Risk measures
Proofs and additional remarks
Christian Y. Robert
ISFA - Université Lyon 1
October 2011
⊲ COMONOTONIC RISKS DEFINITIONS
1. A set S in R2 is said to be comonotonic, if, for all (y1, y2) and (z1, z2) in this set, yi < zi for some i implies yj ≤ zj for j = i.
◦ Notice that a comonotonic set is a ‘thin’ set, in the sense that it is contained in a one-dimensional subset of R2.
2. When the support of a random vector is a comonotonic set, the random vector itself and its joint distribution are called comonotonic.
PROPOSITION :
1. A random vector (X, Y ) is comonotonic if and only if X and Y may be written as non-decreasing functions of the same random variable.
2. A random vector (X, Y ) is comonotonic if and only if
P(X ≤ x, Y ≤ y) = min{P(X ≤ x),P(Y ≤ y)}
for all x, y ∈ R.
3. A random vector (X, Y ) is comonotonic if and only if (X, Y ) =d (FX−1(U), FY−1(U))
where FX−1 stands for the quantile function of X (see below), and U is a random variable that is uniformly distributed over the unit interval (0,1).
⊲ COMPARISONS OF RISKS DEFINITION
Let Y be a set of univariate distribution functions. The binary relation is a partial order on Y if for any elements X with df FX, Y with df FY and Z with df FZ in Y, the following properties hold :
(i) If XY and Y Z then XZ (transitivity).
(ii) XX (reflexivity).
(iii) If XY and Y X then X = Y (antisymmetry).
If, in addition, for any given pair X and Y of elements of either XY or Y X holds, then is said to be a total order.
Remark : We write XY but we actually mean FXFY . In other words, when we say that a risk X is smaller than a risk Y for the stochastic order relation, we assert that this ordering holds for the respective dfs of these risks. Therefore, the joint distribution of X and Y is irrelevant ; only their marginal distributions are important.
⊲ First-order stochastic dominance
PROP : Risk Y dominates risk X stochastically at first order if and only if there exist random variables X′ =d X and Y ′ =d Y such that P(X′ ≤ Y ′) = 1.
Remark : If Y DS1 X, then FY (d) ≤ FX(d), ∀d ∈ R.
If moreover FX and FY are increasing df, we have FY−1(FY (d) = d, FX(X) =d U where U is uniformly distributed over the unit interval (0,1) and FY−1(U) =d Y (see below).
Therefore X = X′ and Y ′ = FY−1(FX(X)) are desirable random variables since X ≤ Y ′ a.s.
by the previous relation.
PROP : Risk Y dominates risk X stochastically at first order if and only if E[u(−X)] ≥ E[u(−Y )]
for all non-decreasing functions u (such that the expectations exist).
PROOF : First note that, letting v(x) = −u(−x), the condition : E[u(−X)] ≥ E[u(−Y )]
for all non-decreasing functions u is equivalent to the condition : E[v(X)] ≤ E[v(Y )]
for all non-decreasing functions v.
The ⇐ part is obvious since F¯X(z) = E[I
{X>z]] and the function x → I
{X>z] is non-decreasing for any z. To get the converse implication, it suffices to invoke the previous proposition and to write
E[v(X)] = Ev(X′) ≤ Ev(Y ′) = E[v(Y )].
PROP : If X and Y have density propobability functions such that there exists a real number c and :
fX (d) ≥ fY (d) for d ∈] − ∞, c) fX (d) ≤ fY (d) for d ∈ [c,∞[.
then Y DS1 X.
PROOF : For x < c, we get FX(x) =
x
−∞fX (u)du ≥
x
−∞ fY (u) du = FY (x) For x > c, we get
FX(x) = 1 −
∞
x fX (u)du ≥ 1 −
∞
x fY (u)du = FY (x) and this concludes the proof.
⊲ Second-order stochastic dominance
PROP : Risk Y dominates risk X stochastically at second order if and only if E[u(−X)] ≥ E[u(−Y )]
for all non-decreasing and concave function u (such that the expectations exist).
PROOF : First note that, letting v(x) = −u(−x), the condition : E[u(−X)] ≥ E[u(−Y )]
for all non-decreasing concave functions u is equivalent to the condition : E[v(X)] ≤ E[v(Y )]
for all non-decreasing convexe functions v.
The ⇐ implication is obvious since the function x → (x − t)+ is convex for all t ∈ R. To get the converse, note that every continuous function v convex is the limit
of an increasing sequence of functions : vn(x) = α1 + α2x +
n j=0
β(n)j
x − t(n)j
+
with β(n)j ≥ 0. It allows us to write E[vn(X)] = α1 + α2E[X] +
n j=0
β(n)j E
X − t(n)j
+
≤ α1 + α2E[Y ] +
n j=0
β(n)j E
Y − t(n)j
+
= E[vn(Y )]
for every n. Taking the limit (Monotone convergence theorem) yields E[v(X)] ≤ E[v(Y )].
PROP : Risk Y dominates risk X stochastically at second order if and only if there exists a random variable D such that :
X + D =d Y and E[D|X] ≥ 0 a.s.
PROOF : The ⇐ implication is derived by using the conditional Jensen’s inequality E[(Y − d)+] = E[(X + D − d)+] = EX[E[ (X + D − d)+
+|X
≥ EX[(X + E[D|X] − d)+ ≥ E[(X − d)+].
