I am self-studying and encountered the following problem:

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My idea was to calculate the price of the put using a replicating portfolio, then use the formula:

$$Pe^{\gamma h} = S\Delta e^{\alpha h} + \beta e^{rh}$$ to solve for $\gamma$, where $P$ is the put premium, $\alpha$ is the continuously compounded return on the stock, $\beta$ is the amount lent in the replicating portfolio, and $\gamma$ is the continuously compounded return on the option.

In this case $$\Delta = \frac{P_u - P_d}{S(u - d)}e^{-\delta h} = \frac{0 - 11.84485}{60.41285 - 33.15522}e^{0\cdot1} = -0.4345506$$ and

$$\beta = \frac{uP_d - dP_u}{u - d}e^{-rh} = \frac{1.40495(11.84485) - 0.77105(0)}{1.40495 - 0.77105}e^{-0.04\cdot1} = 26.22307,$$

giving a put premium of $$P = \Delta\cdot{}S + \beta = -0.4345506\cdot43 + 26.22307 = 7.53739.$$

Since I did not arrive at the same put premium as the textbook, I stopped here. I'm not sure where I am making my mistake at.

I know my formula for $\beta$ is correct because:

A successful replicating portfolio must satisfy: $P_d = \Delta S_d e^{\delta h} + \beta e^{rh}$ and $P_u = \Delta S_u e^{\delta h} + \beta e^{rh}$.

Then $\Delta = \frac{(P_d - \beta e^{rh})}{S_d}e^{-\delta h}$ and $\Delta = \frac{(P_u - \beta e^{rh})}{S_u}e^{-\delta h}$.

Therefore $(P_d - \beta e^{rh})e^{-\delta h} S_u = (P_u - \beta e^{rh})e^{-\delta h} S_d$.

Noting that $S_u = S_0\cdot u$ and $S_d = S_0 \cdot d$, we can eliminate $S_0$ and write

$P_d e^{-\delta h} u - \beta e^{rh}e^{-\delta h} u = P_u e^{-\delta h}d - \beta e^{rh - \delta h} d$

This implies that $P_d u e^{-\delta h} - P_u d e^{-\delta h} = \beta(e^{rh}e^{-\delta h}u - e^{rh}e^{-\delta h}d)$.

Hence $\beta = \frac{(P_d u - P_u d)e^{-\delta h}}{(u - d)e^{rh}e^{-\delta h}} = \frac{P_d u - P_u d}{u - d}e^{-rh}$

  • $\begingroup$ What is $\delta$ in the formula for $\Delta$? $\endgroup$ Commented Sep 13, 2016 at 22:47
  • $\begingroup$ $\delta$ is the continuously compounded dividend yield. $\endgroup$ Commented Sep 15, 2016 at 14:57

1 Answer 1


Your computation of $\Delta$ is correct. However, your computation of the cash amount is wrong. You choose the cash amount $\beta$ that you need to initially lend or borrow such that in the up state, the following holds

\begin{equation} P_u = \Delta S_u e^{\delta h} + \beta e^{r h}. \end{equation}

We get

\begin{eqnarray} \beta & = & \left( P_u - \Delta S_u e^{\delta h} \right) e^{-r h}\\ & = & 0.4345506 \cdot 60.41285 \cdot e^{-0.04}\\ & = & 25.223067. \end{eqnarray}


\begin{equation} P = \Delta S + \beta = 6.537391 \end{equation}

  • $\begingroup$ I understand your solution and why it provides the correct $\beta$. One thing that I still don't understand is why my formula, $$\beta = \frac{uP_d - dP_u}{u - d}e^{-rh}$$ does not apply here and give the same $\beta$ that you calculated. That formula is what the textbook provided. $\endgroup$ Commented Sep 15, 2016 at 14:58
  • $\begingroup$ I showed you how to obtain the correct expression for $\beta$ and your formula for it looks different. So the question is not really why your formula doesn't apply but rather how you obtained it since it seems wrong. $\endgroup$ Commented Sep 15, 2016 at 15:00
  • $\begingroup$ Do you understand how the textbook derived it? Is it in a different context / sure it applies here? Don't just take "formulas from the textbook" for granted. $\endgroup$ Commented Sep 15, 2016 at 16:38
  • $\begingroup$ The value of the replicating formula at time $h$, with stock price $S_h$, is $\Delta S_h + e^{rh}\beta$. At the prices $S_h = S_d$ and $S_h = S_u$, a replicating portfolio must satisfy: $\Delta \cdot S_d \cdot e^{\delta h} + (\beta \cdot e^{rh}) = P_d$ and $\Delta \cdot S_u \cdot e^{\delta h} + (\beta \cdot e^{rh}) = P_u$. The two unknowns $\Delta$ and $\beta$ allow us to solve for $$\beta = e^{-rh} \cdot \frac{u P_d - d P_u}{u - d}.$$ $\endgroup$ Commented Sep 16, 2016 at 0:47
  • 1
    $\begingroup$ You are right - my mistake. But at least now I finally spotted what went wrong in the first place. You have the correct formula for $\beta$ but forget to apply discounting when computing the value - i.e. you don't take into account the $e^{-r h}$ term. $\endgroup$ Commented Sep 16, 2016 at 15:17

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