b = 0$. If $b \neq 0$, then $Y$ is a displaced Gamma random variable whose density function is moved rightwards by $b$ units. Note that $X$ and $Y$ have the same shape parameter. It is not possible for two Gamma random variables with different shape parameters to be perfectly positively correlated. On the other hand, no two Gamma random variables can have Pearson correlation coefficient $-1$. If $X \sim \Gamma(t,\theta)$, then $Y = -aX+b$ takes on negative values with positive probability, and thus cannot have a Gamma distribution.

A somewhat more general definition of Gamma random variables allows $\theta$ to take on negative values, and the random variable takes on positive values only or negative values only according as $\theta > 0$ or $\theta < 0$. Thus, for $t > 0$ and $\theta \neq 0$, we have that $X \sim \Gamma(t,\theta)$ has density function $$f_X(x) = \frac{1}{|\theta| \cdot \Gamma(t)}\left(\frac{x}{\theta}\right)^{t-1}\exp\left(-\frac{x}{\theta}\right)\mathbf 1_{\{x\colon \operatorname{sgn}(x) = \operatorname{sgn}(\theta)\}}\tag{2}$$ so that $-X \sim \Gamma(t,-\theta)$. With this characterization, two perfectly correlated Gamma random variables $X$ and $Y$ necessarily have the same shape parameter while their scale parameters must have the same sign or opposite sign according as $\rho = +1$ or $\rho = -1$, and it must be that $Y = \rho a X$ where $a > 0$.


A different generalization of Gamma random variables is not considered here.

$\endgroup$
4
  • $\begingroup$ The problem becomes subtler and more interesting when you use a slightly more general notion of Gamma variable that includes negative scale parameters. That would rule out your argument, requiring a (slightly) more penetrating analysis of the situation. A more challenging way to understand the question would be to obtain upper and lower bounds for the correlations as a function of the two shape and two scale parameters. $\endgroup$
    – whuber
    Commented May 14, 2015 at 21:03
  • 1
    $\begingroup$ @whuber Could you provide a reference for the slightly more general notion of Gamma random variables? Wikipedia's version has positive-valued random variables. If negative scale parameters mean a reflected Gamma distribution (RV takes on negative values only), then $X \sim \Gamma(t,\lambda)$ makes $-aX$ a $\Gamma(t, -\lambda/a)$ random variable perfectly negatively correlated with $X$. $\endgroup$ Commented May 14, 2015 at 21:54
  • $\begingroup$ To understand this generalization, which you have done correctly, requires no reference, so references are of interest only insofar as they indicate that the generalization has potential use. To that end, inter alia, Bieniek and Szynal, Limit Distribution of Differences and Quotients of Non-adjacent kth Record Values (Probability and Mathematical Statistics, vol. 23, Fasc. 1, 2003) employ this generalization. (They use a different terminology, though: to make all scale parameters positive they consider Gammas with negative scales to be "Negative Gamma" distributions.) $\endgroup$
    – whuber
    Commented May 14, 2015 at 22:13
  • $\begingroup$ just what I was asking for! thanks (also for the intuition behind the result!) $\endgroup$ Commented May 17, 2015 at 8:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.