The other implication is difficult to prove.
PROP : If E[X] ≤ E[Y ] and if there exists a real number c such that FX (d) ≤ FY (d) for d ∈] − ∞, c)
FX (d) ≥ FY (d) for d ∈ [c,∞[
then Y DS2 X. PROOF : Note that
E[ (X − d)
+] =
∞
d (x − d)dFX (x)
= −[(x − d)(1 − FX (x))]∞d +
∞ d
F¯X(x)dx =
∞ d
F¯X(x)dx π′X(d) = −(1 − FX(d)) = −F¯X(d).
Moreover limd→∞E[ (X − d)
+] = 0 and since E[ (X − d)
+] +d = E[max(X, d)]
d→−∞lim
E[ (X − d)
+] + d = E[X].
Let us consider the function φ(d) = πY (d) − πX(d). We have limd→−∞ φ(d) = E[Y ] − E[X] ≥ 0, limd→∞ φ(d) = 0 and φ′(d) = FY (d) − FX(d).
PROP : If Y DS1 X, then Y DS2 X. PROOF : Y DS1 X if and only if
E[u(−X)] ≥ E[u(−Y )]
for all non-decreasing function u (such that the expectations exist). Hence the in- equality holds if u is a non-decreasing and concave function. So it is clear that Y DS2 X.
Note that two risks may be stochastically comparable at second order but not at first order.
⊲ PROPERTIES OF RISK MEASURES
PROP : Π satisfies the properties 1) Monotonicity, 2) Objectivity, iff it satisfies the Invariance by first-order stochastic dominance property.
PROOF : The ⇐ implication is derived by noting that if P(X ≤ Y ) = 1 then P(X > d) ≤ P(Y > d) for all d ∈ R and hence the Monotonicity property is satisfied. Moreover if X =d Y then X DS1 Y and Y DS1 X. By the first-order stochastic dominance property, we deduce that Π(X) = Π(Y ).
The ⇒ part is proven by noting that X DS1 Y if and only if there exist random variables X′ =d X and Y ′ =d Y such that P(X′ ≤ Y ′) = 1. By Monotonicity property Π(X′) ≤ Π(Y ′) and by the Objectivity property Π(X) ≤ Π(Y ).
PROP : If Π satisfies the Invariance by second-order stochastic dominance property, then it satisfies the Invariance by first-order stochastic dominance property.
PROOF : The proposition is proven by noting that, if Y DS1 X, then Y DS2 X. Indeed, assume that X DS1 Y , then Y DS2 X and by the Invariance by second- order stochastic dominance property, we deduce that Π(X) ≤ Π(Y ). Therefore we have shown that
X DS1 Y ⇒ Π(X) ≤ Π(Y ).
PROP : Assume that Π is a risk measure that satisfies the Positive homogeneity property. Π satisfies the Convexity property iff it satisfies the Subadditivity property.
PROOF : Assume that for any positive constant c and for all risks X, Π(cX) = cΠ(X).
i) ⇒ part : take α = 1/2 in the Convexity property, 1
2Π(X + Y ) = Π
1
2X + 1 2Y
≤ 1
2Π(X) + 1
2Π(Y ) to derive the Subadditivity property.
ii) ⇐ part : for α ∈ [0,1],
Π(αX + (1 − α)Y ) ≤ Π(αX) + Π((1 − α)Y ) = αΠ(X) + (1 − α)Π(Y ).
PROP : Assume that Π is a risk measure that satisfies the Convexity property and Π(0) = 0. Π satisfies the Positive homogeneity property iff it satisfies the Subaddi- tivity property.
PROOF : If Π is a risk measure that satisfies the Convexity property, then t → Π(tX)/t is a non-decreasing function for t > 0, since by taking 0 < t1 < t2, and α = t1/t2,
Π(t1X) = Π(αt2X + (1 − α) × 0) ≤ αΠ(t2X) + (1 − α)Π(0) = t1
t2Π(t2X).
i) ⇒ part : obvious since
Π(X + Y ) = 2Π
1
2X + 1 2Y
≤ Π(X) + Π(Y ).
ii) ⇐ part : for k ∈ N and k ≥ 2, Π(kX) ≤ kΠ(X), i.e. Π(kX)/k ≤ Π(X). But since Π is a risk measure that satisfies the Convexity property, t → Π(tX)/t is a non-decreasing and therefore it must be constant for t > 0.
PROP : If Π satisfies the Monotonicity property and the “No unjustified loading”
property, then it satisfies the “Non-excessive loading” property.
PROOF : Since P(X ≤ max[X]) = 1, we deduce that
Π(X) ≤ Π(max[X]) = max[X] by the “No unjustified loading” property.
PROP : If Π satisfies the properties 1) “Non-excessive loading”, 2) Convexity, then it satisfies the Monotonicity property.
PROOF : For α ∈ (0,1)
Π(αX) = Π
αY + (1 − α) α
(1 − α)(X − Y )
≤ αΠ(Y ) + (1 − α)Π
α
(1 − α)(X − Y )
.
If P(X ≤ Y ) = 1, then max[X − Y ] ≤ 0, Π α(1 − α)−1(X − Y ) ≤ 0 and Π(αX) ≤ αΠ(Y ).
Let α ր 1 to conclude.
PROP : If Π satisfies the properties 1) Objectivity, 2) “No unjustified loading”, 3) Convexity, 4) Convergence in distribution, then it satisfies the “Non-negative loading”
property.
