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

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?

• I believe it’s correct. – dm63 Dec 16 '19 at 5:25
• @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? – Idonknow Dec 16 '19 at 7:16
• 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. – Alex C Dec 16 '19 at 23:51

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!
• 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. – Idonknow Dec 16 '19 at 6:36