Appendix


We need to determine if inequality (1.30) holds. That is if:

with ± > 0. Using the simple Matlab code below, we may determine the curve of the above function. The exact shape of the curve depends on the value of the reversion speed ± . Figure 1.2 shows the curve for ± = 0.15. The above inequality does not hold for all values of . The interval on which 4 ± e ˆ’± ˆ’ 3 + 3 e ˆ’± 0 depends on the value of ± .

 alpha = 0.15;   y_ans = [];   x = [];   for T = 0 : 50     T_2 = T * 0.1;     x = [x, T_2];     y=4*T_2 * alpha * exp(-alpha * T_2)       -3+3*exp(-alpha*T_2);     y ans = [y_ans, y];    end    plot(x, y_ans), xlabel('   '), ylabel('4    e   -    +3  e   -    - 3') 
click to expand
Figure 1.2: Shape of curve 4 ± e ˆ’± + 3 e ˆ’± ˆ’ 3for ± = 0.15

We require to show that inequality (1.31) holds. That is:

The Matlab code below determines the shape of this function on the interval ˆˆ [0, 5]. Again the exact shape of the curve depends on the choice of ± . We can show that the above inequality hold for all values of ; however, ± must be positive. ± < 0 implies negative reversion speed and causes a contradiction of the inequality. Figure 1.3 shows the curve for ± = 0.45.

 alpha = 0.45;   y_ans = [];   x = [];   for T = 0 : 50     T_2 = T * 0.1;     x = [x, T_2];     y = 4*T_2 * alpha * exp(-alpha * T_2)-2 * T_2 * alpha       *exp(-2 * alpha *T_2)-3 + 4 * exp(-alpha * T_2)       -exp(-2 * alpha * T_2);     y _ans = [y _ans, y];    end    plot(x,y_ans), xlabel('   '), ylabel('4    e   -    -2    e   -   2    - 3+4  e   -    -  e  -2   ') 
click to expand
Figure 1.3: Shape of curve 4 ± e ˆ’± ˆ’ 2 ± e ˆ’ 2 ± ˆ’ 3 + 4 e ˆ’± ˆ’ e ˆ’ 2 ± for ± = 0.45



Interest Rate Modelling
Interest Rate Modelling (Finance and Capital Markets Series)
ISBN: 1403934703
EAN: 2147483647
Year: 2004
Pages: 132

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