PROOF : Let X, X1, X2... be iid random variables. First note that by the convexity property
Π
1
2X1 + 1 2X2
≤ 1
2Π (X1) + 1
2Π (X2) = Π (X).
Let X¯k = (X1 + ... + Xk)/k. We show by induction and the convexity property that, for any k ∈ N, Π X¯2k ≤ Π (X) since
X¯2k+2 = k
k + 1X¯2k + 1 k + 1
1
2X2k+1 + 1
2X2k+2
. Now by the law of large numbers X¯k →P,d E[X] and
Π X¯2k → Π (E[X]) = E[X] and hence Π (X) ≥ E[X].
PROP : If Π satisfies the properties 1) Objectivity, 2) Monotonicity, 3) Convexity, 4) if (Xn) converges in distribution to X then Π(Xn) → Π(X), then the risk measure does not depend on risks.
PROOF : Fix two numbers a < b. Define X0 = b and for n = 1,2,3, ... let Xn = a + 2n(b − a)I
U∈[0,2−n)
Xn′ = a + 2n(b − a)I
U∈[2−n,2−n+1)
where U is a random variable that is uniformly distributed over the unit interval (0,1). Thus Xn =d Xn′ ,
Xn = 1
2 Xn+1 + Xn+1′ and Xn →d a. Convexity and law-invariance imply
Π(Xn) = Π
1
2 Xn+1 + Xn+1′ ≤ 1
2Π(Xn+1) + 1
2Π(Xn+1′ ) = Π(Xn+1).
Thus n → Π(Xn) is an increasing sequence. Therefore monotonicity and conver- gence in distribution of the risk measure imply
Π(a) = lim
n→∞Π(Xn) ≥ Π(X0) = Π(b) ≥ Π(a).
Thus Π(a) = Π(b) = π and by monotonicity we get for any X with a ≤ X ≤ b that Π(X) = π.
As a and b are arbitrary, this implies that Π is constant on the set of all bounded random variables.
Finally approximate an unbounded random variable Y by the sequence Yn = (Y ∧ n)∨ −n of bounded random variables to extend to unbounded random variables too.
Remark : Xn →d a but max[Xn] →d ∞ = a!
PROP : If Π satisfies the properties 1) Objectivity, 2) Comonotonic additivity, 3) Invariance by second-order stochastic dominance, then it satisfies the Subadditivity property.
PROOF : Let X and Y be two random variables, and U be a random variable that is uniformly distributed over the unit interval (0,1). Set
Xc = FX−1(U) and Y c = FY−1(U).
For d = d1 + d2, we have
(x + y − d)+ = ((x − d1) + (y − d2))+ ≤ ((x − d1)+ + (y − d2)+)+
= (x − d1)+ + (y − d2)+. Let us now choose
d1 = FX−1c(FXc+Y c(d)) and d2 = F−1
Y d (FXc+Y c(d)) and note that, for any d where FXc+Y c is increasing,
d1 + d2 = (FX−1c + FY−1c )(FXc+Y c(d)) = FX−1c+Y c(FXc+Y c(d)) = d.
It follows that
E[(X + Y − d)+]
≤ E[(X − d1)+] + E[(Y − d2)+]
= E[(Xc − F−1
Xc(FXc+Y c(d)))+] + E[(Y c − F−1
Y d (FXc+Y c(d)))+]
= E[(F−1
X (U) − FX−1c(FXc+Y c(d)))+] + E[(F−1
Y (U) − F−1
Y d (FXc+Y c(d)))+]
= E[(F−1
X + FY−1)(U) − (FX−1c + F−1
Y d )(FXc+Y c(d)))+]
= E[(Xc + Y c − d)+] and then
X + Y DS2 Xc + Y c.
If the risk measure Π is invariant by second-order stochastic dominance and is additive for comonotonic risks, then
Π(X + Y ) ≤ Π(Xc + Y c) = Π(X) + Π(Y ), which proves the stated result.
⊲ VaR
There are basically two ways to define a generalized inverse for a distribution function.
DEFINITION
Given a df F, we define the inverse functions F−1 and F−1+ of F as F−1(α) = inf{x ∈ R : F(x) ≥ α} = sup{x ∈ R : F(x) < α}
and
F−1+(α) = inf{x ∈ R : F(x) > α} = sup{x ∈ R : F(x) ≤ α}
for α ∈ [0,1], where, by convention, inf ∅ = +∞ and sup∅ = −∞.
One can check that :
1. F−1 and F−1+ are both non-decreasing (they are continuous everywhere, except on an at most countable set of points) ;
2. F−1 is left-continuous while F−1+ is right-continuous ;
3. F−1(α) = F−1+(α) if, and only if, α does not correspond to a ‘flat part’ of F or equivalently, if, and only if, F−1 is continuous at α.
LEMMA : For all x ∈ R and for all α ∈ (0,1)
(i) F−1(α) ≤ x ⇔ α ≤ F(x)
(ii) F−1+(α) ≥ x ⇔ α ≥ F(x−) = P(X < x)
PROOF : We only prove (i) ; (ii) can be proven in a similar way. The ‘⇒’ part is proven if we can show that
α > F(x) ⇒ F−1(α) > x
Assume that α > F(x). Then there exists an ǫ > 0 such that α > F(x + ǫ). From the sup-definition of V aR[X; α], we find that x + ǫ ≤ F−1(α), which implies that
F−1(α) > x.
