EONIA Jumps and Proper Euribor Forwarding

Synthetic Deposits ON Curve with Jumps Final Results Summary
EONIA Jumps and Proper Euribor Forwarding
The Case of Synthetic Deposits in Legacy Discount-Based Systems
Ferdinando M. Ametrano
[email protected]
[email protected]
Paolo Mazzocchi
[email protected]
[email protected]
QuantLib User Meeting, D¨
usseldorf, 5 December 2014
https://speakerdeck.com/nando1970/eonia-jumps-and-proper-euribor-forwarding
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 1 / 66

View Slide

Outline
1 Synthetic Deposits
The problem
A ﬁrst solution
Residual problems
2 ON Curve with Jumps
Jumps calculation
3 Final Results
Forward rate curve using Synthetic Deposits

View Slide

Outline
The problem
A ﬁrst solution
Residual problems
Jumps calculation
3 Final Results

View Slide

To obtain proper forward rate curves in legacy discount-based systems it is
very important to construct in a good way the ﬁrst section, the one for
maturities shorter than the ﬁrst market pillar.
FRA equation
It is useful to write down the relation between the value of a generic FRA
contract, on x tenor rate, and pseudo-discount factors (P(d), d = date):
1 + Fx (d, d + x) · τ =
P(d)
P(d + x)
= e d+x
d
fx (s)ds (1)
Therefore we have that this contract depends on the values of two
pseudo-discount factors: P(d) and P(d + x).

View Slide

Example
We consider the Euribor 6M curve. The first instrument, in order of
increasing maturities, that we can find on market is the 0x6 FRA (e.g.
FRA over today, FRA over tomorrow, index fixinga)
Since P(0) = 1, using equation (1), we have:
P(6M) =
1
1 + F6M(0, 6M) · τ
= e− 6M
0
f6M (s)ds = e−z6M (6M)τ (2)
With z6M(6M) we indicate the zero rate at 6M.
aEuribor fixing = 0x6 FRA, therefore rfix (6M) = F6M
(0, 6M)

View Slide

Example
Imagine that we now want to insert the 1x7 FRA; using equation (1), we
have:
P(7M) =
P(1M)
1 + F6M(1M, 7M) · τ
=
e− 1M
0
f6M (s)ds
1 + F6M(1M, 7M) · τ
(3)
We need to know 1M
0
f6M(s)
We could interpolate P(1M) between P(0) = 1 and P(6M).
This produces very bad result: without any other information we easily
underestimate or overestimate the proper values of the 1M
pseudo-discount factor.

View Slide

Example
These kind of errors lead to an incorrect calculation of forward rates: when
market quotes become less dense (i.e. where the FRA’s strip ends) the
curve shows an hump:
Figure: Forward Euribor 6M Curve

View Slide

Example
It is helpful to take a look at the instantaneous forward rates, since
everything else is obtained through their integration. In the ﬁrst 6 months,
they have been obtained in an arbitrary way, leading to oscillations:
Figure: Instantaneous Forward Ratea on Euribor 6M Curve (3 years)
awe calculate them as ON forward rate on 6 months curve

View Slide

Outline
The problem
A ﬁrst solution
Residual problems
Jumps calculation
3 Final Results

View Slide

Synthetic Deposits construction
The model
To solve this problem we need a reliable way to prescribe the shape of the
instantaneous forward on the short end. We can use ON instantaneous
forward rate plus a spread, δx (t):
fx (t) = fon(t) + δx (t) ∀ 0 ≤ t ≤ x (4)

View Slide

The model
In generala, even with t2 not equal to t1 + x:
1 + Fx (t1, t2)τx = e
t2
t1
fx (s)ds
= e
t2
t1
(fon(s)+δx (s))ds = e
t2
t1
fon(s)ds
· e
t2
t1
δx (s)ds
= [1 + Fon(t1, t2)τx ] · e∆x (t1,t2)
Where:
∆x (t1, t2) =
t2
t1
δx (s)ds
Fon(t1, t2) =
1
τx
Pon(t1)
Pon(t2)
− 1
aτx
= x-months discrete forward rate year fraction

