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Then we have R1 (V, 2) = 9V 2 . 70) On the other hand, in the scenario S1 = d = 12 , we have R1 V, 1 2 π2 3 −V 1− 4 2 2 = 1 min V 2 (1 + π2 )2 + 2 π2 ∈R = 1 min 4V 2 (5π22 + 4π2 + 8) + 12V (π2 − 2) + 9 .

In our case, the first step of the DP algorithm consists of computing R1 (V, S) for S ∈ 2, 1, 12 . We have R1 (V, 2) = = = min E P [U (V + 2αμ2 , 2(1 + μ2 ))] min EP min 1 4 + + (3 − V − 2α) + (1 − V )+ = (1 − V ) , 3 3 α∈[−V /2,V ] α∈[−V /2,V ] α∈[−V /2,V ] + (2(1 + μ2 ) − 1) − (V + 2αμ2 ) + the minimum being attained at α2 = V. 7 Solved problems 53 the minimum being attained at α2 = 2V. 66) Finally, we have R1 V, 1 2 = = min α∈[−2V,4V ] EP U V + ⎡⎛ αμ2 1 + μ2 , 2 2 ⎢⎜ 1 + μ2 ⎜ −1 EP ⎢ ⎣⎝ 2 α∈[−2V,4V ] min ⎞+ ⎤ + − V + ⎥ αμ2 ⎟ ⎟ ⎥ = 0, ⎠ ⎦ 2 ≥0 =0 and the minimum is attained at any point α2 ∈ [−2V, 4V ].

3 7 75 5 Similarly, in the scenario S11 = m1 we get 4 α21 = − , 3 α22 = 27 , 14 β2 = 56 , 75 and finally, in the scenario S11 = d1 , we have α21 = 0, α22 = 0, β2 = 16 . 44. 42) for which i (1 + μi (hn )), Sni = Sn−1 n = 1, . . d. with values in {1, 2, 3} and ⎧ ⎪ ⎨ui i 1 + μ (h) = mi ⎪ ⎩ di i = 1, 2, if h = 1, if h = 2, if h = 3. 1 1 1 Choosing u1 = 73 , u2 = 22 9 , m1 = m2 = 1, d1 = 2 , d2 = 3 and r = 2 , we have that the unique equivalent martingale measure Q is defined by Q (hn = 1) = q1 = 1 , 2 Q (hn = 2) = q2 = 1 , 6 Q (hn = 3) = q3 = 1 .

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