We now prove the ‘⇐’ part. If α ≤ F(x) then we find that α ≤ F(x + ǫ) for all ǫ > 0. From the inf-definition of F−1(α), we can conclude that F−1(α) ≤ x + ǫ for all ǫ > 0. Taking the limit for ǫ ↓ 0, we obtain F−1(α) ≤ x.
PROPOSITION : Let X be an rv. For any 0 < α < 1, the following equalities hold : (i) If t is non-decreasing and continuous then Ft(X−1)(α) = t FX−1(α).
(ii) If t is non-decreasing and continuous then Ft(X)−1+(α) = t FX−1+(α).
PROOF : We only prove (i) ; (ii) can be proven in a similar way. By application of the previous lemma, we find that the following equivalences hold for all real x :
Ft(X−1)(α) ≤ x ⇔ α ≤ Ft(X)(x) ⇔ α ≤ FX(t−1+(x))
⇔ FX−1(α) ≤ t−1+(x) ⇔ t FX−1(α) ≤ x
Note that the above proof only holds if t−1+ is finite. But one can verify that the equivalences also hold if t−1+(x) = ±∞.
Remark : The continuity assumption put on the function t can be relaxed as follows : in (i) it is enough for t to be left-continuous, whereas in (ii) it is enough for t to be right-continuous.
PROPOSITION :
(i) If an rv X has a continuous df F, then F(X) ∼ U ni(0,1).
(ii) Let X be an rv with df F, not necessarily continuous. If U ∼ U ni(0,1), then X =d F−1(U) =d F−1+(U).
PROOF :
(i) For all 0 < u < 1,
P(F(X) ≥ u) = P(X ≥ F−1(u)) = 1 − F(F−1(u)) = 1 − u from which we conclude that F(X) ∼ U ni(0,1).
(ii) We see from the lemma that
P(F−1(U) ≤ x) = P(U ≤ F(x)) = F(x).
The other statement has a similar proof.
PROP : VaR satisfies the “Non-excessive loading” property.
PROOF : Since X ≤ max[X] we have that V aR[X;α] ≤ max[X] whatever α, so that VaR is indeed no-ripoff.
PROP : VaR does not satisfy the “Non-negative loading” property.
PROOF : Let us define α∗ = F(E[X]). It is clear that VaR does not exceed the expected loss X for probability levels less than α∗.
PROP : VaR satisfies the “No unjustified loading” property.
PROOF : It is easy to see that for any probability level α > 0, V aR[c;α] = c.
PROP : VaR satisfies the Objectivity property.
PROOF :This is a direct consequence of the definition of VaR, since it only depends on the df of X.
PROP : VaR satisfies the Translativity property.
PROOF : VaR possesses the very convenient stability property that the VaR of a non- decreasing function t of some rv X is obtained by applying the same function to the initial VaR. Let us consider the function t : x → x +c, we deduce that VaR has the translativity property.
PROP : VaR fails to be subadditive.
i) A counter-example
Let us consider two independent risks with unit Pareto distribution X ∼ P ar(1,1) and Y ∼ P ar (1,1), i.e.
P (X > t) = P (Y > t) = 1
1 + t, t > 0.
On the one hand,
V aR [X;α] = V aR [Y ;α] = 1
1 − α − 1.
On the other hand, one can show that
P (X + Y ≤ t) = 1 − 2
2 + t + 2ln (1 + t) (2 + t)2 .
Since
P (X + Y ≤ 2V aR [X;α]) = α − (1 − α)2 2 ln
1 + α 1 − α
< α, we get
V aR[X;α] + V aR[Y ;α] < V aR[X + Y ;α]
and, in such a case, diversification will lead to more risk being reported.
ii) Elliptical distributions et subadditivity of VaR DEFINITION :
1. A random vector X = (X1, . . . , Xn) has spherical distribution, if for every ortho- gonal map U ∈ Rn×n (i.e. U′U = UU′ = Id),
U X =d X.
◦ The multivariate standard Gaussian distribution is a spherical distribution since fUX(x) = fX(U−1x) = 1
(2π)n/2 exp
−1
2(U−1x)′(U−1x)
= 1
(2π)n/2 exp
−1
2x′UU−1x)
= fX(x)
PROP : The following are equivalent.
(i) X is spherical
(ii) There exists a function ψ such that, for all t ∈ Rd E[eit′X] = ψ(t2).
(iii) For every a ∈ Rd
a′X =d aX1.
(iv) X =d RS where S is uniformly distributed on the unit sphere Sn−1 = {t ∈ Rd : t2 = 1} and R ≥ 0 is a radial random variable, independent of S.
ψ is called the characteristic generator of the spherical distribution and we write X ∈ Sn(ψ).
2. A random vector X = (X1, . . . , Xn) has an elliptical distribution (X ∈ E(µ, A, ψ)) if there exist µ ∈ Rn, A ∈ Rn×d and Y ∈ Sd(ψ) such that
X =d µ + AY.
◦ It follows that any random vector with components that are linear combinations of the components of an elliptical distribution is again an elliptical distribution with the same characteristic generator.
◦ The Gaussian and the Student distributions are examples of elliptical distributions.
◦ Any multivariate elliptical distribution with mutually independent components and finite variance must necessarily be multivariate normal.
