CURRENCY ARBITRAGE
4.3 DEFINITION AND MEASUREMENT OF EXPOSURE TO FOREIGN EXCHANGE RISK
TABLE 4.3 VAR of majorbanks.
Country Foreign exchange risk (% of Total market risk (% of
capital) capital)
Australia 0.02 0.08
Canada 0.02 0.12
Germany 0.15 0.71
Japan 0.07 0.31
Netherlands 0.02 0.20
UK 0.03 0.16
USA 0.03 0.15
Source: Reserve Bank of Australia (2000)
figures in other countries. In Germany, for example, the corresponding figure was 0.15%. The 1999 figures are shown in Table 4.3.
4.3 DEFINITION AND MEASUREMENT OF EXPOSURE TO
dVx
E= (4.14)
dS
and in discrete form as DVx
E= (4.15)
DS
in which case the exposure is measured in terms of currency y (hence, the amount exposed, or at risk). Alternatively, the expression can be written in terms of the (percentage) rates of change, which gives
E =Vx
(4.16) S
It is obvious from equation (4.16) that exposure in this case is measured without units, more or less like an elasticity. But in all cases, the exposure is the slope of the line (or curve) representing the relationship between Vx and S.
Notice, in general, that
Vy = f1(S) (4.17)
and
Vx = f2(S) (4.18)
Therefore
Vx = SVy = Sf1(S) (4.19)
If, for example, Vx = +a ES+e, then the exposure coefficient, E, is calculated as s(V S)
(4.20) E= x ,
s2( ) S such that
s2(Vx ) =E2 s2(S) +s2(e) (4.21)
× and s( , ) e
where s2( ) × × are the variance and covariance respectively, and s2( ) captures the residual variability that is independent of exchange rate move
ments. In what follows, we consider some possibilities for f1(S) and f2(S).
The effect of the exchange rate on the y-denominated value
We will consider four cases involving linear relationships between Vy and S, some of which produce nonlinear relationships between Vx and S. The list of possibilities presented here is not exhaustive. Because the relationship between Vy and S can take several shapes and forms, foreign exchange expo
sure can be zero, constant or variable. We will now consider the four cases by looking at an asset (that is, a long exposure), but the same description applies to a liability (that is, a short exposure).
Case 1: Vy is independent of the exchange rate
If Vy is constant, assuming the same value at any level of the exchange rate, then
Vy = K (4.22)
Hence
Vx = KS (4.23)
and d d
V S
x =K (4.24)
which means that the exposure does not change with the level of the exchange rate. This case is represented diagrammatically in Figure 4.3. The upper part of the diagram shows that Vy is independent of S, and this is why the relation
ship is represented by a horizontal line. The bottom part of the diagram shows the derived relationship between Vx and S (equation 4.23). As the exchange rate rises, Vy is unaffected but Vx rises. Thus, the relationship between Vx and Sis represented by the line Vx = KS, and the exposure is represented by the slope of this line, which is constant at K. Notice that Vx is represented in the upper part of the diagram by the area under the line Vy = K, whereas in the bottom part it is measured on the vertical axis.
Case2:Vy isinverselyproportionaltotheexchangerate
In this case Vy falls (rises) when the exchange rate rises (falls) such that the change is proportional (equal percentage changes in opposite directions). This means that Vy and S are related as
Vy = K/S (4.25)
which is a rectangular hyperbola indicating that the product of Vy and S (which is the area under the curve) is constant. Hence
Vx = K (4.26)
and dVx
= 0 (4.27)
dS
which means that the exposure is zero (there is nothing at risk or that Vx is insensitive to changes in S). This is because any change in the exchange rate is completely offset by a change (in the opposite direction) in Vy, leaving Vx unchanged. This case is represented by Figure 4.4, where a rise in the exchange rate leads to a proportional fall in Vy, leaving Vx unchanged. Hence the expo
sure is zero because the slope of the line Vx = K is zero.
Vy
S
S K Vy =
KS Vx = V x
FIGURE 4.3 Exposure when Vy is independent of the exchange rate.
Case 3: Vy is negatively and linearly related to the exchange rate If Vy is a decreasing linear function of the exchange rate then
Vy = a – bS (4.28)
where b> 0. Equation (4.28) is represented by a downward-sloping line in the upper part of Figure 4.5. In this case Vx is a nonlinear function of S that can be written as
Vy
S
S S
Vy = K
K Vx = V x
FIGURE 4.4 Exposure when Vy is Inversely proportional to the exchange rate.
Vx = aS – bS2 (4.29)
as shown in the bottom part of Figure 4.5 (the lowest value on the vertical axis on which Vx is measured is a, not 0) . Starting from a low level, as the exchange rate rises Vx also rises, as represented by a larger area under the line Vy = a– bS. But as the exchange rate rises further, Vx reaches a maximum and starts to decline. This is because as S rises, the product of S and Vy may rise or fall, depending on its initial value.