View Slide

The model
Isolating ∆x (t1, t2), we have:
∆x (t1, t2) = ln
1 + Fx (t1, t2)τx
1 + Fon(t1, t2)τx
= ln [1 + Fx (t1, t2)τx ] − ln [1 + Fon(t1, t2)τx ]
We can approximate the previous equation neglecting higher order terms:
∆x (t1, t2) ≈ Fx (t1, t2)τx − Fon(t1, t2)τx ≈ Bx (t1, t2)τx (5)
Where:
Bx (t1, t2) = Fx (t1, t2) − Fon(t1, t2) (6)
is the simply compounded basis.
∆x (t1, t1 + x) and Bx (t1, t1 + x) are observable from market quotes

View Slide

Synthetic Deposits
We can construct synthetic spot instruments, FSynth
x (0, t) with t ≤ x, for
the bootstrapping of the forward curve of tenor x between 0 and x.
P(t) =
1
1 + FSynth
x (0, t) · τx
= e− t
0
fx (s)ds
Therefore in equations (5) and (6) we consider the case:
t1 = 0
t2 = t ≤ t1 + x = x

View Slide

Synthetic Deposits equation
With these hypothesis:
FSynth
x
(0, t) =
[1 + Fon(0, t)τx ] · e∆x (0,t) − 1
τx
(7)
Remark:
∆x (0, t), with t ≤ x, is not observable on the market;
∆x (ti , ti + x), with ti = 0M, 1M, 2M, ... and/or
ti = 0M, IMM1, IMM2, ... is observable from Fx (ti , ti + x) and
Fon(ti , ti + x);

View Slide

Basis calculation
∆x (0, t) =
t
0
δx (s)ds
We parametrize δx (s) using an n degree polynomial calibrated to market
available quotes, Fx (ti , ti + x), and Fon(ti , ti + x) calculated from the ON
curve.

View Slide

1 - Flat parametrization
fx (t) = fon(t) + δx (t) ∀ 0 ≤ t ≤ x
δx (t) = αx
To ﬁx the value of αx we need:
x
0
δx (s)ds = αx τx = ∆x (0, x)
= ln[1 + Fx (0, x)τx ] − ln[1 + Fon(0, x)τx ]
We have the following equation:
∆x (0, x) = αx τx

View Slide

2 - Linear parametrization
δx (t) = αx + βx τx (0, t)
To determine the value of αx , βx we integrate the previous equation
between a generic ti and ti + x:a
ti +x
ti
δx (s)ds = αx (τti +x − τti
) +
1
2
βx (τ2
ti +x
− τ2
ti
) = ∆x (ti , ti + x)
aτt
= τx
(0, t)
∆x (ti , ti + x) is observable for ti = 0M,1M,... and/or ti = IMM1,IMM2,...

View Slide

2 - Linear parametrization
We need 2 equations to calibrate αx and βx :











∆x (0, 0 + x) = ln[1 + Fx (0, 0 + x)τx (0, x)]
− ln[1 + Fon(0, 0 + x)τx (0, x)]
∆x (t1, t1 + x) = ln[1 + Fx (t1, t1 + x)τx (t1, t1 + x)]
− ln[1 + Fon(t1, t1 + x)τx (t1, t1 + x)]
We have the following system of 2 equations:
∆x (0, 0 + x) = αx τx + 1
2
βx τ2
x
∆x (t1, t1 + x) = αx (τt1+x − τt1
) + 1
2
βx (τ2
t1+x
− τ2
t1
)

View Slide

3 - Quadratic parametrization
δx (t) = αx + βx τx (0, t) + γx τx (0, t)2
To determine the value of αx ,βx ,γx as before:
ti +x
ti
δx (s)ds = αx (τti +x − τti
) +
1
2
βx (τ2
ti +x
− τ2
ti
)+
1
3
γx (τ3
ti +x
− τ3
ti
)
= ∆(ti , ti + x)
∆x (ti , ti + x) is observable for ti = 0M,1M,... and/or ti = IMM1,IMM2,...