PROPOSITION : Let X ∈ E(µ, A, ψ) and M = {L : L = λ0 + λ′X}. For any L1, L2 ∈ M, α ≥ 0.5
V aR[L1 + L2;α] ≤ V aR[L1;α] + V aR[L2;α].
PROOF : Let L1 = λ0,1 + λ′1X and L2 = λ0,2 + λ′2X. We have V aR[L1 + L2;α]
= λ0,1 + λ0,2 + V aR[(λ1 + λ2)′X;α]
= λ0,1 + λ0,2 + V aR[(λ1 + λ2)′µ + (λ1 + λ2)′AY1;α]
= λ0,1 + λ0,2 + (λ1 + λ2)′µ + (λ1 + λ2)′A V aR[Y1;α]
If α ≥ 0.5, then V aR[Y1;α] ≥ 0 and V aR[L1 + L2;α]
≤ λ0,1 + λ0,2 + (λ1 + λ2)′µ + λ′1A + λ′2A V aR[Y1;α]
= V aR[L1;α] + V aR[L2;α].
PROP : VaR satisfies the Comonotonic additivity property.
PROOF : For all non-decreasing (left-continuous) functions h and g,
V aR[h(X) + g(X);α] = V aR[(h + g)(X);α] = (h + g)(V aR[X;α])
= h(V aR[X;α]) + g(V aR[X;α])
= V aR[h(X); α] + V aR[g(X);α]
PROP : VaR satisfies the Positive homogeneity property.
PROOF : Let us consider the function t : x → λx with λ > 0, we deduce that VaR has the translativity property.
PROP : VaR satisfies the Monotonicity property.
PROOF : Clearly, if P(X ≤ Y ) = 1 holds then FX(x) ≥ FY (x) is true for any x.
Therefore, V aR[X;α] ≤ V aR[Y ;α] holds in such a case for any probability level α (by symmetry with respect to the main diagonal).
PROP : VaR satisfies the Invariance by first-order stochastic dominance property.
PROOF : It is easy to show that
X DS1 Y ⇔ P(Y ≤ d) ≤ P(X ≤ d) ∀d ∈ R
⇔ V aR[X;α] ≤ V aR[Y ;α] ∀α ∈ (0,1).
PROP : VaR does not satisfy the Invariance by second-order stochastic dominance property.
PROOF : Since X DS2 Y X DS1 Y , the Invariance by second-order stochastic dominance property may not be satisfied.
PROP : VaR does not satisfy the Convexity property.
PROOF : VaR is not subadditive and satisfies the Positive homogeneity property.
PROP : VaR does not satisfy the Iterativity property.
PROOF : Let
X Y
∼ N
µX µY
,
σ2X ρσXσY ρσXσY σ2Y
The conditional distribution of X given Y = y is X|Y = y ∼ N
µX + ρσX
σY (y − µY ), σ2X(1 − ρ2)
.
Hence
V aR[X;α] = µX + σXΦ−1(α) V aR[X|Y ;α] = µX + ρσX
σY (Y − µY ) + σX
(1 − ρ2)Φ−1(α) V aR[V aR[X|Y ;α];α] = µX + σX (1 − ρ2) + ρ
Φ−1(α) and we deduce that V aR[V aR[X|Y ;α];α] = V aR[X;α] only if ρ = 0.
PROP : VaR satisfies the Convergence in distribution.
PROOF : It is well known that the weak convergence of the dfs ensures the same type of convergence for the quantile functions.
PROP : VaR does not satisfy the Stability by mixing property.
PROOF : Consider for example the case where X =d 1
2δ0 + 1
2N (0,1). It is easily seen that
V aR[X;α] = 1
2Φ−1(α)
PROP : Let (X, Y ) be a random vector with pdf f(., .), then
∂
∂γV aR[X + γY ;α] = E[Y |X + γY = V aR[X + γY ;α]]
PROOF : It may be found for example in Gouriéroux C., J.P. Laurent and O. Scaillet (2000). Sensitivity Analysis of Values at Risk. Journal of Empirical Finance, 7, 225- 245.
PROP : Let (X, Y ) be a random vector with pdf f(., .), then
∂2
∂γ2V aR[X + γY ;α] = ∂
∂sV[Y |X + γY = s]
s=V aR[X+γY;α]
+ V[Y |X + γY = s] ∂
∂sfX+γY(s)
s=V aR[X+γY;α]
PROOF : It may be found for example in Gouriéroux C., J.P. Laurent and O. Scaillet (2000). Sensitivity Analysis of Values at Risk. Journal of Empirical Finance, 7, 225- 245.
⊲ TVaR and associated risk measures PROP : For any α ∈ (0,1)
T V aR[X;α] = V aR[X;α] + 1
1 − αES[X;α]
CT E[X;α] = V aR[X;α] + 1
1 − F(V aR[X;α])ES[X;α]
CV aR[X;α] = 1
1 − F(V aR[X;α])ES[X;α]
PROOF : The first expression follows from ES[X;α] =
1
0 (V aR[X;ξ] − V aR[X;α])+ dξ
=
1
α V aR[X;ξ]dξ − V aR[X;α](1 − α).
The second and third expression follow from
ES[X;α] = E[X − V aR[X;α]|X > V aR[X;α]]P(X > V aR[X; α]).
