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.

  • $\begingroup$ What is going on with first line (double equal sign?) and 3rd line (how did you just factor $S_t$ but get $S_{T_2}^{-1}$ inside?). Use 1_{} for indicators $\endgroup$
    – Makina
    May 19, 2019 at 11:54
  • $\begingroup$ Fixed line 1. Thanks! In the third line, instead of using bonds to get the neutral-risk probabilities in order to calculate the replicating portfolio ($V_t$), I use an asset (in this case, the underlying asset). This is: $$ Vt=S_t * E_{R} [X * S_{T} | F_{t}] $$ Where $X$ is the function of the derivative, $t$ is any time of valuation prior to the expiration date, and $T$ is the time of expiration. $\endgroup$
    – BorisD
    May 19, 2019 at 12:40
  • 3
    $\begingroup$ $E\left(e^{-rT_2}(S_{T_1}-K)^+\right) = e^{-r(T_2-T_1)}E\left(e^{-rT_1}(S_{T_1}-K)^+\right)$. $\endgroup$
    – Gordon
    May 19, 2019 at 12:41
  • $\begingroup$ This is still unclear. The payoff formula does not use S(T2), as claimed in the text. $\endgroup$
    – dm63
    May 19, 2019 at 13:19
  • $\begingroup$ Sorry. Fixed. Payoff formula does not use $S_{T_2}$. $\endgroup$
    – BorisD
    May 19, 2019 at 13:21

1 Answer 1


@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: $V_t=e^{-r(T_2-1)}(S_{T_1} - K )^+$

  • 1
    $\begingroup$ You've missed a minus sign in front of $r$, which you can also see by the fact that if $r>0$ then by your formula $V_t>C_{BS}(S_t,t;T_1)$ which is inconsistent with positive rates as it's always preferable to receive a payment at $T_1$ than at $T_2>T_1$ (as you can earn the rate $r$ between $T_1$ and $T_2$ if you've been paid at $T_1<T_2$). $\endgroup$ May 19, 2019 at 14:15

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