A Note about Characterization of Calendar Spread Arbitrage in eSSVI Surfaces ()
1. Introduction
Volatility surface models are vital tools for derivative pricing and risk hedging in financial markets. Among these models, the eSSVI proposed by Hendriks and Martini (2019) has attracted quite a bit of attention in the research community, and not only given that a financial data and software company like Factset provides an implementation of the algorithm described in Corbetta, Cohort, Laachir & Martini (2019) which calibrates eSSVI surfaces to market data. (see Akhundzadeh, Huber, Hyatt, & Schneider, 2020 )
The eSSVI surface is an offspring of SVI model for the implied variance smile at a fixed maturity. According to Gatheral & Jacquier (2014a) , the latter was devised at Merrill Lynch in 1999 and was subsequently publicly disseminated at the Global Derivatives & Risk Management conference in Madrid ( Gatheral, 2004 ). In order to motivate the introduction of the eSSVI model and to explain the contribution of the present paper, it will be convenient to review the main properties of the SVI first. According to the latter, for a given fixed maturity date the implied variance skew
can be represented through the function
where
is the natural logarithm of the ratio between the strike and the forward price of the underlying, and where
,
,
,
and
are parameters governing the shape and position of the smile. The following appealing features of the SVI parametrization are well known:
1) the SVI smiles are asymptotically linear in k as
and therefore consistent with Roger Lee's moment formula given in Lee (2004) .
2) and the large maturity limit of the implied variance smile of a Heston model with correlation parameter ρ is SVI with the same value of ρ ( Gatheral & Jacquier, 2014b ).
However, it is also well known that the SVI parametrization, in its full generality, is not arbitrage free. For example, it is easy to see that
and hence it should always be required that
. However, the latter condition is not enough to rule out butterfly arbitrage as can be seen from a well known counterexample of Axel Vogt (see Gatheral & Jacquier (2014a) ). Moreover, fitting the SVI parametrization to more than a single maturity date may produce slices that cross over each other which is equivalent to the existence of calendar spread arbitrage opportunities. To overcome these issues, Gatheral & Jacquier (2014a) introduced the SSVI parametrization which is a global parametrization for the whole implied total variance surface where the fixed maturity slices are restricted to a subfamily of the SVI parametrization (the first “S” in front of SSVI stands for “surface”). In the SSVI parametrization, the implied total variance surface is given by
(1)
where
is the ATM implied total variance at maturity t,
and
is a smooth function from
to
such that the limit
exists in
. According to Theorems 4.1 and 4.2 in Gatheral & Jacquier (2014a) , the SSVI surface (1) is free of calendar spread arbitrage if and only if
C1)
for all
C2) and
for all
(the upper bound is infinite when
)
and it is free of butterfly arbitrage if for all
the following two conditions are both satisfied:
B1)
,
B2')
.
Mingone (2022) , using results given in Martini & Mingone (2022) , strengthens this result by showing that absence of butterfly arbitrage is equivalent to B1 and
B2)
, where
with
Now we are finally ready to introduce the eSSVI. In order to make the SSVI surface more flexible, Hendriks and Martini (2019) made the ρ-parameter maturity dependent as well and called the resulting implied total variance surface model eSSVI surface (the “e” in front of SSVI stands for “extended”). Proposition 3.1 in Hendriks & Martini (2019) provides necessary and sufficient conditions for the absence of calendar spread arbitrage between two time slices. In order to state these conditions, we indicate the parameters of two slices with
,
and
, where the subscript i takes on the value 1 or 2 according to whether the closer (
) or farther (
) maturity date is referred to. Proposition 3.1 in Hendriks & Martini (2019) says that two eSSVI slices do not cross over each other only if
N')
and
and that condition N along with condition S below is sufficient to rule out the existence of crossing points:
S)
or
However, the statement about sufficiency is wrong and this is where the contribution of the present paper comes in. In fact, as can be seen from Proposition 13 in Section 2 below,
· when
there are no crossing points if and only if either (i)
and
or (ii)
and
.
· and when
there are no crossing points if and only if condition S holds jointly with condition
N)
and
.
Almost all of the proof of Proposition 13 in Section 2 is built on the main ideas from the proof of Proposition 3.1 in Hendriks & Martini (2019) . As far as I know the only novelty are the result about tangency points in Lemma 10 and the two final Lemmas 11 and 12.