Remark :
1. If F has a positive probability distribution function, for any α ∈ (0,1), T V aR[X;α] = CT E[X;α].
2. For any α ∈ (0,1)
minπ (E[(X − π)+] + (1 − α)π)
= E[(X − V aR[X;α])+] + (1 − α)V aR[X;α]
= (1 − α)T V aR[X; α].
EXAMPLES :
1. Consider a random variable X ∼ N(µ, σ2) which is normally distributed with mean µ and variance σ2. We have
V aR[X;α] = µ + σΦ−1(α) T V aR[X;α] = µ + σϕ(Φ−1(α))
1 − α CT E[X;α] = µ + σϕ(Φ−1(α))
1 − α CV aR[X;α] = σ
ϕ(Φ−1(α))
1 − α − Φ−1(α)
ES[X;α] = σϕ(Φ−1(α)) − σΦ−1(α)(1 − α)
2. Consider a random variable that is lognormally distributed, i.e. ln X ∼ N(µ, σ2).
We have
V aR[X;α] = eµ+σΦ−1(α)
T V aR[X;α] = eµ+σ2/2Φ(σ − Φ−1(α)) 1 − α
CT E[X;α] = eµ+σ2/2Φ(σ − Φ−1(α)) 1 − α
CV aR[X;α] = µ+σ2/2Φ(σ − Φ−1(α))
1 − α − eµ+σΦ−1(α)
ES[X;α] = eµ+σ2/2Φ(σ − Φ−1(α)) − eµ+σΦ−1(α)(1 − α)
PROP : TVaR satisfies the “Non-excessive loading” property.
PROOF : This comes from the fact that VaR is known to be no-ripoff, so that T V aR[X;α] = 1
1 − α
1
α V aR[X;ξ]dξ ≤ 1 1 − α
1
α max[X]dξ.
PROP : TVaR satisfies the “Non-negative loading” property.
PROOF : This is again an immediate consequence of the corresponding properties for VaRs, since
T V aR[c;α] = 1 1 − α
1
α V aR[c;ξ]dξ = 1 1 − α
1
α cdξ = c.
PROP : TVaR satisfies the “No unjustified loading” property.
PROOF : If U ∼ Uni(0,1) then
E[X] = E[F−1(U)] =
1
0 F−1(u)du = T V aR[X; 0].
The claimed property will hold if we are able to show that TVaR is non-decreasing in the probability level. We clearly have that
T V aR[X;α] = 1 1 − α
E[X] −
α
0 V aR[X;ξ]dξ
. Therefore, we can write
d
dαT V aR[X;α] = 1
1 − α (T V aR[X;α] − V aR[X;α]). Since α → V aR[X;α] is non-decreasing,
T V aR[X;α] = 1 1 − α
1
α V aR[X;ξ]dξ ≤ 1 1 − α
1
α V aR[X;α]dξ = V aR[X;α]
which gives
d
dαT V aR[X;α] ≥ 0.
We conclude
T V aR[X;α] ≥ T V aR[X; 0] = E[X]
so that TVaR induces a non-negative loading whatever the probability level α.
PROP : TVaR satisfies the Objectivity property.
PROOF : Knowing α → T V aR[X;α] is equivalent knowing α → V aR[X;α] since by definition
T V aR[X;α] = 1 1 − α
1
α V aR[X;ξ]dξ.
and
V aR[X;α] = T V aR[X;α] − (1 − α) d
dαT V aR[X;α].
Hence TVaR satisfies the Objectivity property.
PROP : TVaR satisfies the Translativity property.
PROOF : This is immediate from the corresponding properties of the VaRs T V aR[X + c;α] = 1
1 − α
1
α V aR[X + c;ξ]dξ
= 1
1 − α
1
α V aR[X;ξ]dξ + c = T V aR[X;α] + c PROP : TVaR satisfies the Subadditivity property.
PROOF : First note that
T V aR[X;α] = min
π
π + 1
(1 − α)E[(X − π)+]
.
We thus have for any 0 < λ < 1 that T V aR[λX + (1 − λ)Y ;α]
≤
π + 1
(1 − α)E[(λX + (1 − λ)Y − π)+]
π=λV aR[X;α]+(1−λ)V aR[Y ;α]
= λV aR[X;α] + (1 − λ)V aR[Y ;α]
+ 1
(1 − α)E[(λX + (1 − λ)Y − (λV aR[X;α] + (1 − λ)V aR[Y ;α]))+
≤ λV aR[X;α] + (1 − λ)V aR[Y ;α]
+ λ
(1 − α)E[(X − V aR[X;α])+] + (1 − λ)
(1 − α)E[(Y − V aR[Y ;α])+]
= λT V aR[X;α] + (1 − λ)T V aR[Y ;α].
Hence the TVaR is convexe and since it is positive homogeneous, it is subadditive.
PROP : TVaR satisfies the Comonotonic additivity property.
PROOF : This is immediate from the corresponding properties of the VaRs.
PROP : TVaR satisfies the Positive homogeneity property.
PROOF : This is immediate from the corresponding properties of the VaRs.
PROP : TVaR satisfies the Monotonicity property.
PROOF : This is immediate from the corresponding properties of the VaRs.
PROP : TVaR satisfies the Invariance by first-order stochastic dominance property.
PROOF : TVaR satisfies the Invariance by second-order stochastic dominance property and so it satisfies the Invariance by first-order stochastic dominance property..
