Valuing a Minimum of Current and Future Asset Prices
Summary
The document recasts a claim paying the lower of an asset’s intermediate and maturity prices as the later price minus a call payoff. This decomposition connects the claim to a forward-start option and clarifies how the minimum operation affects valuation. The call component is valued conditionally at the intermediate date, when the asset price is known, then averaged back to the initial date.
Under the answer’s Black-Scholes assumptions, the normalized at-the-money call value scales with the intermediate asset price. The resulting expectation yields a compact expression for the minimum-price payoff in the zero-rate, zero-dividend case. This is a derivation strategy rather than empirical evidence. The answer simplifies by setting rates and dividends to zero and focuses on valuation from time zero; it does not fully reconcile that setup with the question’s stated general-rate formula or explicitly derive the requested value at an arbitrary time.
Key ideas
- The minimum of two asset prices can be expressed as the later price less a call payoff.
- The option component is a forward-start call struck at the intermediate asset price.
- Conditioning on the intermediate price allows the option value to be averaged back to the earlier date.
- The presented result assumes constant volatility and zero interest rates and dividends.
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Full text
# Value of contingent claim at a given time
# Value of contingent claim at a given time
Consider a contingent claim whose value at maturity T is given by
$\min(S_{T_0}, S_T)$
where $T_0$ is some intermediate time before maturity, $T_0 < T$, and $S_T$ and $S_{T_0}$ are the asset price at $T$ and $T_0$, respectively. Assuming the usual Geometric Brownian process for the price of the underlying asset that pays no dividend, show that the value of the contingent claim at time $t$ is given by
$C(T-T_0, s) = S[1-N(d_1)+e^{r(T-T_0)}N(d_2)]$
where S is the asset price at time $t$ and
$d_1 = \frac{\left( r+ \frac{\sigma^{2}}{2} \right)(T-T_0)}{\sigma \sqrt{T-T_0}}, d_2 = d_1 - \sigma \sqrt{T-T_0}$
I asked a similar question like this before but I am confused now when the min function is applied. Any suggestions is greatly appreciated.
## Answer by user34971 (score 8, accepted)
https://quant.stackexchange.com/a/68991
This is related to the valuation of a forward start option. Let's assume Black-Scholes world (constant vol) and for simplicity byt without loss of generality I'll set $r=q=0$.
Notice that $$ \min(S_T,S_t) = S_T - (S_T - S_t)_+ $$ The value of this claim at time $0$ is \begin{align} E_0 \left[ \min(S_T,S_t) \right] &= E_0\left[S_T \right] - E_0 \left[(S_T - S_t)_+\right] \\ &= S_0 - E_0 \left[(S_T - S_t)_+\right] \end{align}
The second term is a Type II ATM forward start option. To value it use the 'conditioning trick': \begin{align} E_0 [\left[(S_T - S_t)_+\right] &= E_0 \left[ E_t \left[(S_T - S_t)_+\right] \right] \\ &= E_0 [C(S_t,S_t,T-t)] \\ &= E_0 [S_t C(1,1,T-t)] \\ &= S_0 C(1,1,T-t) \end{align} with $C(1,1,T-t)$ the Black-Scholes call price function with spot price $1$ and strike $1$.
So, $$ E_0 \left[ \min(S_T,S_t) \right] = S_0 \left(1 - C(1,1,T-t) \right) $$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.