Is there a quick way to see why this claim $C(S, t)$ on $S$ does not satisfy the Black-Scholes PDE?

Question

I'm self-studying for an actuarial exam on financial economics and encountered the below practice exam problem.

An exam problem should typically take 5-6 minutes to complete, so I'm wondering if there is a "quick" way to confirm that answer choice (D) does not satisfy the Black-Scholes PDE.

Assuming for the moment that $C(S, t)$ does not pay dividends (which in my opinion cannot be assumed just from the information provided), the PDE implies that $r = 0.04$, $\delta = 0.02$ and $\sigma = 0.3$.

So I would think that any asset that has these parameters will satisfy the PDE. Let's check:

(A) is the price of a risk-free bond with maturity value 1.

(B) is the price of a cash-or-nothing call that pays 1 when the stock price is above 100.

(C) is the price of a cash-or-nothing put that pays 1 when the stock is below 100.

(E) is the price of an asset-or-nothing put that pays the stock when the stock price is below 100.

By elimination, that leaves (D) as the claim that does not satisfy the PDE.

But what if I wanted to show that (D) cannot satisfy the PDE? I can only think to find $C_s$, $C_{ss}$ and $C_t$. However, this would be messy as $C(S, t)$ would require differentiating $N(d_1)$. Is there an quicker or better way of convincing myself that (D) cannot satisfy the equation?

Answer 1

You already associated the valuation function ins A, B, C and E with the corresponding products. In order to explicitly exclude D, you don't have to compute all the derivatives but just note that

\begin{eqnarray} C_S & = & e^{-0.02 (T - t)} \mathcal{N}' \left( d_1 \right) \frac{\partial d_1}{\partial S}\\ C_t & = & 0.02 \underbrace{e^{-0.02 (T - t)} \mathcal{N} \left( d_1 \right)}_{=C(S, t)} + e^{-0.02 (T - t)} \mathcal{N}' \left( d_1 \right) \frac{\partial d_1}{\partial t}. \end{eqnarray}

From the expression for $C_S$ you can infer that the expression for $C_{SS}$ does not contain a $C(S, t)$-term. Thus, you can conclude that the $C(S, t)$-terms in the PDE coming from $C_t$ and the r.h.s. don't cancel each other out and you are done.