PROP : TVaR satisfies the Invariance by second-order stochastic dominance property.
PROP : For any random pair (X, Y ) we have that X DS2 Y if and only if their respective T V aR’s are ordered :
X DS2 Y ⇔ T V aR[X;α] ≤ T V aR[Y ;α] ∀α ∈ (0,1)
PROOF : First we assume X DS2 Y and let α ∈ (0,1). Consider the function f(d) defined by
f(d) = (1 − α)π + E[(X − d)+].
We have
T V aR[X;α] = f(V aR[X;α])
1 − α ≤ f(V aR[Y ;α]) 1 − α
= V aR[Y ;α] + 1
1 − αE[(X − V aR[Y ;α])+]
≤ V aR[Y ;α] + 1
1 − αE[(Y − V aR[Y ;α])+] = T V aR[Y ;α].
To prove the other implication, we assume that the TVaR’s are ordered for all α ∈ (0,1). Note that for any random variable X, we have that
E[(X − d)+] = E[(F−1
X (U) − d)+]
=
1
FX(d) V aR[X;α]dα − d(1 − FX(d)).
Hence, for d such that 0 < FX(d) < 1, we find
E[(X − d)+] = (T V aR[X;FX(d)] − d) (1 − FX(d))
≤ (T V aR[Y ;FX(d)] − d) (1 − FX(d))
= E[(Y − d)+] +
FY(d)
FX(d) (V aR[Y ;α] − α)dα
Using the equivalence α ≤ FY (d) ⇔ d ≥ V aR[Y ;α], it is straightforward to prove that
FY(d)
FX(d) (V aR[Y ;α] − α)dα ≤ 0.
If FX(d) = 1, we find E[(X − d)+] = 0 ≤ E[(Y − d)+]. Since E[X] ≤ E[Y ] Thus E[(X − d)+] = E[X] ≤ E[(Y − d)+] also holds for d such that FX(d) = 0.
Hence, we have proven that X DS2 Y .
PROP : TVaR satisfies the Convexity property.
PROOF : See the proof for the Subadditivity property.
PROP : TVaR does not satisfy Iterativity property.
PROOF : Consider the case where
X Y
∼ N
µX µY
,
σ2X ρσXσY ρσXσY σ2Y
PROP : TVaR satisfies the Convergence in distribution property if moreover E[Xn] → E[Xn].
PROOF : By using the Objectivity property.
PROP : TVaR does not satisfy the Stability by mixing property.
PROOF : Consider for example the case where X =d 1
2δ0 + 1
2N (0,1).
Remark : Let X and x be such that F(x) > 0. For any event A such that P(A) = F(x),
E[X|A] ≤ E[X|X > x].
It suffices to write
E[X|X > x] = x + E[X − x|X > x, A]P(A|X > x) +EX − x|X > x,A¯P A|X > x¯
≥ x + E[X − x|X > x, A] P(A|X > x)
= x + E[X − x|X > x, A] P(X > x|A)
≥ x + E[X − x|X > x, A] P(X > x|A) +E[X − x|X ≤ x, A]P(X ≤ x|A)
= E[X|A].
It sheds a new light on CTE, which can be represented as a worst-case conditional expectation since
CT E[X;α] = sup E[X|A]|P(A) ≥ F¯(V aR[X; α]) which reduces to
CT E[X;α] = sup{E[X|A]|P(A) ≥ 1 − α}
when F is continuous.
This result is closely related to the notion of scenario or stress testing : the CTE appears as the largest possible expected value of X under the set of all plausible scenarios (that is, those whose probabilities exceed 1 − α).
PROP : Let (X, Y ) be a random vector with pdf f(., .), then
∂
∂γT V aR[X + γY ;α] = E[Y |X + γY ≥ V aR[X + γY ;α]]
PROOF : See for example Scaillet, O., 2004. Nonparametric Estimation and Sensitivity Analysis of Expected Shortfall. Mathematical Finance, 14 : 115-129.
PROP : Let (X, Y ) be a random vector with pdf f(., .), then
∂2
∂γ2T V aR[X + γY ;α] = 1
1 − αV[Y |X + γY = V aR[X + γY ;α]]
×fX+γY (V aR[X + γY ;α]).
PROOF : See for example Scaillet, O., 2004. Nonparametric Estimation and Sensitivity Analysis of Expected Shortfall. Mathematical Finance, 14 : 115-129.
⊲ RISK MEASURES BASED ON EXPECTED UTILITY THEORY
Remark : u may be chosen such that u(0) = 0, u′(0) = 1 and u′′(0) = −a ≤ 0.
PROP : Π(.) satisfies the “Non-excessive loading” property (no ripoff).
PROOF : Because u is non-decreasing and X ≤ max[X] a.s., we have 0 = E[u(Π(X) − X)] ≥ u(Π(X) − max[X])
so that Π(X) ≤ max[X] holds, and the zero-utility premiums satisfy the no-ripoff condition.
PROP : Π(.) satisfies the “Non-negative loading” property.
PROOF : If u is concave then Jensen’s inequality ensures that 0 = E[u(Π(X) − X)] ≤ u(Π(X) − E[X])
so that Π(X) ≥ E[X] and the zero-utility premiums contain a non-negative loading.
PROP : Π(.) satisfies the “No unjustified loading” property.
PROOF : Note that
0 = E[u(Π(c) − c)] = u(Π(c) − c).
