Market – Nathan McDonald

Derivatives Markets, McDonald

Thanks in advance for any help, and please tell me if there's anything I can do to make things clearer.

I am having trouble understanding appendix 10.A to Derivatives Markets by Robert L. McDonald. This book is the primary recommended study material for the Society of Actuaries Exam MFE. My own attempts to resolve this issue follow, but any justification for the computations on the above page would be appreciated.

My misunderstanding is probably due to an algebra mistake, but I may be messing up notation or "problem set-up." In addition to the definitions and assumptions given in the above page, I am concluding the following:

(1) $[S_1 -\tau_g (S_1-S_0) + \delta S_1(1-\tau_d)]\Delta \\ + [1+r_h(1-\tau_i)]B\\ = \phi_1 (S_1) - \tau_O [\phi_1 (S_1) - \phi_0 (S_0)]$,

both when ($S_1=S_1^+, \phi_1=\phi_1^+$) and when ($S_1=S_1^-, \phi_1=\phi_1^-$).

(2) $\phi_0 (S_0) = B + \Delta \cdot S_0$.

I am not finding much help in the book in understanding the notation, but I hope that most of it is fairly self-explanatory (with $\phi$ representing an option with an underlying asset whose price is S, and which can take one of two values in period 1).

On the basis of these assumptions, I did the following:

-used Cramer's rule on the pair of equations specified in (1) above (equation 10.25 in the above appendix), solving for $\Delta$ and $B$;

-substituted (2) into the resulting equation for $B$;

-got the following variant of McDonald's conclusion:

$\Delta = \frac{1-\tau_O}{1-\tau_g} \frac{\phi_1^+ (S_1^+)-\phi_1^- (S_1^-)}{S_1^+ - S_1^-} $;

$B=\frac{1}{1+r_h\frac{1-\tau_i}{1-\tau_O}} (\frac{u\phi_1 (S_1^-)-d\phi_1 (S_1^+)}{u-d} - \frac{\Delta}{1-\tau_O} S_0 [\tau_g-\tau_O +\delta(1-\tau_d)])$.

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