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')
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 ')