# Verifying value of claim as an expectation

Background:

We have so far taken the bond B to be deterministic for simplicity, but some reflection shows that this is not in any way necessary. Everything works out the same way with a stochastic bond $B_1(u) \neq B_1(d)$ (except the algebra takes a little more work), as we now describe. The equations defining the hedging portfolio now become $$\phi S(u) + \psi B(u) = X(u) \ \ \ \ \ (1.11)$$ $$\phi S(d) + \psi B(d) = X(d) \ \ \ \ \ (1.12)$$ where we have temporarily dropped the subscript “1" on $S_1$ and $B_1$ for convenience. Assuming that the determinant $\Delta = S(u)B(d) − S(d)B(u)$ is nonzero, the unique solution is $$\phi = \frac{B(d)X(u) − B(u)X(d)}{\Delta} \ \ \ \ (1.13)$$

$$\psi = \frac{S(u)X(d) − S(d)X(u)}{\Delta} \ \ \ \ (1.14)$$ The claim value is again forced by the assumption of no arbitrage to be $$V_0(X) = \phi S_0 + \psi B_0$$ If we define the discount factor (now stochastic) as $\beta(\cdot) = B_0/B(\cdot)$, we obtain $$V_0(X) = E_Q[\beta X] \ \ \ \ \ (1.15)$$ where $Q$ is a candidate probability measure defined by $$Q(u) = \frac{−B(u)S(d) + B(u)B(d)(S0/B0)}{\Delta} \ \ \ \ \ (1.16)$$ $$Q(d) = \frac{−B(d)B(u)(S0/B0) + S(u)B(d)}{\Delta} \ \ \ \ \ (1.17)$$

Question:

Verify formulas (1.15),(1.16), and (1.17)

Partial solution:

We have

\begin{align*} V_0(X) = \phi S_0 + \psi B_0 &= \left(\frac{B(d)X(u) − B(u)X(d)}{\Delta}\right)S_0 + \left(\frac{S(u)X(d) − S(d)X(u)}{\Delta}\right)B_0\\ &= \frac{B(d)X(u)S_0 - B(u)X(d)S_0 + S(u)X(d)B_0 - S(d)X(u)B_0}{\Delta}\\ &= \frac{(B(d)S_0 - S(d)B_0)X(u) + (S(u)B_0 - B(u)S_0)X(d)}{\Delta}\\ &= \left(\frac{B(d)S_0 - S(d)B_0}{\Delta}\right)X(u) + \left(\frac{S(u)B_0 - B(u)S_0}{\Delta}\right)X(d) \end{align*}

I believe this is where I need to incorporate the $\beta(\cdot)$ discount factor but I am really sure how to do that, I don't really understand the discount factor. Any suggestions is greatly appreciated.

You are almost there. Note that, to have an expectation of the form $(1.15)$, you need to treat $X\beta$ as a random variable together. That is, the respective probabilities $Q(u)$ and $Q(d)$ should apply to the corresponding realizations $X(u)\beta(u)$ and $X(d)\beta(d)$. Specifically, continuing from your last step, \begin{align*} V_0(X) &=\left(\frac{B(d)S_0 - S(d)B_0}{\Delta}\right)X(u) + \left(\frac{S(u)B_0 - B(u)S_0}{\Delta}\right)X(d) \\ &=\left(\frac{B(d)B(u)S_0/B_0 - S(d)B(u)}{\Delta}\right)X(u)\frac{B_0}{B(u)} \\ &\quad+ \left(\frac{S(u)B(d) - B(u)B(d)S_0/B_0}{\Delta}\right)X(d)\frac{B_0}{B(d)}\\ &=Q(u)\big[X(u)\beta(u)\big] + Q(d)\big[X(d)\beta(d)\big], \end{align*} which is $(1.15)$.