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The following is taken from Mark Joshi's Concepts and Practice of Mathematical Finance, second edition, exercise $5.6$.

Question: Show that $Ae^{rt}$ is a solution of the Black-Scholes equation. Why should this be so?

Recall that the Black-Scholes equation is $$\frac{\partial V}{\partial t} + rS\frac{\partial V}{\partial S} + \frac{1}{2} S^2\sigma^2 \frac{\partial^2 V}{\partial S^2} = rV$$ where $V=V(t,S_t)$ is either European call or put option value, $r$ is risk-free interest rate and $\sigma$ is volatility.

It can be verified easily that $Ae^{rt}$ satisfies the equation above. In terms of explanation on why this is so, I guess we need to concoct a European option whose payoff is $Ae^{rt}$ to justify it.

Since $S_t$ follows geometric Brownian motion with respect to risk-neutral probability measure, so $$S_t = S_0 e^{(r-\frac{1}{2}\sigma^2)t + \sigma W_t}.$$ I think in this case, we take $\sigma = 0$ to obtain that $$S_t = S_0 e^{r t}.$$ So $A=S_0.$ Therefore, $V(t,S_t)=S_0 e^{rt} = S_t$ is a European call option with zero strike price. This justifies why $Ae^{rt}$ satisfies the Black-Scholes equation.

Is it correct?

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  • $\begingroup$ I believe it’s correct. $\endgroup$ – dm63 Dec 16 '19 at 5:25
  • $\begingroup$ @dm63 Thanks for your comment.Am I right to say that as long as something cab be expressed as European call option with appropriate strike price, then it should satisfy Black-Scholes equation? $\endgroup$ – Idonknow Dec 16 '19 at 7:16
  • $\begingroup$ A "European call option with zero volatility and zero strike" is commonly called a bond (specifically a zero coupon bond). The reason $A e^{r T}$ satisfies the B-S Equation is that a bond is one (trivial) example of a security which can be replicated by a dynamic mixture of stock and bond, stock is another and call option is another. $\endgroup$ – Alex C Dec 16 '19 at 23:51
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While your approach is correct, generally what people would do is that find derivative of the equation for example $V(t,S_t)=Ae^{rt}$. \begin{eqnarray} &\frac{dV}{dt}=rA e^{rt}\\ &\frac{dV}{dS}=0 \\ &\frac{d^2V}{dS^2}=0 \end{eqnarray} Then you plug in the derivatives above to the left hand side of your Black-Scholes equation. Then you will get $rAe^{rt}=rV$ which is equals to the right hand side of your Black-Scholes equation. Therefore, $V(t,S_t)=Ae^{rt}$ is a solution to the Black Scholes equation!

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  • $\begingroup$ Yes, indeed. You answer first part of the question. However, I am more interested in the second part of the question, that is, explain qualitatively why $Ae^{rt}$ satisfies Black-Scholes partial differential equation. $\endgroup$ – Idonknow Dec 16 '19 at 6:36
  • $\begingroup$ Oh, oops my bad for not seeing that. Your answer is correct! $\endgroup$ – Crushh Dec 16 '19 at 6:43
  • $\begingroup$ Thanks. Am I right to say that as long as something cab be expressed as European call option with appropriate strike price, then it should satisfy Black-Scholes equation? $\endgroup$ – Idonknow Dec 16 '19 at 7:16
  • $\begingroup$ Hmmm not quite, Black scholes equation is actually derived from something called "discounted asset prices" where the asset prices follows geometric brownian motion and it can be dynamically perfect hedged. It has nothing to do with call option, unless you set V=C where C is the call price. From what you asked, let me rephrase it, "as long as something can be expressed as European call option where it's underlying asset follows GBM and can be dynamically hedged, then it satisfies the following Black Scholes equation". The key point is actually the GBM :) $\endgroup$ – Crushh Dec 16 '19 at 7:30
  • $\begingroup$ Noted. Thanks a lot. $\endgroup$ – Idonknow Dec 16 '19 at 7:35

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