Why Black-Scholes Values Depend on the Terminal Payoff

# Black-Scholes PDE & Terminal Condition


# Black-Scholes PDE & Terminal Condition












Just a quick question I was hoping someone could shed light on.

- So far I am familiar with the Black-Scholes PDE with the terminal condition at time $T$ been $V(t=T,S)=(S-K)^+$.

- I also understand that the Black-Scholes PDE does not contain $S(T)$ and therefore is independent of the terminal condition.

As such, if the terminal condition was to be $(S^2 - K)^+$ the PDE for the call option remains the same - at least that is what I am told.

Intuitively I don't understand the logic behind this?

For example, if $K$ is 50 and $S$ ends up been 100;

- $(S - K) = \$50$, where as.

- $(S^2 - K) = 10,000 - 50 = \$9,950$

Surely the $(S^2 - K)+$ option must be worth a lot more?

But apparently the PDE for both these options is the same and therefore the time $t$ value is also the same?

Could anyone please explain?

## Answer by aw80 (score 2, accepted)

https://quant.stackexchange.com/a/27445

The PDE will be the same but because the terminal condition is different the solutions will not be the same. The different boundary condition will give different values at $t=T$. Then the equation is marched backwards in time in both cases using the same equation but because the terminal condition is different the solutions will not agree

## Answer by user16651 (score 0)

https://quant.stackexchange.com/a/27448

It is reasonable and PDE approach is not suitable. In the Black scholes model we have $$d\ln {{S}_{T}}=\,(r-\frac{1}{2}{{\sigma }^{2}})dt+\sigma d{{W}_{t}}$$ so $$d\ln {{S}_{T}^2}=\,(2r-{{\sigma }^{2}})dt+2\sigma d{{W}_{t}}$$ as a result $$\ln {{S}_{T}^2}=\ln{{S}_{t}^2}\,+(2r-{{\sigma }^{2}})(T-t)+2\sigma (W_T-W_t)$$ let $$Y(t)=(2r-{{\sigma }^{2}})(T-t)+2\sigma [W(T)-W(t)]$$ it is clear \begin{align} & \,\,\,\,{{E}^{\mathbb{Q}}}\left[ Y(t) \right]=(2r-{{\sigma }^{2}})(T-t) \\ & {{\operatorname{var}}^{\mathbb{Q}}}\left[ Y(t) \right]=4{{\sigma }^{2}}(T-t) \\ \end{align} in the other words we can say $$Y\overset{d}{\mathop{=}}\,\ N\left( (2r-{{\sigma }^{2}})(T-t)\,,4{{\sigma }^{2}}\left( T-t \right) \right)$$ For simplicity we let ${\widetilde{r}}=(2r-{{\sigma }^{2}})$ and $\tau =(T-t)$ so $Y\overset{d}{\mathop{=}}\,\ N\,(\overset{\tilde{\ }}{\mathop{r}}\,\tau \,,{4{\sigma }^{2}}\tau )$ as a result

$$\frac{Y-\overset{\tilde{\ }}{\mathop{r}}\,\tau }{2\sigma \sqrt{\tau }}\overset{\,\,d}{\mathop{\,=}}\,\ N(0,\,1)$$

we have $$\Pi (t)=e^{-{r}\tau}{{E}_{t}}^{Q}\left[ {{({{S}_{T}}^{2}-K)}^{+}} \right]=e^{-{r}\tau}{{E}_{t}}^{Q}\left[ {{({{S}_{t}}^{2}{{e}^{{{Y}_{t}}}}-K)}^{+}} \right] $$ $$\Pi (t)=e^{-{r}\tau}\int\limits_{-\infty }^{\infty }{{{({{S}_{t}}^{2}\,{{e}^{y}}-K,\,0)}^{+}}\,{{f}_{Y}}}(y)\,dy$$ Finally we should do same procedure with $(S_T-K)^+$, then $$\Pi (t)={{S}_{t}}{{\,}^{2}}N\left[ {{d}_{1}} \right]-K{{e}^{-r\,\tau}}\ N\left[ {{d}_{2}} \right]$$ where \begin{align} & {{d}_{1}}=\frac{(2r+3{{\sigma }^{2}})\tau +\ln \left( \frac{{{S}_{t}}^{2}}{K} \right)}{2\sigma \sqrt{\tau }} \\ & {{d}_{2}}={{d}_{1}}-2\sigma \sqrt{\tau } \\ \end{align}