# 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 $$t), using "R" (risk-neutral) probabilities. I tried to solve by doing: $$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.

Thanks in advance to whoever can provide some assistance.

• 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 – Makina May 19 '19 at 11:54
• 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. – BorisD May 19 '19 at 12:40
• $E\left(e^{-rT_2}(S_{T_1}-K)^+\right) = e^{-r(T_2-T_1)}E\left(e^{-rT_1}(S_{T_1}-K)^+\right)$. – Gordon May 19 '19 at 12:41
• This is still unclear. The payoff formula does not use S(T2), as claimed in the text. – dm63 May 19 '19 at 13:19
• Sorry. Fixed. Payoff formula does not use $S_{T_2}$. – BorisD May 19 '19 at 13:21

At time time $$T_2$$ the holder receives $$X=(S_{T_1}-K)^+$$. According to Risk Neutral Valuation the value at time $$t$$ $$(t is $$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 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: $$V_t=e^{-r(T_2-1)}(S_{T_1} - K )^+$$
• 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 $T_1<T_2$). – Daneel Olivaw May 19 '19 at 14:15