The following is the Black Scholes equation for a function $V(S,t)$:

$$\frac{∂V}{∂t} +\frac{1}{2}σ^2S^2\frac{∂^2V}{∂S^2}+ (r − y) S\frac{∂V}{∂S} − r V = 0$$

The question asks me to show that if $V(S,t)$ is a solution then so is $V(bS,t)$, where $b>0$. The equation holds $\forall S>0, t<T$, so substituting in $K=bS$ gives still a valid equation from which the result follows.

However, my problem is the following: If we set $bS=K$, then $V(bS,t)=V(K,t)$ and we'll have $\frac{\partial (V(K,t))}{\partial K}=\frac{\partial (V(S,t))}{\partial S}$ (similar results for the second partial derivative) and plugging that in doesn't results in an equality since $K\neq S$. Where is my mistake?


Remember that you are supposed to show that: $$\frac{\partial}{\partial t}V(bS,t)+\frac{1}{2}\sigma^2S^2\frac{\partial^2 }{\partial S^2}V(bS,t)+(r-y)S\frac{\partial }{\partial S}V(bS,t)-rV(bS,t)=0$$ So you need to figure out how to write $\partial_S V(bS,t)$ in terms of $K$, and not $\partial_S V(S,t)$ as you have done. But you can do this using the chain rule: $\partial_S V(bS,t) = \partial_K V(K,t) \cdot \partial_S K=b\partial_KV(K,t)$. Doing this for the entire equation yields:

$$\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2S^2b^2\frac{\partial^2 V}{\partial K^2}+(r-y)Sb\frac{\partial V}{\partial K} - rV=\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2K^2\frac{\partial^2 V}{\partial K^2}+(r-y)K\frac{\partial V}{\partial K}-rV$$ This is the original Black Scholes equation, for which we know $V(K,t)$ is a solution. So: \begin{align} \frac{\partial}{\partial t}V(bS,t)+\frac{1}{2}\sigma^2S^2\frac{\partial^2 }{\partial S^2}V(bS,t)+&(r-y)S\frac{\partial }{\partial S}V(bS,t)-rV(bS,t)\\ &=\frac{\partial V(K,t)}{\partial t}+\frac{1}{2}\sigma^2K^2\frac{\partial^2 V(K,t)}{\partial K^2}+(r-y)K\frac{\partial V(K,t)}{\partial K}-rV(K,t)\\ &= 0 \end{align}

  • $\begingroup$ Don't I need to have the partial derivative with respect to $bS$, rather than $S$ since $V$ is now a function of $bS$? $\endgroup$ – asdf Feb 21 '18 at 13:05
  • $\begingroup$ @asdf You want to show that $V(bS,t)$ is a solution of the Black-Scholes equation if $V(S,t)$ is a solution. In other words, if you plug in $V(bS,t)$ instead of $V(S,t)$, the left hand side should still end up being equal to zero. In the second equation line, I have rewritten the equation in terms of partial derivatives with respect to $bS$, so I am not sure what you are referring to. $\endgroup$ – bames Feb 21 '18 at 13:08
  • $\begingroup$ I was thinking that the BS equation is changing with respect to the first argument - that is if we have $V(S,t)$ then we'll have to take partial derivatives with respect to $S$, while if we have $V(bS,t)$ then the partial derivative should be $bS$, that is we should have $(bS)\frac{\partial V}{\partial (bS)}$, rather than $S\frac{\partial V}{\partial S}$ $\endgroup$ – asdf Feb 21 '18 at 13:10
  • $\begingroup$ @asdf No, the BS equation is the same. You need to show that the new function $V(bS,t)$ satisfies the old BS equation. $\endgroup$ – bames Feb 21 '18 at 13:12
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    $\begingroup$ I see, thanks a lot! $\endgroup$ – asdf Feb 21 '18 at 13:12

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