Risk-Neutral Pricing of a Delayed-Payment Call Payoff
Summary
The document considers a derivative that pays at a later date the positive part of the underlying asset price observed at an earlier date minus a strike. Under risk-neutral valuation with constant rates, the supplied answer prices the claim before the observation date by taking the ordinary call value expiring on that date and discounting it further to the payment date. Once the observation date has passed, the payoff is known, so its value is the known amount discounted to maturity.
The discussion identifies the key timing distinction: the underlying call payoff is determined at the earlier date, while cash is delivered later. The original questioner's proposed formula and parameters are uncertain, and the answer relies on the Black–Scholes call formula and a constant risk-free rate. The displayed formula for the post-observation case appears to contain a typographical error in the discounting interval; the stated logic implies discounting from the payment date back to the current time.
Key ideas
- The payoff is determined by the underlying price at the earlier observation date and paid at the later maturity date.
- Before observation, value the ordinary call payoff at the observation date and discount that value to the payment date.
- After observation, the payoff is measurable and its known value is discounted to maturity.
- The formula assumes risk-neutral valuation with a constant rate and uses Black–Scholes for the call component.
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# Pricing under risk-neutral probabilities for weird derivatives?
# Pricing under risk-neutral probabilities for weird derivatives?
I would really appreciate some help to value a weird derivative that I've found in an assignment:
$$ X=(S_{T_1}-k)^{+} = \max(S_{T_{1}}-k;0) $$
which expires at time $T_{2}$ and uses the price at time $T_{1}$ (therefore $t<T_1<T_2$), using "R" (risk-neutral) probabilities. I tried to solve by doing: $$ V_t=S_t \times E_R [(S_{T_1}-k)\times1_{(S_{T_1}>k)}\times S_{T_2}^{-1} | F_t] $$ where $1_{(S_{T_2}>k)}$ is a function that takes a value of 1 if the condition is met and 0 if it's not, and $F_t$ is the information set at $t$. Solved it assuming $S_t=S_0\times e^{(r+\sigma^{2}/2)\times t+\sigma\times W_t}$ where $W_t$ is a Brownian Motion process, and got the expression:
$$ V_t=S_t \times N(d_1) - k\times N(d_2) $$
where $d_1=\frac{ln(K)+(r+\frac{\sigma2}{2})\times(T_{2}-T_{1})}{\sigma \times \sqrt{T_{2}-T_{1}}}$ and $d_2=\frac{ln(K)+(r+\frac{\sigma2}{2})\times(T_{2}-t)}{\sigma \times \sqrt{T_{2}-t}}$ but I'm not sure this is even close to being correct.
Then I'm asked to price the same derivative under $Q$ (risk neutral probabilities) given that $T_1<t<T_2$.
Thanks in advance to whoever can provide some assistance.
## Answer by Sanjay (score 3)
https://quant.stackexchange.com/a/45688
@Gordon has already given the answer but here is a little more notes to it...
At time time $T_2$ the holder receives $X=(S_{T_1}-K)^+$. According to Risk Neutral Valuation the value at time $t$ $(t<T_1<T_2)$ is $$V_t = e^{-r(T_2-t)}E_t[(S_{T_1}-K)^+] = \\ e^{-r(T_2-t+T_1-T_1)}E_t[(S_{T_1}-K)^+]=\\ e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+] $$
$e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]$ is the value of a Call Option at time $t$ with expiration at time $T_1$. This is simply given by the Black-Sholes formula so $e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=C_{BS}(S_t,t;T_1)$
$$ V_t=e^{-r(T_2-T_1)}e^{-r(T_1-t)}E_t[(S_{T_1}-K)^+]=e^{-r(T_2-T_1)}C_{BS}(S_t,t;T_1) $$
For $T_1<t<T_2$ then $(S_{T_1} - K )^+$ is measurable so $E_t[(S_{T_1} - K )^+]=(S_{T_1} - K )^+$. This means you know exactly what you get and only have to discount the pay-off: $V_t=e^{-r(T_2-1)}(S_{T_1} - K )^+$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.