2. The Correct and Sharper Statement of the Hendriks-Martini Proposition and Its Proof
Consider two eSSVI slices which we shall denote by
As in the previous section, assume that the subscript
refers to the closer maturity date. Then there is absence of calendar spread arbitrage if and only if
for all
.
Note that
so that
for all
. Since
if and only if
, we conclude that
By combining this result with the fact that
, we see that absence of calendar spread arbitrage implies
(2)
Another necessary condition may be obtained by considering the asymptotes of the two slices. Since
We conclude that absence of calendar spread arbitrage also implies
(3)
The latter condition is satisfied if and only if
(4)
Of course, in the argument leading to the necessary condition (2) we are tacitly assuming that
,
and
are all strictly positive and in this case it follows from (2) that
, i.e. that
. Note that if
and/or
, then
for all
, and in this case we have absence of calendar spread arbitrage if and only if
or
according to whether
is also zero or not. On the other hand, if
, then we have
for all
, and in this case it follows from the asymptotic behavior of
that we have absence of calendar spread arbitrage if and only if
and
. In what follows we rule out these trivial cases by assuming that
and
are well defined (i.e. that their denominators are strictly positive) and that
and
.
Lemma 1. If
,
and
are all strictly positive, then there is absence of calendar spread arbitrage only if conditions (2) and (3) are both satisfied.
Now it raises the question whether the conditions (2) and (3) are sufficient as well. To answer this question we look for conditions under which the graphs of
and
have at least one point in common. We will proceed as in Hendriks & Martini (2019) , but we will expand on some details. So let
,
and note that the two eSSVI slices do have points in common if and only if the equation
has real solutions. Squaring twice yields the quartic polynomial
where we have omitted the independent variable x on the RHS. Note that every root of
must satisfy one (and only one) of the following conditions:
(5)
Of course, a root of
is a point where the two slices intersect if and only if it satisfies condition c). To explore the existence of such roots we first observe that
, where
(6)
Note that
(which is a root of
) is an intersection point if and only if
, i.e. if and only if
(in fact,
if and only if
). Assuming that this is the case, we will now find conditions under which the two slices do cross over each other. To this aim we consider their derivatives. With
(i.e. with
) we obtain
To rule out the possibility that the two slices cross over in
, we must therefore impose
(7)
If either one of these conditions fails, the two slices cross over in
. Since we are assuming that the
’s and
’s are all strictly positive, the conditions (7) can be jointly satisfied only if
· either
and
, in which case it is easy to verify that
for all
;
· or
and
, in which case the constant term and the coefficient of x in the polynomial
do both vanish, and hence the two slices have no intersection points other than
.
These considerations prove the following lemma:
Lemma 2. Assume that Φ and Θ are well defined and that
. If
, there is no calendar spread arbitrage if and only if either (i)
and
or (ii)
and
.
Note that conditions (2) and (3) do not imply condition (i) or (ii) of the previous lemma (take for example
,
and
) and the former are therefore not strong enough to rule out calendar spread arbitrage even if we restrict to the case where
.
Consider now what happens when
. In this case
and
is therefore not an intersection point. To investigate the existence of intersection points we analyze the polynomial
. We begin with the following lemma:
Lemma 3. Assume that
and
. Then
is of second degree if and only if
(8)
and in this case its discriminant is given by
(9)
Proof. The coefficient of
in
vanishes if and only if either
The second condition implies
, and hence we conclude that
is of second degree if and only if condition (8) holds. In this case the discriminant of
can be written as in expression (9).¨
From the previous lemma we know that
must have real roots if condition (8) holds jointly with
(10)
Since we are only concerned with the case where the necessary conditions (2) and (4) hold, we must consider the set of
-pairs where the equality in (10) holds and the set
-pairs where
We denote these two sets by
and
, respectively:
It will be useful to visualize these two sets. Since we have already dealt with the case
, and since we are assuming the necessary condition (4), we need to consider only the case where
and
. Figure 1 shows all possible shapes of
and
. The set
is the graph of a hyperbola. It is always symmetric with respect to both axes of the
-plane and its prolongation always goes through the four vertices of the square
. Moreover,
· if
,
does not intersect the
axis and intersects the
axis in
;
· if
,
reduces to the straight lines
.
