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I have been given this formula for upper tail dependence and read that tail dependence depends on the copula and not the marginals: $$ \lambda_U = \lim_{a \to 1} \Pr[Y>F_Y^{-1}(a)\mid X>F_X^{-1}(a)] . $$

Would the inverse functions $F_X^{-1}(a)$ and $F_Y^{-1}(a)$ be copula functions ?

I have an intuitive understanding of what couplas are, but my knowledge of stats is quite weak, I would appreciate if you provide simplistic answers where possible.

QuantIbex
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AfterWorkGuinness
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  • The limit as $a \to 1$ is missing in your formula for $\lambda_U$. – QuantIbex Apr 22 '14 at 12:45
  • Also, the conditionning should be on $X > F_x^{-1}(a)$, not $Y > F_x^{-1}(a)$ . – QuantIbex Apr 22 '14 at 12:47
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    This [question](http://stats.stackexchange.com/q/72999/27403) provides the link between the upper tail dependence coefficient and the copula of $X$and $Y$. – QuantIbex Apr 22 '14 at 12:54
  • Hi @QuantIbex from the post you linked to above, I see that the functions in the formula for $lambda_u$ are not the coupla, but I don't understand from that post where the coupla fits in – AfterWorkGuinness Apr 22 '14 at 14:15
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    The link I provided gives the expression $$ \lambda_U = \lim_{u \to 1} \frac{1 - 2u + C(u,u)}{1-u}, $$ where $C$ is the copula (not "coupla") associated with $X$ and $Y$. Obviously, $\lambda_U$ depends on the copula, but not on the margins. @ocram's answer shows a way to derive this expression from the definition of $\lambda_U$. Please clarify if you need more information. – QuantIbex Apr 22 '14 at 15:21

1 Answers1

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\begin{align} \Pr & \big[ Y > F_Y^{-1}(a) \,\big| X > F_X^{-1}(a) \big] \\[1em] & = \frac{\Pr \big[ Y > F_Y^{-1}(a), \, X > F_X^{-1}(a) \big]}{\Pr \big[ X > F_X^{-1}(a) \big]} \\[1em] & = \frac{\Pr \big[ Y < F_Y^{-1}(a), \, X < F_X^{-1}(a) \big] + \Pr \big[ X > F_X^{-1}(a) \big] + \Pr \big[ Y > F_Y^{-1}(a) \big] - 1}{1 - \Pr \big[ X < F_X^{-1}(a) \big]} \\[1em] & = \frac{C\big(a, a\big) + 1 - 2a}{1 - a} \end{align}

Thus, ''tail dependence [$\lambda_U$] depends on the copula and not the marginals.´´

Notes:

$$ \Pr \big[ A \,\big| B \big] = \frac{\Pr[A,\, B]}{\Pr[B]} $$

$$ \Pr[X > x] = 1 - \Pr[X < x] $$

$$ \Pr[X > x, Y > y] = \Pr[X < x, Y < y] + \Pr[X > x] + \Pr[Y > y] - 1 $$

ocram
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