aa r X i v : . [ s t a t . O T ] N ov Quantile of a Mixture
Carole Bernard ∗ and Steven Vanduffel †‡ November 19, 2014
Abstract
In this note, we give an explicit expression for the quantile of a mixture of tworandom variables. We carefully examine all possible cases of discrete and continuousvariables with possibly unbounded support. The result is useful for finding boundson the Value-at-Risk of risky portfolios when only partial information is available(Bernard and Vanduffel (2014)). ∗ Carole Bernard, Department of Statistics and Actuarial Science at the University of Waterloo (email: [email protected] ). † Corresponding author : Steven Vanduffel, Department of Economics and Political Sciences at VrijeUniversiteit Brussel (VUB). (e-mail: [email protected] ). ‡ C. Bernard gratefully acknowledges support from the Natural Sciences and Engineering ResearchCouncil of Canada, the Humboldt Research Foundation and the hospitality of the chair of mathematicalstatistics of Technische Universit¨at M¨unchen where the paper was completed. S. Vanduffel acknowledgesthe financial support of the BNP Paribas Fortis Chair in Banking. uantile of a Mixture Let X and Y be two random variables. We denote by F X and F Y their respectivemarginal distributions. For p ∈ (0 , , the quantile of X at level p is defined as F − X ( p ) = inf { x ∈ R | F X ( x ) > p } . (1)By convention, inf {∅} = ∞ and inf { R } = −∞ , so that the quantile is properly definedby (1) for all p ∈ [0 , X as a loss variable inwhich case F − X ( p ) can be broadly interpreted as the maximum loss (“Value-at-Risk”) onecan observe with p − confidence. Value-at-Risk computations are at the core of settingcapital requirements for banks and insurance companies.Consider a sum S = I X + (1 − I ) Y, where I is a Bernoulli distributed random variablewith parameter q and where the components X and Y are independent of I . Our objectiveis to find an explicit expression for the quantiles of S as a function of the quantiles ofits components X and Y . A direct application is to find bounds on Value-at-Risk in thecase when partial information is available (Bernard and Vanduffel (2014)). Theorem 1 (Quantile of a mixture) . Consider a sum S = I X + (1 − I ) Y, where I is aBernoulli distributed random variable with parameter q and where the components X and Y are independent of I . Define α ∗ ∈ [0 , by α ∗ := inf (cid:26) α ∈ (0 , | ∃ β ∈ (0 , n qα + (1 − q ) β = pF − X ( α ) > F − Y ( β ) (cid:27) and let β ∗ = p − qα ∗ − q ∈ [0 , . Then, for p ∈ (0 , ,s p := F − S ( p ) = max (cid:8) F − X ( α ∗ ) , F − Y ( β ∗ ) (cid:9) (2) This maximum can be computed explicitly by distinguishing along the four following casesfor F ( · ) and for G ( · ) : Case 1: F is continuous in s p and for all z < s p , F ( z ) < F ( s p ) Case 2: F is continuous in s p and there exists z < s p , F ( z ) = F ( s p ) Case 3: F is discontinuous in s p and for all z < s p , F ( z ) < F ( s − p ) Case 4: F is discontinuous in s p and there exists z < s p , F ( z ) = F ( s − p ) Case a: G is continuous in s p and for all z < s p , G ( z ) < G ( s p ) Case b: G is continuous in s p and there exists z < s p , G ( z ) = G ( s p ) Case c: G is discontinuous in s p and for all z < s p , G ( z ) < G ( s − p ) Case d: G is discontinuous in s p and there exists z < s p , G ( z ) = G ( s − p ) We have summarized the computations of s p in Table 1 for the sixteen possible com-binations. α ∗ = F ( s p ) β ∗ = G ( s p ) s p = F − X ( α ∗ ) s p = F − Y ( β ∗ ) α ∗ = F ( s p ) β ∗ = G ( s p ) s p = F − X ( α ∗ ) s p > F − Y ( β ∗ ) α ∗ = F ( s p ) s p = F − X ( α ∗ ) s p = F − Y ( β ∗ ) α ∗ = F ( s p )if F S ( s − p ) < p,s p = F − X ( α ∗ ) s p = F − Y ( β ∗ )if F S ( s − p ) = p,s p = F − X ( α ∗ ) s p > F − Y ( β ∗ )(2) α ∗ = F ( s p ) β ∗ = G ( s p ) s p = F − Y ( β ∗ ) s p > F − X ( α ∗ ) Impossible α ∗ = F ( s p ) s p = F − Y ( β ∗ ) s p > F − X ( α ∗ ) α ∗ = F ( s p ) s p = F − Y ( β ∗ ) s p > F − X ( α ∗ )(3) β ∗ = G ( s p ) s p = F − X ( α ∗ ) s p = F − Y ( β ∗ ) β ∗ = G ( s p ) s p = F − X ( α ∗ ) s p > F − Y ( β ∗ ) s p = F − X ( α ∗ ) s p = F − Y ( β ∗ ) if F S ( s − p ) < p,s p = F − X ( α ∗ ) s p = F − Y ( β ∗ )if F S ( s − p ) = p,s p = F − X ( α ∗ ) s p > F − Y ( β ∗ )(4) β ∗ = G ( s p )if F S ( s − p ) < p,s p = F − Y ( β ∗ ) s p = F − X ( α ∗ )if F S ( s − p ) = p,s p = F − Y ( β ∗ ) s p > F − X ( α ∗ ) β ∗ = G ( s p ) s p = F − X ( α ∗ ) s p > F − Y ( β ∗ ) if F S ( s − p ) < p,s p = F − X ( α ∗ ) s p = F − Y ( β ∗ )if F S ( s − p ) = p,s p = F − Y ( β ∗ ) s p > F − X ( α ∗ ) ImpossibleTable 1: Summary of all cases for the quantiles of a mixture where s p = F − S ( p ). In allcases, α ∗ is defined as (3) and β ∗ = p − qα ∗ − q G ( s p ), α ∗ = p − (1 − q ) β ∗ q > F ( s p ). Proof.
