# 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?

• Assuming $S(T)$ is the price of the stock at the terminal time $T$, then in your notation, $S=S(T)$, right? – P.Windridge Jun 5 '16 at 12:59
• Yes that is correct S(T) is the stock price at the terminal time T. – Tejay Lovelock Jun 5 '16 at 13:02
• So the terminal condition is $(S_T - K)^+$ ... or $(S_T^2 - K)^+$ for the option on the squared price. – P.Windridge Jun 5 '16 at 13:17
• Wish I knew how to write the math code like you, yes ordinarily the terminal condition is as you stated. But I have been asked to show that the PDE is the same (** and therefore I presume the option price is also the same **) for the terminal condition: (ST^2 - K)+ Which I believe would compare as follows at ST For example, if K is 50 and ST ends up been 100; (ST - K) = $50, where as. (ST^2 - K) = 10,000 - 50 =$9,950 So how can both options have the same t value? Naturally I am expecting to be proven drastically wrong with my thinking here... – Tejay Lovelock Jun 5 '16 at 13:21
• Yes the option prices are different. Note that the payoff function $f(S_T)$ (e.g. $f(s) = (s^2 - K)^+$) does appear in the (generalised) BS PDE. Also you can write $\LaTeX$ math equations enclosed in dollar signs. – P.Windridge Jun 5 '16 at 21:07

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
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}