Transaction Cost Analysis A-Z

Transaction Cost Analysis A-Z — November 2008

IV. Estimating Transaction Costs with Pre-Trade Analysis

movement. We can derive from it target prices for all the periods in the trading horizon. Three models can be used for this purpose. To present them and facilitate understanding, we refer to the following numerical example. Suppose a security that is currently traded at € 100 ( P 0 ) and is expected to increase to € 125 ( P T ) after one year. If we consider fifteen-minute periods, we have 8,500 trading intervals ( n ) over the forecasted horizon. 13 We can first determine the expected price change over each trading interval and then calculate the expected price at the end of two days (sixty-eight trading intervals ( m )). Linear model This model assumes constant price changes regardless of current prices. Applying this model to our example, we get the constant price change ( Δ P ) and the expected price ( P t ) as follows:

changes by a constant percentage of the current price. Applying this model to our example, we get the exponential price change ( g ) and the expected price ( P t ) as follows:

P T

g = 1 n

1 8500

⎛ ⎝⎜

⎞ ⎠⎟ =

ln 125 100 ⎛ ⎝⎜

⎞ ⎠⎟ = 0.0026%

ln

P 0

= P 0

e ( m × g ) = 100 × e ( 68 × 0.000026 ) = 100.18

P t

Return and growth models are interchangeable since they deliver similar results for any short-term trading horizon. (b) Cost estimates Consistent with the implementation shortfall approach and if we apply the linear price change model, the transaction cost component due to price appreciation is estimated as follows. 14 P 0 = security price at the beginning of trading Δ P = linear price change forecast P j = forecasted price in period j ; X = order size; X>0 for buys and X<0 for sells x j = number of shares traded in period j ; x j = X i = 1 n ∑ U(X) = cost of price appreciation for order X

13 - We consider that there are thirty-four intervals of fifteen minutes in a usual trading day (9:00AM to 5:30PM) and 250 trading days in a year. 14 - The method is similar when we consider return and growth models.

Δ P = 1

1 8500

(

) =

− P 0

(125 − 100 ) = 0.0029

n P T

= P 0

+ m Δ P = 100 + 68 × 0.0029 = 100.20

P t

Return model This model assumes that the price changes by a constant percentage of the current price. Applying this model to our example, we get the percentage price change ( r ) and the expected price ( P t ) as follows:

n ∑

U( X ) =

x

P j

− XP 0

j

j = 1

n ∑

n ∑

U( X ) =

x

P 0

x

j Δ P − XP 0

+

1 n − 1 =

1 8500 − 1 = 0.0026%

j

j

P T

125

⎛ ⎝⎜

⎞ ⎠⎟

⎛ ⎝⎜

⎞ ⎠⎟

j = 1

j = 1

r =

P 0

100

n ∑

U( X ) = XP 0

x

j Δ P − XP 0

+

j

j = 1

= P 0 (1 + r ) m = 100 × (1 + 0.000026 ) 68 = 100.18

P t

n ∑

U( X ) =

x

j Δ P

j

Growth model This model also assumes that the price

j = 1

43 An EDHEC Risk and Asset Management Research Centre Publication