3
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ 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 $\endgroup$
    – nbbo2
    Jun 13, 2017 at 17:54
  • 3
    $\begingroup$ 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. $\endgroup$
    – will
    Jun 13, 2017 at 21:12

3 Answers 3

5
$\begingroup$

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

$\endgroup$
1
  • 3
    $\begingroup$ $S \mathcal{N}' \left( d_1 \right) - K e^{-r T} \mathcal{N}' \left( d_2 \right) = 0$ is very useful to know, comes up repeatedly $\endgroup$
    – nbbo2
    Jun 13, 2017 at 18:39
4
$\begingroup$

One interesting property among the variables in the Black-Scholes formula is $$ S_0 \varphi(d_1) = K e^{-rT} \varphi(d_2), $$ where $\varphi(x) = \Phi'(x)$ is the normal distribution PDF.

This is because $$ d_1^2 - d_2^2 = (A+B)^2 - (A-B)^2 = 4AB = 2\log(S_0\,e^{rT}/K) $$ where $$A = \frac{\log(S_0\,e^{rT}/K)}{\sigma\sqrt{T}} \quad\text{and}\quad B = \frac{\sigma\sqrt{T}}{2}.$$

So the last line of @Milan's derivation is simplified to $$ \frac{\partial C}{\partial K} = K^2 \left( -e^{-rT} \frac{\varphi(d_2)}{\sigma \sqrt{T}} \right) + K \left( S \frac{\varphi(d_1)}{\sigma \sqrt{T}}\right) -e^{-rT}\Phi(d_2) = -e^{-rT}\Phi(d_2), $$ which is negative.

BTW, this property is also useful in the derivation of the Black-Scholes delta. See my answer.

$\endgroup$
1
  • $\begingroup$ Oops, I realize that @LocalVolatility already mentioned the same property from Equation (1.29) in Wystup (2006). Consider my answer a proof for the property. $\endgroup$
    – Najee
    Jun 5, 2022 at 14:20
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.