Also for the set
there are essentially only two possible shapes:
· If
(since we are assuming
, this implies
), the set
is given by the stripe
The stripe reduces to the line
if
.
· If
(since we are assuming
, this implies
), the set
is the union of the two parallel and disjoint stripes
and
The two stripes reduce to the lines
if
.
From the description of the sets
and
we see immediately that
when
. Next we prove that
Figure 1. The red areas show all possible shapes of the
when
and
. The black line is the graph of the hyperbola
. The two dashed lines are the asymptotes
of
.
except for this special case the intersection is empty.
Lemma 4. If
, then it follows that
.
Proof. If
we must have
From the latter equality we obtain
The quantity on the RHS is positive because
and because we are assuming that
. Hence we conclude that
, because otherwise we should have
¨
From now on we consider roots of
as functions of
and
. Any root of
will be denoted by
. Of course the subset of the
-plane where such a root exists depends on
and
. We denote this set with
. Since we are only interested in
values within the interval
, we consider
as a subset of the open square
. Note that
must be a continuous function of
and that
only if condition (10) holds.
As we have already seen, a root
must be of one of the three types in (5) and only roots of type c) are intersection points. In order to determine which type applies, we will need the following functions:
and
Note that these functions must all be continuous functions of
.
Lemma 5. Assume that
and
. Then
only if
.
Proof. The proof is essentially the same as the proof of Lemma A.2 in Hendriks & Martini (2019) . Since
must satisfy one of the two equalities
(otherwise
is not a root of
and hence neither a root of
) and since
for all
, we conclude that
if and only if
and
do both vanish. In this case we must have
(11)
and we must also have
(12)
Subtracting Equation (11) from Equation (12) yields
Since
must be different from zero (otherwise we would have
contrary to our assumption), this implies that
which is equivalent to
.¨
Corollary 1. Assume that
,
and that A is a connected subset of
which does not intersect the set
Then it follows that function
does not change sign on A.
The previous corollary will be useful to distinguish whether a given root
is of type a) rather than of type b) or c). Once we know that it is not of type a), we will apply the next lemma in order to find out whether it is of type b) or c).
Lemma 6. Let A be a connected subset of
such that
for all
. Then it follows that the function
does not change sign on A.
Proof. Under the assumptions of the lemma
must be a root of either type b) or c) in (5). Hence we must have either
or
The conclusion of the lemma follows now from the fact that
is continuous and that
for all
.¨
Now we are finally ready to investigate about the existence of intersection points. We start from the special cases where
or
.
Lemma 7. Assume that
and that either
or
. Then
implies
for all
.
Proof. If
, we must have
and thus there exist intersection points only if the polynomial
has real roots different from
. Now, consider first the case
. Since we are assuming that
, it follows that
(see the description of the set
for the special case where
). However, it can be verified that in this case we must have
which has no roots at all.
Next, consider the case
. In this case we must have
(see the description of the set
for the special case where
). Substituting
and
in the coefficients of
shows that
which has no roots at all.¨
Next, we deal with the case where Φ is strictly smaller than 1 and different from
(i.e.
). The inequality
, which is necessary by condition (4), allows then only for values of Φ in the range
. Note that for
the necessary condition (2) is already implied by condition (4) and therefore we do not need to assume condition (2) explicitly.
Lemma 8. Assume that
,
and
(this implies
). Then
implies
for all
.
Proof. Once again, if
we must have
and there exist points
where
if and only if intersection points exist, i.e. if and only if the polynomial
has at least one real root which satisfies condition c) in (5). From Lemma 3 we know that
must have roots if
,
,
and if
belongs to the interior of
. Since under the present conditions
is connected and does not intersect
(see Lemma 4), we may apply Corollary 1 to check whether the roots in
are of type a). This will be the case if there exists a single
such that
. Under the present conditions the origin belongs to
. Hence we use
as test point. Of course,
. Moreover, it is easy to check that
so that
We conclude that
must be of type a) whenever
.
In the previous lemma we have assumed that
which forces
. To apply the same method of proof for the case where
we must however assume that
.
Lemma 9. Assume
,
and
(note that
and
implies
). Then
implies
for all
.
