Deriving Risk-Neutral Claim Valuation with a Stochastic Bond
Summary
This note extends a one-period, two-state replication argument to a bond whose future value differs between the up and down states. Solving the two hedge equations gives stock and bond holdings when the payoff matrix is nonsingular, meaning the two assets have distinct state payoffs relative to one another. The initial claim value is then the cost of that replicating portfolio.
The worked algebra rewrites that cost as a probability-weighted expectation of the payoff multiplied by the state-specific discount factor. The probabilities are obtained from the asset and bond values and match the formulas stated in the source. The explanation clarifies that each state probability weights the product of that state's payoff and discount factor; it does not weight the payoff alone. This is a two-state replication result, conditional on a nonzero determinant and a valid candidate probability measure. The note does not discuss broader multi-period pricing or how to establish the probability measure's validity in general.
Key ideas
- A two-state claim can be replicated using stock and a bond with state-dependent future values.
- The hedge equations have a unique solution when their payoff matrix has a nonzero determinant.
- The claim's initial value equals the cost of its replicating portfolio.
- The expectation applies state probabilities to payoff times the corresponding stochastic discount factor.
Tags
Full text
# Verifying value of claim as an expectation
# 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.
## Answer by Gordon (score 1, accepted)
https://quant.stackexchange.com/a/30590
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 $(1.15)$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.