View Slide

3 - Quadratic parametrization
We need 3 equations to calibrate αx ,βx ,γx :





















∆x (0, 0 + x) = ln[1 + Fx (0, 0 + x)τx (0, x)]
− ln[1 + Fon(0, 0 + x)τx (0, x)]
− ln[1 + Fon(t1, t1 + x)τx (t1, t1 + x)]
− ln[1 + Fon(t2, t2 + x)τx (t2, t2 + x)]
We have the following system of 3 equations:





∆x (0, 0 + x) = αx τx + 1
2
βx τ2
x
+ 1
3
γx τ3
x
∆x (t1, t1 + x) = αx (τt1+x − τt1
) + 1
2
βx (τ2
t1+x
− τ2
t1
) + 1
3
γx (τ3
t1+x
− τ3
t1
)
∆x (t2, t1 + x) = αx (τt2+x − τt2
) + 1
2
βx (τ2
t2+x
− τ2
t2
) + 1
3
γx (τ3
t2+x
− τ3
t2
)

View Slide

To ﬁnd ∆x (ti , ti + x) = ti +x
ti
δx (s)ds, we use:
market available quotes on x-months tenor Euribor, like:
index ﬁxing, Futures, FRA, IRS.
equivalent ON discrete forward rates, built using the ON curve (in
the Euro market they are equivalent to forward OIS contract).

View Slide

Basis Calculation
Example
Let’s consider the 6 Months Euribor curve.
Until 2 years ago, to calculate ∆6M(0, 6M) we had:
FRA over today and FRA over tomorrow;
Today they are not quoted anymore, therefore we need to find a different
way to calculate ∆6M(0, 6M). The only contract that we can use is the
index fixing.
After that instruments, to calculate ∆6M(ti , ti + x), we have:
FRA up to 2 years.
We calculate ON discrete forward rates insisting on the same set of dates
as the 6M market quotes and we calculate the difference between these
two values.

View Slide

Basis Calculation
Example
Figure: ∆6M
(ti , ti
+ 6M) = ln[1 + F6M
(ti , ti
+ 6M)] − ln[Fon
(ti , ti
+ 6M)]

View Slide

Basis Calculation
Example
∆6M(ti , ti + 6M) =
ti +6M
ti
δx (s)ds
Figure: Continuously compounded basis

View Slide

Synthetic Deposits Calculation
Example
It is important to note that the integral of δ6M(s) in the first 6 months it
is the same for all of the parametrization used, but with different shape:
∆x (0, 6M) =
6M
0
δx (s)ds =





6M
0
α6Mds flat
6M
0
(ˆ
α6M + ˆ
β6Mτs)ds linear
6M
0
(˜
α6M + ˜
β6Mτs + ˜
γ6Mτ2
s
)ds quad

View Slide

Synthetic Deposits Calculation
Example
We can build Synthetic Deposits for every 0 ≤ t ≤ 6M:
FSynth
6M
(0, t) =
[1 + Fon(0, t)τ6M] · e∆6M (0,t) − 1
τ6M
Usually we construct Synthetic Deposits to match the start date of market
FRAs or Futures.

View Slide

Example
Not using 1M, 2M, 3M, 4M, 5M, 6M Synthetic Deposits for the 6M curve.
Figure: Forward Euribor 6M Curve without Synthetic Deposits

View Slide

Example
Using 1M, 2M, 3M, 4M, 5M, 6M Synthetic Deposits for the 6M curve.
Figure: Forward Euribor 6M Curve with Synthetic Deposits

View Slide

Outline
The problem
A ﬁrst solution
Residual problems
Jumps calculation
3 Final Results