Since u′(0) > 0, we deduce that Π(c) = c.
PROP : Π(.) satisfies the Objectivity property.
PROOF : If X and Y have the same distribution,
0 = E[u(Π(X) − X)] = E[u(Π(Y ) − Y )] = E[u(Π(X) − Y )]
and hence Π(X) = Π(Y ).
PROP : Π(.) satisfies the Translativity property.
PROOF : We have
0 = E[u(Π(X +c)−(X +c))] = E[u((Π(X +c)−c)−X)] = E[u(Π(X)−X)]
and then Π(X + c) = Π(X) + c.
PROP : Π(.) does not satisfy the Subadditivity property.
PROOF : Consider the case where
X Y
∼ N
0 0
,
1 ρ ρ 1
and u(x) = −e−αx for α > 0. Then Π(X) = 1
α lnEeαX = α
2 and Π(X + Y ) = (1 + ρ)α and therefore Π(X + Y ) = Π(X) + Π(Y ).
Note that Π(X) satisfies the Additivity for independent risks property iff u(x) =
−e−αx or u(x) = x (up to a linear relation).
PROP : Π(.) does not satisfy the Comonotonic additivity property.
PROOF : Π(.) does not satisfy the Positive homogeneity property and therefore it does not satisfy the Comonotonic additivity property.
PROP : Π(.) does not satisfy the Positive homogeneity property.
PROOF : Consider the case where X ∼ N (0,1). Then Π(λX) = λ 1
λα lnE[eλαX] = λαλ
2 = λ2Π(X).
PROP : Π(.) satisfies the Monotonicity property.
PROOF : Π(.) satisfies the Invariance by first-order stochastic dominance property and therefore it satisfies the Monotonicity property.
PROP : Π(.) satisfies the Invariance by first-order stochastic dominance property.
PROOF : Π(.) satisfies the Invariance by second-order stochastic dominance property and therefore it satisfies the Invariance by first-order stochastic dominance property.
PROP : Π(.) satisfies the Invariance by second-order stochastic dominance property.
PROOF : Assume that X DS2 Y . We have
E[u(Π(X) − X)] ≥ E[u(Π(X) − Y )].
But
0 = E[u(Π(X) − X)] = E[u(Π(Y ) − Y )] ≥ E[u(Π(X) − Y )]
and it follows that Π(Y ) ≥ Π(X).
PROP : Π(.) satisfies the Convexity property if u′′ < 0.
PROOF : Consider two risks X and Y and define
g(t;X, Y ) = Π(X + tV ) where V = Y − X.
Assume that g(t) = g(t;X, Y ) is convexe for all X and Y and for α ∈ (0,1). Then Π(αX + (1 − α)Y ) = Π(X + (1 − α)V ) = g((1 − α))
≤ αg(0) + (1 − α)g(1) = αΠ(X) + (1 − α)Π(Y ).
It is now enough to show that
g′′(0;X, Y ) ≥ 0 for all X and Y since
g′′(0;X + tV, Y ) = (1 − t2)g′′(t;X, Y ).
But
g′′(0;X, Y ) = −E[u′′(Π(X) − X)(g′′(0;X, Y ) − V )2] E[u′(Π(X) − X)] .
PROP : Π(.) satisfies the Iterativity property iff u(x) = −e−αx or u(x) = x (up to a linear relation).
PROOF : The ‘⇐’ part is proven if we can show that Π(X) = Π(Π(X|Y )) for u(x) = x (obvious) and for u(x) = −e−αx. But
Π (X) = 1
α lnE[eαX] = 1
α lnE[E[eαX|Y ]]
= 1
α lnE
eαα1 lnE[eαX|Y ]
= Π (Π(X|Y )).
The ‘⇒’ part is proven the following way. Let z > 0 and y1, y2 ∈ [0,1]. Define : Y = 1
2δy1 + 1
2δy2 and X|Y = y =d Xy,z =d (1 − y)δ0 + yδz. On the other hand
X =d 1
2Xy1,z + 1
2Xy2,z =d Xq,z with q = 1
2(y1 + y2).
We will use the following notation
π(yi) = Π (X|Y = yi) which satisfies
yiu (π(yi) − z) + (1 − yi)u(π(yi)) = 0.
Differentiating one time with respect to yi
∂
∂yi (yiu (π(yi) − z) + (1 − yi)u(π(yi))) = 0 and letting yi tends to 0 leads to
u(−z) = −π′(0).
By iterativity, we have that Π(X) = Π (Π (X|Y )) i.e. : 1
2u (π(q) − π(y1)) + 1
2u (π(q) − π(y2)) = 0
Differentiate two times with respect to yi
∂2
∂yi2
1 2u
π
y1 + y2 2
− π(y1)
+ 1 2u
π
y1 + y2 2
− π(y2)
= 0.
And choosing y1 = y2 = y, we get
π′′(y) + aπ′(y)2 = 0 with π(0) = 0, π(1) = z.
If a = 0, π(y) = y. If a > 0, π(y) = a−1 log(1 − y + yeaz) and u(−z) =
−π′(0) = −a−1 (eaz − 1).
PROP : Π(.) satisfies the Convergence in distribution property if limn→∞ E[u(−Xn)] = E[u(−X)].
PROOF : Obvious.
PROP : Π(.) does not satisfy the Stability by mixing property.
PROOF : By choosing an appropriate counter-example.