Black-Scholes方程解的尺度不变性证明：V(bS,t)解验证疑问

Alright, let's walk through this step by step to confirm that if ( V(S,t) ) solves the Black-Scholes equation, then ( V(bS, t) ) (for ( b>0 )) does too. The key here is using the chain rule to compute the partial derivatives of the scaled function, then plugging them back into the original equation.

First, let's restate the original Black-Scholes equation for clarity:
$$\frac{∂V}{∂t} +\frac{1}{2}σ^2S2\frac{∂^2V}{∂S2}+ (r − y) S\frac{∂V}{∂S} − r V = 0$$
This holds for all ( S>0 ) and ( t<T ).

Let's define our scaled function as ( \hat{V}(S,t) = V(bS, t) ). We need to show ( \hat{V} ) satisfies the equation above. To do this, we'll compute each partial derivative of ( \hat{V} ) using the chain rule, then substitute them into the equation.

Step 1: Compute partial derivatives of ( \hat{V} ) #

Let ( K = bS ) (so ( \hat{V}(S,t) = V(K,t) )). Now calculate each derivative:

• Partial derivative with respect to ( t ):
  Since ( t ) isn't modified by the scaling, this is straightforward:
  $$\frac{\partial \hat{V}}{\partial t} = \frac{\partial V}{\partial t}$$
• First partial derivative with respect to ( S ):
  Use the chain rule (derivative of ( V ) with respect to ( K ), times derivative of ( K ) with respect to ( S )):
  $$\frac{\partial \hat{V}}{\partial S} = \frac{\partial V}{\partial K} \cdot \frac{\partial K}{\partial S} = b \cdot \frac{\partial V}{\partial K}$$
• Second partial derivative with respect to ( S ):
  Apply the chain rule again to the first derivative:
  $$\frac{\partial^2 \hat{V}}{\partial S^2} = \frac{\partial}{\partial S} \left( b \cdot \frac{\partial V}{\partial K} \right) = b \cdot \frac{\partial^2 V}{\partial K^2} \cdot \frac{\partial K}{\partial S} = b^2 \cdot \frac{\partial^2 V}{\partial K^2}$$

Step 2: Substitute into the Black-Scholes equation #

Now plug these derivatives into the left-hand side of the original equation for ( \hat{V} ):
$$\frac{\partial \hat{V}}{\partial t} +\frac{1}{2}σ^2S2\frac{\partial^2 \hat{V}}{\partial S^2}+ (r − y) S\frac{\partial \hat{V}}{\partial S} − r \hat{V}$$

Replace each term with the derivatives we calculated:
$$\frac{\partial V}{\partial t} +\frac{1}{2}σ^2S2(b^2 \cdot \frac{\partial^2 V}{\partial K^2})+ (r − y) S(b \cdot \frac{\partial V}{\partial K}) − r V(K,t)$$

Now recall that ( K = bS ), so ( S = K/b ). Substitute ( S = K/b ) into the equation to rewrite everything in terms of ( K ):
$$\frac{\partial V}{\partial t} +\frac{1}{2}σ^{2\left(\frac{K}2}{b^2}\right)(b2 \cdot \frac{\partial^2 V}{\partial K^2})+ (r − y) \left(\frac{K}{b}\right)(b \cdot \frac{\partial V}{\partial K}) − r V(K,t)$$

Step 3: Simplify and verify #

Notice the ( b^2 ) terms cancel in the second term, and the ( b ) terms cancel in the third term. This simplifies to:
$$\frac{\partial V}{\partial t} +\frac{1}{2}σ^2K2\frac{\partial^2 V}{\partial K^2}+ (r − y) K\frac{\partial V}{\partial K} − r V(K,t)$$

But wait—this is exactly the original Black-Scholes equation, but with ( S ) replaced by ( K ). Since the original equation holds for all ( S>0 ), and ( K = bS >0 ) (because ( b>0 ) and ( S>0 )), this expression equals 0.

That means ( \hat{V}(S,t) = V(bS,t) ) satisfies the original Black-Scholes equation!

Addressing your question about ( bS = K ) #

When you set ( bS = K ), ( V(bS,t) = V(K,t) ) is just the original function evaluated at ( K ) instead of ( S ). The key isn't renaming variables—it's showing that when we scale ( S ), the partial derivatives adjust in a way that the equation still balances out. The chain rule lets us translate the derivatives of the scaled function back to the original function's derivatives, which we already know satisfy the equation.

内容的提问来源于stack exchange，提问作者asdf