View Slide

Problem
Until about two years ago this approach was satisfactory: it produced
smooth forward rates.
Recently it began to show some problems, because of two crucial issues:
instruments available to calculate the basis, Fx (t, t + x)a
shape of the ON term structure, Fon(t, t + x)
aFRA over today/tomorrow are not available anymore
∆x (t, t + x) = ln[1 + Fx (t, t + x)τx ] − ln[1 + Fon(t, t + x)τx ]

View Slide

First Problem
The ﬁrst problem is the correct selection of market instruments used to
determine the values of the basis: an illiquid product will produce an
incorrect estimation.
E.g.: the index ﬁxing is not an optimal choice. Its value remains constant
during the day, insensitive to market changes.
Solution
A solution is to interpolate/extrapolate only on liquid values of
Bx (ti , ti + x)a, the simply compounded basis; then we calculate the
interpolated/extrapolated values of Fx , for the illiquid ones and we use
them to determine ∆x .
a∆x
(ti , ti
+ x) ≈ Bx
(ti , ti
+ x)τ cannot be directly interpolated because τ is
not exactly constant (business days adjustment)

View Slide

Basis Calculation
Example
B6M(ti , ti + 6M) = F6M(ti , ti + 6M) − Fon(ti , ti + 6M)
Figure: Interpolated Simply Compounded Basis

View Slide

Second Problem
The second critical point is the ON term structure: to obtain a good
approximation of the basis, it is fundamental to build a good ON curve.
Figure: ON interest rate curve with log-cubic interpolation on Discount factor

View Slide

Problem
The second critical point is the ON term structure: to obtain a good
approximation of the basis, it is fundamental to build a good ON curve.
Figure: ON interest rate curve with log-linear interpolation on Discount factor

View Slide

Outline
The problem
A ﬁrst solution
Residual problems
Jumps calculation
3 Final Results

View Slide

How many Jumps?
The Eonia ﬁxing has a jump at least at every end of month.
Figure: Eonia ﬁxings, last 6 months

View Slide

Jump Size estimation
We show how the jump estimation improves the quality of the curve with
a concrete example:
Figure: ON interest rate curve: starting point

View Slide

We show how the jump estimation improves the quality of the curve with
a concrete example:
Figure: ON interest rate curve with the ﬁrst jump

View Slide

Jumps size estimation
To calculate Jumps’ size we can follow an approach similar to the one used
by Burghardt [3] to estimate turn of year jumps:
1 Construct an ON curve using all liquid market quotes using a flat
interpolation on forward rate
2 Estimate the first jump assuming that a segment out of line with
preceding and following segments can be put back in line dumping
the difference into the jump1:
[Foriginal (t1, t2) − Finterp(t1, t2)] · τ(t1, t2) = Jump · τJump
3 Clean the curve from the jump at point 2
Iterate ad libitum 2 and 3 on next jump date
1τJump
is the year fraction between jump business day and the next business day

View Slide

Figure: ON interest rate curve: starting point

View Slide

Figure: ON interest rate curve with the ﬁrst jump

View Slide

Figure: ON interest rate curve with the second jump

View Slide

Figure: ON interest rate curve with the third jump

View Slide

Figure: ON interest rate curve with jumps

View Slide

Figure: ON interest rate curve with log-cubic interpolation on Discount factor

View Slide

Outline
The problem
A ﬁrst solution
Residual problems
Jumps calculation
3 Final Results

View Slide

Synthetic Deposits Construction
We had 2 residual problems for construction of Synthetic Deposits:
sub-optimal instruments selection
wrong ON curve calculation
Modelling ON jumps we ﬁxed the ON curve and obtained reliable B(0, x)
to be interpolated/extrapolated.

