# Attempt of an analytical proof that a call price decreases as its strike increases

I'm stuck trying to analytically prove that a partial derivative of a specific, lower defined function $C$ is negative. The context of this problem is actually a Black-Scholes market situation, where a price of a call option decreases as its strike increases.

For given positive constants $S, K, r, \sigma$ and $T$, we have: $$C(S,K,r, \sigma,T)=S \Phi(d_1)-Ke^{-rT}\Phi(d_2),$$ where $$d_1=\frac{\ln \frac{S}{K}+(r+\frac{1}{2}\sigma^2)T}{\sigma \sqrt{T}},$$ $$d_2=d_1-\sigma \sqrt{T}.$$

I have to prove that the function $C$ is decreasing if $K$ is increasing. First, I calculate the partial derivation: \begin{align} \frac{\partial C}{\partial K}&=S\frac{d \Phi(d_1)}{d (d_1)} \frac{\partial d_1}{\partial K}-e^{-rT}\Phi(d_2)-Ke^{-rT}\frac{d \Phi(d_2)}{d (d_2)}\frac{\partial d_2}{\partial K}\\ & = S \varphi(d_1)\frac{K}{\sigma \sqrt{T}}-e^{-rT}\Phi(d_2)-Ke^{-rT}\varphi(d_2)\frac{K}{\sigma \sqrt{T}}\\ & = K^2 \left( -e^{-rT} \frac{\varphi(d_2)}{\sigma \sqrt{T}} \right) + K \left( S \frac{\varphi(d_1)}{\sigma \sqrt{T}}\right) -e^{-rT}\Phi(d_2). \end{align} where $\Phi(x)$ is the standard normal cumulative, and $\varphi(x)$ standard normal density function.

This is more or less what I've got. I understand that I should somehow prove that the last expression is always non-negative, so I've tried calculating the determinant of the quadratic function of $K$, and I got $$D= \frac{S^2 (\varphi(d_1)^2)}{\sigma^2 T}-4 \frac{e^{-2rT }\varphi(d_2)\Phi(d_2)}{\sigma \sqrt{T}}.$$

And now I should prove that it is positive. No idea how? There is also a chance I misunderstood something and am leaving out some necessary conditions, I'm not sure.

Thanks for an insights on this, I really appreciate it.

• I have in my notes that $\frac{\partial C}{\partial K}= -e^{-r T} N(d_2)$ which is a heck of a lot simpler than what you have – noob2 Jun 13 '17 at 17:54
• Even easier still, if you write the value of the option as $\int_k^\infty \phi (s) (s-k) \mathrm{d}s$ then it's pretty easy to show. – will Jun 13 '17 at 21:12

Something went wrong in the third equality of the equation where you compute $\partial C_0 / \partial K$. Starting from the second equality, you can use that

\begin{equation} S_0 \mathcal{N}' \left( d_1 \right) = K e^{-r T} \mathcal{N}' \left( d_2 \right), \end{equation}

see e.g. Equation (1.29) in Wystup (2006). Alternatively, you could use the homogeneity result

\begin{equation} C_0 = S_0 \frac{\partial C_0}{\partial S_0} + K \frac{\partial C_0}{\partial K} \end{equation}.

see Equation (1.36) in Wystup (2006). This immediately yields the result as

\begin{equation} \frac{\partial C_0}{\partial K} = \frac{C_0 - S_0 \partial C_0 / \partial S_0}{K} = -e^{-r T} \mathcal{N} \left( d_2 \right). \end{equation}

The homogeneity result actually holds for all models with constant returns to scale, not just geometric Brownian motion, see Theorem 9 in Merton (1973).

Yet another approach to show that $\partial C_0 / \partial K < 0$ in a model-free setting is to note that the portfolio which is long a call with strike $K + \Delta$ and short a call with strike $K$ has a payoff equal to

\begin{equation} C_T = \begin{cases} -\Delta < 0 & \text{if } S_T > K + \Delta\\ K - S_T < 0 & \text{if } K + \Delta \geq S_T > K\\ 0 & \text{otherwise} \end{cases}. \end{equation}

Since the portfolio payoff is non-positive everywhere but strictly negative for some $S_T$, its initial value $C_0$ must be strictly negative if $\mathbb{P} \left\{ S_T > K \right\} > 0$. Now divide by $\Delta$, take the limit as $\Delta \downarrow 0$ and you have

\begin{equation} \frac{\partial C_0}{\partial K} = \lim_{\Delta \downarrow 0} \frac{C_0(K + \Delta) - C_0(K)}{\Delta} < 0. \end{equation}

References

Merton, Robert C. (1973) "Theory of Rational Option Pricing," Bell Journal of Economics and Management Science, Vol. 4, No. 1, pp. 141-183

Wystup, Uwe (2006) FX Options and Structured Products, Wiley Finance

• $S \mathcal{N}' \left( d_1 \right) - K e^{-r T} \mathcal{N}' \left( d_2 \right) = 0$ is very useful to know, comes up repeatedly – noob2 Jun 13 '17 at 18:39

Ok, I think I've figured it out.

\begin{align} \frac{\partial C}{\partial K}& = S \varphi(d_1)\frac{K}{\sigma \sqrt{T}}-e^{-rT}\Phi(d_2)-Ke^{-rT}\varphi(d_2)\frac{K}{\sigma \sqrt{T}}\\ & = -e^{-rT}\Phi(d_2)+\frac{K}{\sigma \sqrt{T}} \left[ S \varphi(d_1)-K e^{-rT} \varphi(d_2) \right] \\ & = -e^{-rT}\Phi(d_2)+\frac{K}{\sigma \sqrt{T}} \left[ e^{\ln S} \frac{1}{\sqrt{2 \pi}}e^{-\frac{d_1^2}{2}}-e^{\ln K} e^{-rT} \frac{1}{\sqrt{2 \pi}}e^{-\frac{d_2^2}{2}}\right] \\ & = -e^{-rT}\Phi(d_2)+\frac{K}{\sigma \sqrt{2\pi T}} \left[ e^{\ln S-\frac{d_1^2}{2}}- e^{\ln K-rT-\frac{d_2^2}{2}}\right] . \end{align} Now, we can try proving that \begin{align} \ln S-\frac{d_1^2}{2} & \stackrel{?}{=} \ln K-rT-\frac{d_2^2}{2} \\ \ln S-\frac{d_1^2}{2} & \stackrel{?}{=} \ln K-rT-\frac{(d_1-\sigma \sqrt{T})^2}{2} \\ \ln S-\frac{d_1^2}{2} & \stackrel{?}{=} \ln K-rT-\frac{(d_1^2-2d_1\sigma \sqrt{T}+\sigma^2 T)}{2} \\ \ln \frac{S}{K} & \stackrel{?}{=} -rT + d_1 \sigma \sqrt{T}-\frac{\sigma^2 T}{2} \\ \ln \frac{S}{K} +(r+\frac{\sigma^2 }{2})T & \stackrel{?}{=} d_1 \sigma \sqrt{T} \\ d_1 & = \frac{\ln \frac{S}{K} +(r+\frac{\sigma^2 }{2})T}{\sigma \sqrt{T}}. \end{align} Thus we are left with $$\frac{\partial C}{\partial K} = -e^{-rT}\Phi(d_2),$$ implying that the partial derivative is always negative.