Background Information:

The price of a portfolio at time $t$ ($t = 0 ,1$) is $$V_t(\pi) = \phi S_t + \psi B_t$$ The portfolio $\pi$ is a perfect hedge for the claim $X$ if $V_1(\pi) = X$ a.s. as random variables.

Given claim $X$, to have a perfect hedge then $\phi$ and $\psi$ must satisfy \begin{equation} \phi S_1(u) + \psi B_1 = X(u) \end{equation} \begin{equation} \phi S_1(d) + \psi B_2 = X(d) \end{equation} We have solving for this that $$\phi = \frac{X(u) - X(d)}{S_1(u) - S_1(d)}$$ $$\psi = B_1^{-1}(X(u) - \phi S_1(u)) = B_1^{-1}(X(d) - S_1(d))$$ Thus the resulting value of $X$ at $t=0$ is $$V_0(X) = V_0(\pi) = \phi S_0 + \psi B_0$$


Let $\beta = B_0 B_1^{-1}$ be the discount factor. Show that

$$V_0(X) = \beta\left[\left(\frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)}\right)X(u) + \left(\frac{S_1(u) - \beta^{-1}S_0}{S_1(u) - S_1(d)}\right)X(d)\right]$$

I have tried this three times now and I still not getting the result what I do is this if you want to see my attempted work let me know. Otherwise it would be great if someone could give me a good start to this. There must be something that I am not seeing in regards to some algebra trick.


1 Answer 1


Using the values for $\phi$ and $\psi$ that you have derived, \begin{align*} V_0(X) &= \phi S_0 + \psi B_0\\ &= \frac{X(u) - X(d)}{S_1(u) - S_1(d)} S_0 + B_1^{-1}\left(X(u) - \frac{X(u) - X(d)}{S_1(u) - S_1(d)}S_1(u)\right) B_0\\ &=\beta\left(\frac{X(u) - X(d)}{S_1(u) - S_1(d)} \beta^{-1}S_0 + \frac{X(d) S_1(u) - X(u)S_1(d)}{S_1(u) - S_1(d)} \right)\\ &=\beta\left(\frac{\beta^{-1}S_0 - S_1(d)}{S_1(u) - S_1(d)} X(u) + \frac{S_1(u) - \beta^{-1}S_0 }{S_1(u) - S_1(d)} X(d)\right). \end{align*} What you need is to combine terms with $X(u)$ together, and likewise for terms with $X(d)$.

  • $\begingroup$ I am a little lost on your second to last step to the last step, could you add a little more detail in? $\endgroup$
    – Wolfy
    Commented Oct 11, 2016 at 20:24
  • $\begingroup$ try to put all $X(u)$ terms together and all $X(d)$ terms together and see what you can get. It is just algebra, please try. $\endgroup$
    – Gordon
    Commented Oct 11, 2016 at 20:27
  • $\begingroup$ sorry to keep bothering you but I have a new question that is tied to this one $\endgroup$
    – Wolfy
    Commented Oct 11, 2016 at 22:14

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