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Let $S_u$ be the price of stock in the up-state one period from now. Let $S_d$ be the price of the stock in the down state.

Let $C_u$ be the payoff of a call option at time $1$ in the up-state and similarly for $C_d$.

If we define $\delta = \frac{C_u-C_d}{S_u-S_d}$, then if we purchase $\delta$ shares at time zero and borrow $e^{-r}(\delta S_u - C_u) = e^{-r}(\delta S_d - C_d)$, this portfolio replicates the payoff of the call option.

So at initiation we must have $C(0) = \delta S - e^{-r}(\delta S_u - C_u)$.

substituting for $\delta$ can get us

$$ C = \frac{S-e^{-r}S_d}{S_u-S_d}C_u + \frac{e^{-r}S_u - S}{S_u-S_d}C_d$$

now the author claims in the book I am reading that a "little algebra" quickly implies

$$S = \frac{S-e^{-r}S_d}{S_u-S_d} S_u + \frac{e^{-r}S_u - S}{S_u-S_d} S_d$$ and

$$1 = \frac{S-e^{-r}S_d}{S_u-S_d} e^r + \frac{e^{-r}S_u - S}{S_u-S_d} e^r$$

Now ive been looking at these last two equations for over an hour and I cannot for the life of me see how you algebraically get to them. Can someone help me out?

The above comes from page 12 of "a course in derivative securities" by Kerry Back...to the comment below, I agree, it does seem like a contradiciton but perhaps I am missing something.

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  • $\begingroup$ The last three equations imply that $e^r=\frac{C_u}{C}=\frac{S_u}{S}=\frac{S_d}{S}=\frac{C_d}{C}$ and therefore that $C_u = C_d$ and $S_u = S_d$, which would appear to be contradictions. $\endgroup$ – Attack68 Jan 22 at 16:29
  • $\begingroup$ A very warm welcome to Quant.SE! Could you pls. give the source? $\endgroup$ – vonjd Jan 22 at 17:03
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I think the author is just saying that

\begin{equation} \begin{split} \frac{S - e^{-r}S_d}{S_u - S_d} S_u + \frac{e^{-r}S_u - S}{S_u - S_d} S_d &= \frac{S S_u - e^{-r}S_d S_u + e^{-r}S_uS_d - SS_d}{S_u - S_d} \\ &= \frac{S (S_u - S_d)}{S_u - S_d} \\ &= S \end{split} \end{equation}

and that

\begin{equation} \begin{split} \frac{S - e^{-r}S_d}{S_u - S_d} e^r + \frac{e^{-r}S_u - S}{S_u - S_d} e^r &= \frac{S e^r - S_d + S_u - S e^r}{S_u - S_d} \\ &= \frac{S_u - S_d}{S_u - S_d} \\ &= 1 \end{split} \end{equation}

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