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# EONIA Jumps and Proper Euribor Forwarding

EONIA Jumps and Proper Euribor Forwarding: the Case of Synthetic Deposits in Legacy Discount-Based Systems.

Presented at the 2014 QuantLib User Meeting in Düsseldorf ## Ferdinando M. Ametrano

December 05, 2014

## Transcript

1. 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

2. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 2 / 66

3. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 3 / 66

4. Synthetic Deposits ON Curve with Jumps Final Results Summary
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).
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 4 / 66

5. Synthetic Deposits ON Curve with Jumps Final Results Summary
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)
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 5 / 66

6. Synthetic Deposits ON Curve with Jumps Final Results Summary
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.
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 6 / 66

7. Synthetic Deposits ON Curve with Jumps Final Results Summary
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.
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 6 / 66

8. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 7 / 66

9. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 8 / 66

10. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 9 / 66

11. Synthetic Deposits ON Curve with Jumps Final Results Summary
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)
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 10 / 66

12. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 11 / 66

13. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 12 / 66

14. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 13 / 66

15. Synthetic Deposits ON Curve with Jumps Final Results Summary
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);
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 14 / 66

16. Synthetic Deposits ON Curve with Jumps Final Results Summary
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.
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 15 / 66

17. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 16 / 66

18. Synthetic Deposits ON Curve with Jumps Final Results Summary
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,...
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 17 / 66

19. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
)
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 18 / 66

20. Synthetic Deposits ON Curve with Jumps Final Results Summary
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,...
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 19 / 66

21. Synthetic Deposits ON Curve with Jumps Final Results Summary
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)]
∆x (t1, t1 + x) = ln[1 + Fx (t1, t1 + x)τx (t1, t1 + x)]
− ln[1 + Fon(t1, t1 + x)τx (t1, t1 + x)]
∆x (t1, t1 + x) = ln[1 + Fx (t2, t2 + x)τx (t2, t2 + 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
)
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 20 / 66

22. Synthetic Deposits ON Curve with Jumps Final Results Summary
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).
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 21 / 66

23. Synthetic Deposits ON Curve with Jumps Final Results Summary
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.
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 22 / 66

24. Synthetic Deposits ON Curve with Jumps Final Results Summary
Basis Calculation
Example
Figure: ∆6M
(ti , ti
+ 6M) = ln[1 + F6M
(ti , ti
+ 6M)] − ln[Fon
(ti , ti
+ 6M)]
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 23 / 66

25. Synthetic Deposits ON Curve with Jumps Final Results Summary
Basis Calculation
Example
∆6M(ti , ti + 6M) =
ti +6M
ti
δx (s)ds
Figure: Continuously compounded basis
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 24 / 66

26. Synthetic Deposits ON Curve with Jumps Final Results Summary
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

α6M + ˆ
β6Mτs)ds linear
6M
0

α6M + ˜
β6Mτs + ˜
γ6Mτ2
s
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 25 / 66

27. Synthetic Deposits ON Curve with Jumps Final Results Summary
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.
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 26 / 66

28. Synthetic Deposits ON Curve with Jumps Final Results Summary
Example
Not using 1M, 2M, 3M, 4M, 5M, 6M Synthetic Deposits for the 6M curve.
Figure: Forward Euribor 6M Curve without Synthetic Deposits
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 27 / 66

29. Synthetic Deposits ON Curve with Jumps Final Results Summary
Example
Using 1M, 2M, 3M, 4M, 5M, 6M Synthetic Deposits for the 6M curve.
Figure: Forward Euribor 6M Curve with Synthetic Deposits
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 28 / 66

30. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 29 / 66

31. Synthetic Deposits ON Curve with Jumps Final Results Summary
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 ]
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 30 / 66

32. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 31 / 66

33. Synthetic Deposits ON Curve with Jumps Final Results Summary
Basis Calculation
Example
B6M(ti , ti + 6M) = F6M(ti , ti + 6M) − Fon(ti , ti + 6M)
Figure: Interpolated Simply Compounded Basis
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 32 / 66

34. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 33 / 66

35. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 34 / 66

36. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 35 / 66

37. Synthetic Deposits ON Curve with Jumps Final Results Summary
How many Jumps?
The Eonia ﬁxing has a jump at least at every end of month.
Figure: Eonia ﬁxings, last 6 months
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 36 / 66

38. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 37 / 66

39. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jump Size estimation
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 38 / 66

40. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jumps size estimation
To calculate Jumps’ size we can follow an approach similar to the one used
by Burghardt  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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 39 / 66

41. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jumps size estimation
To calculate Jumps’ size we can follow an approach similar to the one used
by Burghardt  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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 39 / 66

42. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jumps size estimation
To calculate Jumps’ size we can follow an approach similar to the one used
by Burghardt  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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 39 / 66

43. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jumps size estimation
To calculate Jumps’ size we can follow an approach similar to the one used
by Burghardt  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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 39 / 66

44. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jumps size estimation
To calculate Jumps’ size we can follow an approach similar to the one used
by Burghardt  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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 39 / 66

45. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jump Size estimation
Figure: ON interest rate curve: starting point
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 40 / 66

46. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jump Size estimation
Figure: ON interest rate curve with the ﬁrst jump
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 41 / 66

47. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jump Size estimation
Figure: ON interest rate curve with the second jump
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 42 / 66

48. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jump Size estimation
Figure: ON interest rate curve with the third jump
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 43 / 66

49. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jump Size estimation
Figure: ON interest rate curve with jumps
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 44 / 66

50. Synthetic Deposits ON Curve with Jumps Final Results Summary
Jump Size estimation
Figure: ON interest rate curve with log-cubic interpolation on Discount factor
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 45 / 66

51. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 46 / 66

52. Synthetic Deposits ON Curve with Jumps Final Results Summary
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.
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 47 / 66

53. Synthetic Deposits ON Curve with Jumps Final Results Summary
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
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 48 / 66

54. Synthetic Deposits ON Curve with Jumps Final Results Summary
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 without jumps
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 49 / 66

55. Synthetic Deposits ON Curve with Jumps Final Results Summary
6-Months Euribor Curve
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]

∆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]
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 50 / 66

56. Synthetic Deposits ON Curve with Jumps Final Results Summary
6-Months Euribor Curve
Example
δ6M(t) using index ﬁxing:
Figure: Continuously Compounded Basis δ6M
(t) for 0 ≤ t ≤ 6M
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 51 / 66

57. Synthetic Deposits ON Curve with Jumps Final Results Summary
6-Months Euribor Curve
Example
δ6M(t) extrapolating B6M, instead of index ﬁxing:
Figure: Continuously Compounded Basis δ6M
(t) for 0 ≤ t ≤ 6M
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 52 / 66

58. Synthetic Deposits ON Curve with Jumps Final Results Summary
3-Months Euribor Curve
Example
Interpolating only liquid B3M(ti , ti + 3M), i.e. ﬁxing not included, we
match the ﬁxing level nonetheless!
Figure: Interpolated Simply Compounded Basis
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 53 / 66

59. Synthetic Deposits ON Curve with Jumps Final Results Summary
3-Months Euribor Curve
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]

∆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]
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 54 / 66

60. Synthetic Deposits ON Curve with Jumps Final Results Summary
3-Months Euribor Curve
Example
Figure: Continuously Compounded Basis δ3M
(t) for 0 ≤ t ≤ 3M
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 55 / 66

61. Synthetic Deposits ON Curve with Jumps Final Results Summary
6-Months Euribor Curve
Example
Forward Euribor 6M without synthetic depositsa:
Figure: Forward Euribor 6M Curve
ausing index ﬁxing
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 56 / 66

62. Synthetic Deposits ON Curve with Jumps Final Results Summary
6-Months Euribor Curve
Example
Forward Euribor 6M with synthetic depositsa:
Figure: Forward Euribor 6M Curve
ausing index ﬁxing and without ON jumps
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 57 / 66

63. Synthetic Deposits ON Curve with Jumps Final Results Summary
6-Months Euribor Curve
Example
Forward Euribor 6M with synthetic depositsa:
Figure: Forward Euribor 6M Curve
awithout index ﬁxing, with correct instruments selection and ON jumps
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 58 / 66

64. Synthetic Deposits ON Curve with Jumps Final Results Summary
6-Months Euribor Curve
Example
Instantaneous Forward Euribor 6M without synthetic depositsa:
Figure: Instantaneous Forward Rate on Euribor 6M Curve
ausing index ﬁxing
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 59 / 66

65. Synthetic Deposits ON Curve with Jumps Final Results Summary
6-Months Euribor Curve
Example
Instantaneous Forward Euribor 6M with synthetic depositsa:
Figure: Instantaneous Forward Rate on Euribor 6M Curve
awithout index ﬁxing, with correct instruments selection and ON jumps
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 60 / 66

66. Synthetic Deposits ON Curve with Jumps Final Results Summary
3-Months Euribor Curve
Example
Forward Euribor 3M without synthetic depositsa:
Figure: Forward Euribor 3M Curve
ausing index ﬁxing
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 61 / 66

67. Synthetic Deposits ON Curve with Jumps Final Results Summary
3-Months Euribor Curve
Example
Forward Euribor 3M with synthetic depositsa:
Figure: Forward Euribor 3M Curve
awithout index ﬁxing, with correct instruments selection and ON jumps
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 62 / 66

68. Synthetic Deposits ON Curve with Jumps Final Results Summary
3-Months Euribor Curve
Example
Instantaneous Forward Euribor 3M without synthetic depositsa:
Figure: Instantaneous Forward Rate on Euribor 3M Curve
ausing index ﬁxing
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 63 / 66

69. Synthetic Deposits ON Curve with Jumps Final Results Summary
3-Months Euribor Curve
Example
Instantaneous Forward Euribor 3M with synthetic depositsa:
Figure: Instantaneous Forward Rate on Euribor 3M Curve
awithout index ﬁxing, with correct instruments selection and ON jumps
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 64 / 66

70. Synthetic Deposits ON Curve with Jumps Final Results Summary
Conclusion
ON curve jumps must be taken into account
the smoothness of the basis, not of the forward rates, is the relevant
factor
Ferdinando M. Ametrano, Paolo Mazzocchi version 1.5 65 / 66

71. Synthetic Deposits ON Curve with Jumps Final Results Summary
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