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Assume that I have a random variable $X$ (which I know will have a power law tail).

If I had the CDF for $X$, $G(x)$, then I could easy calculate this tail as something like, $$ \alpha = \lim_{x\to\infty}\left(\frac{x G'(x)}{1 - G(x)}\right) $$ Alternatively, in logs with $X = e^Z$ and a CDF for $z$ instead, F(z), then the calculation should be $$ \alpha = \lim_{z\to\infty}\left(\frac{F'(z)}{1 - F(z)}\right) $$ (you can try these out for $X$ a pareto distribution, which would mean that $Z$ is an exponential distribution from standard probability).

QUESTION: Lets say that I cannot find an expression for $F(z)$ (or $G(x)$ if you prefer, I should be able to convert) Instead, I am able to calculate a (mostly) analytical quantile function $Q(q) = F^{-1}(q)$ for $z$. Assume monotonicity, invetibility, etc.

My question is: From only $Q$ and its derivatives, is there a limit I can take to find $\alpha$ for the underlying variable? (I imagine the limit would be as $q \to 1$)?

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up vote 6 down vote accepted

From the Chain Rule,

$$1 = \frac{d}{dq}\left(q\right) = \frac{d}{dq}\left(F(Q(q))\right) = F^\prime(Q(q)) Q^\prime(q).$$

Letting $0 \lt q \lt 1$, substituting $z = Q(q)$ into the limiting expression for $\alpha$, and solving the preceding equation for $F^\prime(Q(q))$ in terms of $Q^\prime(q)$ gives

$$\alpha = \lim_{z\to\infty}\left(\frac{F'(z)}{1 - F(z)}\right) = \lim_{Q(q)\to\infty}\left(\frac{F'(Q(q))}{1 - F(Q(q))}\right) =\lim_{q\to 1^{-}}\left(\frac{1/Q^{\prime}(q)}{1 - q}\right).$$

Both the numerator and denominator approach zero. Applying L'Hopital's Rule gives the simple useful formula

$$\alpha =\lim_{q\to 1^{-}}\left(\frac{\frac{d}{dq}\left(1/Q^{\prime}(q)\right)}{\frac{d}{dq}(1 - q)}\right) = \lim_{q\to 1^{-}}\frac{Q^{\prime\prime}(q)}{\left(Q^\prime(q)\right)^2}.$$

For a Pareto distribution $F(z) = 1 - \exp(-\alpha z),$ whence $Q(q) = -\frac{1}{\alpha}\log(1-q)$ and differentiating yields $Q^\prime(q) = \frac{1}{\alpha}\frac{1}{1-q}$, $Q^{\prime\prime}(q) = \frac{1}{\alpha}\frac{1}{(1-q)^2}$. We obtain

$$\alpha = \lim_{q\to 1^{-}}\frac{Q^{\prime\prime}(q)}{\left(Q^\prime(q)\right)^2} = \lim_{q\to 1^{-}}\frac{\frac{1}{\alpha}\frac{1}{(1-q)^2}}{\left(\frac{1}{\alpha}\frac{1}{1-q}\right)^2} = \lim_{q\to 1^{-}} \alpha,$$

which checks out beautifully.

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spectacular. Thanks – jlperla Jun 20 '14 at 19:13

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