View Slide

6-Months Euribor Curve
Example
Interpolating only liquid B6M(ti , ti + 6M), i.e. ﬁxing not included, we
match the ﬁxing level nonetheless!
Figure: Interpolated Simply Compounded Basis with jumps

View Slide

Example
Figure: Interpolated Simply Compounded Basis without jumps

View Slide

Example
In order to estimate δ6M(t) we use:
ﬂat parametrization:
∆6M(0, 6) = ln[1 + Fextrap
6M
(0, 6) · τ] − ln[1 + FON(0, 6) · τ]
linear parametrization:
∆6M(0, 6) = ln[1 + Fextrap
6M
(0, 6) · τ] − ln[1 + FON(0, 6) · τ]
∆6M(1, 7) = ln[1 + Finterp
6M
(1, 7) · τ] − ln[1 + FON(1, 7) · τ]
quadratic parametrization:





∆6M(0, 6) = ln[1 + Fextrap
6M
(0, 6) · τ] − ln[1 + FON(0, 6) · τ]
∆6M(1, 7) = ln[1 + Finterp
6M
(1, 7) · τ] − ln[1 + FON(1, 7) · τ]
∆6M(2, 8) = ln[1 + Finterp
6M
(2, 8) · τ] − ln[1 + FON(2, 8) · τ]

View Slide

Example
δ6M(t) using index ﬁxing:
Figure: Continuously Compounded Basis δ6M
(t) for 0 ≤ t ≤ 6M

View Slide

Example
δ6M(t) extrapolating B6M, instead of index ﬁxing:
(t) for 0 ≤ t ≤ 6M

View Slide

Example
Figure: Interpolated Simply Compounded Basis

View Slide

Example
In order to estimate δ3M(t) we use:
ﬂat parametrization:
∆3M(0, 3) = ln[1 + Fextrap
3M
(0, 3) · τ] − ln[1 + FON(0, 3) · τ]
linear parametrization:
∆3M(0, 3) = ln[1 + Fextrap
3M
(0, 3) · τ] − ln[1 + FON(0, 3) · τ]
∆3M(1, 4) = ln[1 + Finterp
3M
(1, 4) · τ] − ln[1 + FON(1, 4) · τ]
quadratic parametrization:





∆3M(0, 3) = ln[1 + Fextrap
3M
(0, 3) · τ] − ln[1 + FON(0, 3) · τ]
∆3M(1, 4) = ln[1 + Finterp
3M
(1, 4) · τ] − ln[1 + FON(1, 4) · τ]
∆3M(2, 5) = ln[1 + Finterp
3M
(2, 5) · τ] − ln[1 + FON(2, 5) · τ]

View Slide

Example
(t) for 0 ≤ t ≤ 3M

View Slide

Example
Forward Euribor 6M without synthetic depositsa:
ausing index ﬁxing

View Slide

Example
Forward Euribor 6M with synthetic depositsa:
ausing index ﬁxing and without ON jumps

View Slide

Example
awithout index ﬁxing, with correct instruments selection and ON jumps

View Slide

Example
Instantaneous Forward Euribor 6M without synthetic depositsa:
Figure: Instantaneous Forward Rate on Euribor 6M Curve

View Slide

Example
Instantaneous Forward Euribor 6M with synthetic depositsa:

View Slide

Example
Forward Euribor 3M without synthetic depositsa:

View Slide

Example

View Slide

Example
Instantaneous Forward Euribor 3M without synthetic depositsa:

View Slide

Example
Instantaneous Forward Euribor 3M with synthetic depositsa:

View Slide

Conclusion
ON curve jumps must be taken into account
the smoothness of the basis, not of the forward rates, is the relevant
factor

View Slide

Bibliography
F.M. Ametrano, M. Bianchetti
Everything you always wanted to know about multiple interest rate
curve bootstrapping but were afraid to ask.
http://ssrn.com/abstract=2219548
SSRN, 2013.
S. Schlenkrich,A. Miemiec
Choosing the right spread.
SSRN, 2014.
G. Burghardt, S. Kirshner
One good turn.
CME Interest Rate Products Advanced Topics. Chicago: Chicago
Mercatile Exchange,2002.

View Slide

EONIA Jumps and Proper Euribor Forwarding

EONIA Jumps and Proper Euribor Forwarding

Ferdinando M. Ametrano

More Decks by Ferdinando M. Ametrano

Featured

Transcript