Vy
S
S bS
a Vy = -
bS2
aS Vx = - V x
FIGURE 4.5 Exposure when Vy is negatively and linearly related to the exchange rate.
Case4:Vy ispositivelyandlinearly relatedto theexchangerate If Vy is an increasing linear function of the exchange rate then
Vy = a+ bS (4.30)
which is represented by an upward-sloping line in the upper part of Figure 4.6. In this case Vx is a nonlinear function of S, which is written as
Vx = aS+ bS2 (4.31)
Vy
S
S bS a Vy = +
bS2
aS Vx = + V x
FIGURE 4.6 Exposure whenVy is positively and linearly related to the exchange rate.
as shown in the bottom part of Figure 4.6 (the lowest value on the vertical axis on which Vx and Vy are measured is a, not 0). As the exchange rate rises, Vx also rises, as represented by a larger area under the line Vy = a + bS. It is important to observe that as the exchange rate rises, Vx rises more proportionately. The curve in the bottom part of Figure 4.6 has an increasing positive slope, implying that exposure increases as the exchange rate rises.
The exposure line
Let us now concentrate on the case when exposure is constant, as in Figure 4.3, expressing the relationship in terms of the changes in Vx and S(DVx and DS) rather than their levels. This relationship is represented diagrammatically in Figure 4.7, where changes in the exchange rate are measured on the horizontal axis and changes in the base currency value of the asset are measured on the vertical axis. The line representing this relationship is called the exposure line.
In this case, the exposure line has a positive slope to indicate a positive rela
tionship between changes in the exchange rate and changes in the base currency value of the asset. The equation of the exposure line is
D
DVx =E S (4.32)
where the slope of the line, E, is the exposure. In the case of a long exposure, as shown in Figure 4.7, E> 0. Notice that there are two lines: a steep line repre
senting high exposure and a shallow line representing low exposure. Hence zero exposure would be represented by a horizontal line, whereas an infinite exposure would be represented by a vertical line.
We now consider the case of exposure to foreign liabilities (short exposure), as shown in Figure 4.8. In this case, a rise in the exchange rate induces a rise in
DVx
High exposure
Low exposure
DS
FIGURE 4.7 Long exposure line (assets).
( / ( / ( /
DVx
D S High exposure
Low exposure
FIGURE 4.8 Short exposure line (liabilities).
the base currency value of liabilities, which entails a loss. This is why the expo
sure line in this case is downward-sloping. It has the same equation as (4.32) except that E< 0.
The relationship between risk and exposure can be determined from equa
tion (4.32), which gives
s2(DVx ) =E2 s2(DS) (4.33)
Equation (4.33) tells us that the variance of changes in the base currency value of foreign assets and liabilities is related to the variance of changes in the exchange rate by a factor that reflects exposure, E2.
Multiple exposure
So far we have dealt with exposure to a single currency, y. In practice, expo
sure to several currencies is normally the case, as international business firms diversify their investment and financing portfolios. Hence a multiple expo
sure model may be written as V
x =a1 S x y1) +a2 S x y2) +…+an S x yn) (4.34) where S x y( / i) is the percentage change in the exchange rate between the base currency, x, and currency yifor i= 1, 2, ..., n. The coefficient ai(i= 1, 2, ..., n)
measures the exposure to currency yi. In the case of assets, the coefficients are positive, whereas in the case of liabilities they are negative.
There are at least two problems with the empirical model of multiple expo
sure that is represented by equation (4.34). The first problem is that the model assumes a linear exposure, which may not be the case. A nonlinear exposure may arise because of nonlinearity in the firm’s price elasticity of demand. The second problem is that exposure may not be constant over time, whereas this model would produce constant exposures if it is estimated by a conventional estimation method, such as OLS. This problem can be circumvented by resorting to an estimation method that allows the estimated coefficients to vary over time. This can be accomplished by specifying a state space model that can be estimated by utilising the recursive method of the Kalman filter (see Chapter 6).
The model represented by equation (4.34) can be used to formulate general multicurrency exposure relationships. If we measure value in terms of currency y1, then we have
( / ) ( / ( /
Vy = b S y x + b S y y2) +…+ b S y yn) (4.35)
1 1 1 2 1 n 1
Vy 2 =c S y1 ( 2/x) +c2 S y( 2 / y1) +…+c S yn ( 2/yn) (4.36) If, for example, xis the US dollar and y1 is the pound, then equation (4.34) relates changes in the US dollar value of the (US) company to changes in various exchange rates. If this US company has British shareholders, whose base currency is the pound, then equation (4.35) relates the pound value of the company (which is what matters for British shareholders) to changes in the exchange rates of the other currencies against the pound. The same interpre
tation can be given to equation (4.36) if y2 is, for example, the Japanese yen.
Adler and Jorion (1992) have shown that the exposure coefficients of equation (4.34) are related to those of equation (4.35) as follows
(b2, b3, ..., bn) = (a2, a3, ..., an) (4.37) and
b1= 1 – (a2+ a3 + ... + an) (4.38)