Proof. The proof is similar to the proof of the previous lemma. However, in the present case we must deal with the fact that the set
is not connected but only the union of the two connected sets S1 and S2. In each one of these two sets we must therefore find a point
such that
and
are of opposite sign. To locate these points, note that the
-axis intersects both sets and hence we choose the
-points with
and
This choice is convenient because it makes the discriminant of
vanish. According to the sign in
, it gives rise to the roots
(13)
which, regardless of the sign, yields
(14)
and
(15)
Note that
regardless of the value of Φ(provided that
), but to make sure that
we need to assume
.¨
Lemma 7, Lemma 8 and Lemma 9 show that the necessary condition (4) along with
and
are jointly sufficient to rule out calendar spread arbitrage. The next lemma deals with the condition
(16)
which allows for values of Φ larger than 1.
Lemma 10. Assume that
and that condition (16) holds (note that these conditions jointly imply the necessary condition (4)). Then it follows that
for all
. Under the assumptions of this lemma there exist tangency points (i.e. values of k where
) if and only if
and condition (16) holds with equality sign. In that case there must exist exactly one tangency point.
Proof. If
and condition (16) holds, we must have
and hence
(otherwise there would not exist any
-pair for which (16) holds). Consider first what happens when
. In this case we must have
and for
we have already proved that
for all
.
Consider now what happens when
. Since we are assuming that
(and hence we must have
), we must have
and the assumed inequality (16) is satisfied if and only if the
-pair belongs to the area between the two stripes S1 and S2 or to one of the inner boundaries of S1 or S2. We indicate this set of
-pairs with S3. Note that S3 must be a proper subset of S since we are assuming that
. Since
implies
, we can apply Lemma 4 and conclude that
. Now it follows from Lemma 3 that
has no real roots when
belongs to the interior of S3, i.e. if the inequality (16) is strict. Since the two slices can intersect only if
has real roots, we conclude that
for all
in this case. On the other hand, if the
-pair belongs to the boundary of S3, then it must also belong to the inner boundary of one of the two stripes S1 or S2. In other words, there must be equality in (16) which means that
and that the discriminant of
must be zero (see Lemma 3). Hence
must have a root and this root must be unique. As usual we write
to indicate the root. Since we are assuming that Θ and Φ are both larger than 1, we can apply Lemma 5 and conclude that
when the
-pair belongs to
and hence that
for all
-pairs which belong to the boundary of S3 where a root
must exist and must be unique. Since S1 and S2 are two disjoint and connected sets,
does not change sign on each of these two sets. We will now show that the sign of
is positive on both sets. This can be done by proving that the sign of
is positive at a single point in each of the two sets (see Corollary 1). As in the proof of Lemma (9) we use the
-pairs with
and
as test points (note that these points also belong to the boundary of S3). With this choice we still get the expressions in (13), (14) and (15) for
,
and
. However, since we are now assuming that
, we see that
and not
as in the proof of Lemma 9 (of course,
remains still negative). We conclude that the roots we are considering now must be either of type b) or c) in (5). Hence we must have
. In order to prove that the roots correspond to intersection points, we first note that the mapping
does not change sign on each of the two sets S1 and S2 (use Lemma 6). However, as far as we know by now, the sign of
might be different according to whether
belongs to S1 or to S2. Thus, if there exists a single point
such that the sign of
is positive, we can conclude that
is of type c) and hence that
for every
(
). Again, we use the two
-pairs with
and
as test points. For these points we get the same expressions of
and
as in equations (13) and (14), while for
we get the expression
Hence we conclude that
This argument shows that every root
with
is an intersection point and that there must be a unique intersection point when
. It is not difficult to show that in the latter case the intersection point must be a tangency point. In fact, if it was a crossing point, there should exist one further crossing point because under our present assumptions the left and right asymptotes of
are both steeper than those of
(recall that we are assuming
and
: on
condition (4) must therefore hold with strict inequality sign).
As far as I know, the results in the next two lemmas are new.
Lemma 11. If
(17)
there must exist exactly two points where the slices
and
cross over each other.