Denote by F ( x ) and G ( x ) the distributions of X resp. Y. Since X and Y areindependent of I we find for the distribution of S = I X + (1 − I ) Y,F S ( x ) = qF ( x ) + (1 − q ) G ( x ) x ∈ R . Let p ∈ (0 ,
1) and denote F − S ( p ) by s p , s p = inf { x ∈ R | qF ( x ) + (1 − q ) G ( x ) > p } . In what follows, when considering α, β ∈ (0 ,
1) we always assume that they satisfy qα +(1 − q ) β = p. Note that we define α ∗ as α ∗ := inf (cid:8) α ∈ (0 , | ∃ β ∈ (0 , / qα + (1 − q ) β = p and F − X ( α ) > F − Y ( β ) (cid:9) (3)and β ∗ = p − qα ∗ − q . The proof consists in verifying that s p can always be expressed as s p = max (cid:8) F − X ( α ∗ ) , F − Y ( β ∗ ) (cid:9) . (4)From Table 1, it is clear that (4) is proved. Let us now make the calculations case bycase to prove Table 1. 3ase 1: F is continuous in s p and for all z < s p , F ( z ) < F ( s p )In this case we always have that s p = F − X ( F ( s p )) . Hence, we only need to show that α ∗ = F ( s p ) (i.e. β ∗ = p − qF ( s p )1 − q ) and that s p = F − X ( α ∗ ) > F − Y ( β ∗ ) as in this case (4) willobviously hold.Since F − S ( p ) = s p then F S ( s − p ) = qF ( s − p )+(1 − q ) G ( s − p ) p F S ( s p ) = qF ( s p )+(1 − q ) G ( s p ). Thus, by continuity of F , qF ( s p ) + (1 − q ) G ( s − p ) p qF ( s p ) + (1 − q ) G ( s p ) . Thus, G ( s − p ) p − qF ( s p )1 − q G ( s p ) (5)(1a): G is continuous in s p and for all z < s p , G ( z ) < G ( s p ) . Then, s p = F − Y ( G ( s p )) . It isalso clear that for α < F ( s p ) and thus β > G ( s p ) , one has that F − X ( α ) < F − Y ( β ) . Hence,as per definition of α ∗ , one has α ∗ = F ( s p ) , β ∗ = G ( s p ) and s p = F − X ( α ∗ ) = F − Y ( β ∗ ) . (1b): G is continuous in s p and there exists z < s p , G ( z ) = G ( s p ) (thus, G is constanton the interval ( z, s p )). Then, F − Y ( G ( s p )) < s p = F − X ( F ( s p ) . However, for α < F ( s p )and thus β > G ( s p ) , one has that F − X ( α ) < F − Y ( β ) . Hence, as per definition of α ∗ ,α ∗ = F ( s p ) , β ∗ = G ( s p ) and s p = F − X ( α ∗ ) > F − Y ( β ∗ ) . Thus, s p = F − X ( α ∗ ) > F − Y ( β ∗ ).(1c): G has a discontinuity in s p and for all z < s p , G ( z ) < G ( s − p ) . From (5), in this case, F − Y (cid:16) p − qF ( s p )1 − q (cid:17) = s p . For α < F ( s p ) and thus β > p − qF ( s p )1 − q , F − X ( α ) < F − Y ( β ) . Hence, asper definition of α ∗ , α ∗ = F ( s p ) , β ∗ = p − qF ( s p )1 − q and s p = F − X ( α ∗ ) = F − Y ( β ∗ ) . (1d): G has a discontinuity in s p and there exists z < s p , G ( z ) = G ( s − p ) so that G isconstant on some interval ( r, s p ) with r < s p . From (5), F − Y (cid:18) p − qF ( s p )1 − q (cid:19) s p . If p − qF ( s p )1 − q > G ( s − p ) (or equivalently, F S ( s − p ) < p ) , then F − Y (cid:16) p − qF ( s p )1 − q (cid:17) = F − X ( F ( s p ) = s p . Clearly, for α < F ( s p ) and thus β > p − qF ( s p )1 − q , one has that F − X ( α ) < F − Y ( β ) . Hence, asper definition of α ∗ , one has α ∗ = F ( s p ) , β ∗ = p − qF ( s p )1 − q and s p = F − X ( α ∗ ) = F − Y ( β ∗ ) . If p − qF ( s p )1 − q = G ( s − p ) (or equivalently, F S ( s − p ) = p ) , then this implies that F − Y (cid:16) p − qF ( s p )1 − q (cid:17) p − qF ( s p )1 − q one has that F − X ( α ) < F − Y ( β ) s p . Hence, asper definition of α ∗ , one has α ∗ = F ( s p ) , β ∗ = p − qF ( s p )1 − q and s p = F − X ( α ∗ ) > F − Y ( β ∗ ) . Case 2: F is continuous in s p and there is a z < s p , F ( z ) = F ( s p ) ( F ( · ) is constant on ( z, s p ))(2a): this case can be obtained from (1b) by changing the role of X and Y .