Proof. From Lemma 3 and Lemma 4 we know that under condition (17) there must exist two roots
. Moreover, from the proof of Lemma 10 we know that both these roots must be intersection points. The chained inequality in (17) says that both asymptotes of
are steeper than those of
. Therefore only two cases can occur: either (i) both intersection points are tangency points, or (ii) both intersection points are crossing points. It is not difficult to see that case (i) is impossibile. In fact, if both intersection points were tangency points, any increase of Θ should lead to
for all
. However, given fixed values of Φ,
and
, a small enough increase of Θ does not lead to a violation of condition (17) which implies the existence of two intersection points.¨
Now, it remains to see what happens when
(18)
i.e. when the left or right asymptote of
is the same as the corresponding asymptote of
.
Lemma 12. Assume condition (18) holds. Then there must exist exactly one point where the slices
and
cross over each other.
Proof. Define
as the subset of the
-plane where the equality in condition (18) holds.
is then the boundary of S, i.e. the subset of the
-plane where
or
On
the polynomial
reduces to
and the only root of
is given by
We will show that this root must be a crossing point. To this aim note that
is the union of two disjoint connected sets which we denote with
. From Lemma 4 it follows that
. Hence we may apply Corollary 1 and conclude that
does not change sign on each one of the two connected components of
. In order to show that
and
are of the same sign, it is therefore sufficient to find a single point
in each of the two sets
and
for which
and
are of the same sign. As test points we choose the points
where
. It is easily seen that these points are
where
. Substituting in the formula for the root we get
Regardless of the sign in
, this root yields
and
Of course
. As for
, its sign depends on the sign of
which is positive whenever
and
(we omit the details of the proof of this assertion). Hence also
and thus we conclude that
is a root of either type b) or c) in (5). To find out which type applies, we must determine the sign of
. It is not difficult to verify that
and hence we get
The numerator in this expression can be written as
and therefore we must have
. By Lemma 6 we conclude that
must be an intersection point for every
.
To complete the proof it remains to show that for every
the corresponding root
is a crossing point. To this aim we apply the argument in the proof of Lemma 11 once again: if
was a tangency point, then by increasing Θ a little bit we should have no intersection points at all. However, if we increase Θ a little bit while leaving Φ,
and
unchanged, we pass from condition (18) to condition (17) which implies the existence of two crossing points.
Combining the statements in Lemma 1, Lemma 2, Lemma 7, Lemma 8, Lemma 9, Lemma 10, Lemma 11 and Lemma 12 yields a corrected and sharper version of Proposition 3.1 in Hendriks & Martini (2019) . The little corrections concern
i) the special case where
and
ii) the fact that Proposition 3.1 in Hendriks & Martini (2019) seems to imply that with
,
and
it would be possible to have absence of calendar spread arbitrage even if
which goes against the necessary condition (4). However, the preprint version of the article contains a slightly different version of the proposition which is not subject to this problem but where the two necessary conditions are a little too strong due to strong inequality signs instead of weak ones (see Proposition 3.5 in Hendriks & Martini, 2017 ).
The sharper (and corrected) statement of the Hendriks-Martini proposition is given below. To make it more concise, the necessary condition (3) will be stated as in (19).
Proposition 13. Assume that
and
are both strictly positive and let
and
. Then, there is absence of calendar spread arbitrage (i.e.
for all
) only if
and
(19)
Moreover,
· when
there is absence of calendar spread arbitrage if and only if either (i)
and
or (ii)
and
;
· when
there is absence of calendar spread arbitrage if and only if condition (19) holds jointly with
· when
and condition (19) holds jointly with
there are no intersection points (i.e.
for all
)
· when
,
and
the two slices have exactly one intersection point which is a tangency point;
· when
,
and
there must exist exactly two points where the slices
and
cross over each other.
· when
,
and
there must exist exactly one point where the slices
and
cross over each other.
3. Conclusion
This paper provides a detailed proof of conditions which characterize calendar spread arbitrage in eSSVI volatility surfaces. Moreover, it gives a full characterization for the case where two eSSVI slices have tangency points without crossing over each other. The motivation for this paper stems from a little error in the statement of a proposition in Hendriks & Martini (2019) where Hendriks and Martini introduced the eSSVI model, and from the fact the correct statement cannot be easily deduced from their proof. From a practical point of view, the conditions given in this paper can be used to check for the presence of calendar spread arbitrage in calibrated eSSVI surfaces and/or they can be incorporated in a calibration algorithm in order to obtain fitted volatility surfaces which are free of calendar spread arbitrage.