(2b): G is continuous in s p and there exists z < s p , G ( z ) = G ( s p ) . Thus G is constant onsome interval ( r, s p ) with r < s p . Hence, F − S ( p ) min( z, z ) < s p which contradicts thedefinition of s p = F − S ( p ). The case (2b) is impossible.(2c): G is discontinuous in s p and for all z < s p , G ( z ) < G ( s − p ) . From (5), in this case, F − Y (cid:16) p − qF ( s p )1 − q (cid:17) = s p > F − X ( F ( s p ). However, for all α > F ( s p ) and thus β < p − qF ( s p )1 − q itholds that F − X ( α ) > F − Y ( β ) . Hence, as per definition of α ∗ , α ∗ = F ( s p ) , β ∗ = p − qF ( s p )1 − q and s p = F − Y ( β ∗ ) > F − X ( α ∗ ) . (2d): G is discontinuous in s p and there exists z < s p , G ( z ) = G ( s − p ) . From (5), F − Y (cid:16) p − qF ( s p )1 − q (cid:17) s p . If p − qF ( s p )1 − q > G ( s − p ) (or equivalently, F S ( s − p ) < p ) , then F − Y (cid:16) p − qF ( s p )1 − q (cid:17) =4 p > F − X ( F ( s p ) . For α > F ( s p ) and thus β < p − qF ( s p )1 − q one has that F − X ( α ) > F − Y ( β ) . Hence, as per definition of α ∗ , α ∗ = F ( s p ) , β ∗ = p − qF ( s p )1 − q and F − X ( α ∗ ) < F − Y ( β ∗ ) = s p . The case that p − qF ( s p )1 − q = G ( s − p ) is excluded as it implies that F − S ( p ) < s p should hold(similar to the case (2b)) which is a contradiction with the definition of s p .Case 3: F has a discontinuity in s p and for all z < s p , F ( z ) < F ( s − p )In this case, s p = F − X ( F ( s p ) . This situation is merely identical to previous cases.(3a): it is the same as (1c) by changing the role of X and Y .(3b): it is the same as (2d) by changing the role of X and Y .(3c): Observe that F − X ( α ) = s p for all F ( s − p ) α F ( s p ) and also that F − Y ( β ) = s p for all G ( s − p ) β G ( s p ) . We also know that F S ( s − p ) p F S ( s p ) hence there exists F ( s − p ) α F ( s p ) and G ( s − p ) β G ( s p ) so that qα + (1 − q ) β = p and F − X ( α ) = F − Y ( β ) = s p . Therefore, F − X ( α ∗ ) = F − Y ( β ∗ ) = s p . (3d): Observe that F − X ( α ) = s p for all F ( s − p ) α F ( s p ) and also that F − Y ( β ) = s p for all G ( s − p ) < β G ( s p ) . We also know that F S ( s − p ) p F S ( s p ) and there are two possibilities:In the case when F S ( s − p ) < p , then there exists α ∈ ( F ( s − p ) , F ( s p )) and β ∈ ( G ( s − p ) , G ( s p )) so that qα + (1 − q ) β = p and F − X ( α ) = F − Y ( β ) = s p . Therefore, F − X ( α ∗ ) = F − Y ( β ∗ ) = s p . In the case when F S ( s − p ) = p, then qF ( s − p ) + (1 − q ) G ( s − p ) = p and one has that F − X ( F ( s − p )) > F − Y ( G ( s − p ) ∗ ) , while for α < F ( s − p ) and β > G ( s − p ) one has that F − X ( α )
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Available at SSRN 2393054 ..