A problem asks whether the following statement is true assuming the Black-Scholes Framework:

The expected return on a call option goes up as the stock price goes up.

The solution is:

The statement is false.

As the stock price increases, the call option becomes less risky, so the expected return on the call option decreases.

As the stock price decreases, the call option becomes more risky, so the expected return on the call option increases.

I was hoping for a more satisfying answer, so I was wondering how I would convince myself of this mathematically.

We know that $\gamma_\text{Call} = \Omega_\text{Call}(\alpha - r) + r$, where $\gamma_\text{Call}$ is the continuously compounded return on the Call and $\Omega_\text{Call}$ is the elasticity of the Call.

We also know that $\Omega_\text{Call} = \frac{\Delta_\text{Call} \cdot S_0}{\text{Call Premium}}$. Now in the Black-Scholes model, $\alpha$ and $r$ are constant.

So I think we only need to consider $\Omega_\text{Call}$.

As $S_0 \rightarrow \infty$, $\Delta_\text{Call} \rightarrow 1$ and $\text{Call Premium} \rightarrow \infty$.

Similarly, as $S_0 \rightarrow 0$, $\Delta_\text{Call} \rightarrow 0$ and $\text{Call Premium} \rightarrow 0$.

So I think we either have an indeterminate of the form $\frac{\infty}{\infty}$ or $\frac{0}{0}$.

For the first indeterminate, applying L'Hôpital's rule (multiple times):

\begin{align*}\lim_{S_0 \rightarrow \infty} \Omega_\text{Call} &= \lim_{S_0 \rightarrow \infty} \frac{\Delta_\text{Call} S_0}{\text{Call Premium}} = \lim_{S_0 \rightarrow \infty} \frac{\Gamma_\text{Call} S_0 + \Delta_\text{Call}}{\Delta_\text{Call}}\\ &= \lim_{S_0 \rightarrow \infty} \frac{\Gamma_\text{Call} S_0 + 1}{1} = \infty.\end{align*}

For the second indeterminate, applying L'Hôpital's rule (multiple times):

\begin{align*}\lim_{S_0 \rightarrow 0} \Omega_\text{Call} &= \lim_{S_0 \rightarrow 0} \frac{\Delta_\text{Call} S_0}{\text{Call Premium}} = \lim_{S_0 \rightarrow 0} \frac{\Gamma_\text{Call} S_0 + \Delta_\text{Call}}{\Delta_\text{Call}}\\ &= \lim_{S_0 \rightarrow 0} \frac{\Gamma_\text{Call}}{\Gamma_\text{Call}} = 1.\end{align*}

So, if my work is correct, as $S_0 \rightarrow 0$, $\Omega_\text{Call} \rightarrow 1$ and as $S_0 \rightarrow \infty$, $\Omega_\text{Call} \rightarrow \infty$.

Then as $S_0 \rightarrow 0$, $\gamma_\text{Call} = \Omega_\text{Call}(\alpha - r) + r \rightarrow \alpha$ and $S_0 \rightarrow \infty$, $\gamma_\text{Call} = \Omega_\text{Call}(\alpha - r) + r \rightarrow \infty$.

This seems to be the opposite of what the author states, so I don't know where I made my mistake.

  • $\begingroup$ This may make more intuitive sense to you if you consider how the option's price appreciates (or depreciates) relative to the stock price as the stock price moves up (or down) relative to the strike. i.e. Going from out-of-the-money to at-the-money to in-the-money or vice versa. $\endgroup$
    – amdopt
    Mar 9, 2017 at 19:52
  • $\begingroup$ It seems to me you are on the right track, but I couldn't follow all the steps. You might plot $\frac{\Delta}{c}$ graphically to see how it varies with S as a check on your algebraic reasoning. $\endgroup$
    – Alex C
    Mar 11, 2017 at 20:37

3 Answers 3


I think you nearly got there but made a few mistakes in the application of l'Hopital's rule.

First Limit

In the first case, you got

\begin{eqnarray} \lim_{S_0 \rightarrow \infty} \Omega & = & \lim_{S_0 \rightarrow \infty} \frac{\Gamma_{\text{call}} S_0 + \Delta_{\text{call}}}{\Delta_{\text{call}}}\\ & = & \lim_{S_0 \rightarrow \infty} \frac{\Gamma_{\text{call}} S_0 + 1}{1} \end{eqnarray}

and you seem to conclude that $\lim_{S_0 \rightarrow \infty} \Gamma_{\text{call}} S_0 = \infty$ and thus $\lim_{S_0 \rightarrow \infty} \Omega = \infty$. This is not true however. Remember that

\begin{equation} \Gamma_{\text{call}} = \frac{\mathcal{N}' \left( d_+ \right)}{S_0 \sigma \sqrt{T}} \end{equation}

and thus

\begin{equation} \lim_{S_0 \rightarrow \infty} \Gamma_{\text{call}} S_0 = \lim_{S_0 \rightarrow \infty} \frac{\mathcal{N}' \left( d_+ \right)}{\sigma \sqrt{T}} = 0. \end{equation}

Consequently $\lim_{S_0 \rightarrow \infty} \Omega_{\text{call}} = 1$ as postulated by the sample solution.

Second Limit

I agree with you up to the point where you have

\begin{equation} \lim_{S_0 \rightarrow 0} \Omega_{\text{call}} = \frac{\Gamma_{\text{call}} S_0 + \Delta_{\text{call}}}{\Delta_{\text{call}}} \end{equation}

which results in a $0 / 0$ situation. However, when you apply l'Hopital's rule again now, you don't apply the chain rule correctly and forgot to differentiate gamma. I get

\begin{equation} \ldots = \lim_{S_0 \rightarrow 0} \frac{\mathcal{S}_{\text{call}} S_0 + 2 \Gamma_{\text{call}}}{\Gamma_{\text{call}}} \end{equation}

where I use $\mathcal{S}_{\text{call}}$ to denote the third derivative w.r.t. the spot (the speed). It is given by

\begin{equation} \mathcal{S}_{\text{call}} = -\frac{\Gamma_{\text{call}}}{S_0} \left( \frac{d_+}{\sigma \sqrt{T}} + 1 \right). \end{equation}

We thus get

\begin{equation} \ldots = \lim_{S_0 \rightarrow 0} \left\{ -\left( \frac{d_+}{\sigma \sqrt{T}} + 1 \right) + 2 \right\}. \end{equation}

Now, since $\lim_{S_0 \rightarrow 0} d_+ = -\infty$, this yields $\lim_{S_0 \rightarrow 0} \Omega_{\text{call}} = \infty$ as postulated by the sample solution.


I have a different take on this: in the Black Scholes framework, the expected return on all tradeable assets is the risk free rate. It doesn't matter what the stock price is. That's because the Black Scholes framework is an example of a risk-neutral pricing system (one in which the underlying asset has a lognormal distribution). Hence , the expected return on any call option is the risk free rate.

  • $\begingroup$ Not true. The expected return of a Call Option is significantly higher since it entails more risk than a stock. The option is priced under the risk neutral measure, but evolves in the real world measure. $\endgroup$
    – user9403
    Mar 11, 2017 at 23:12
  • $\begingroup$ I agree with that, but your question asks about the return of the option "in the Black Scholes model " which I took to mean the expected return in the risk neutral measure. $\endgroup$
    – dm63
    Mar 11, 2017 at 23:41
  • $\begingroup$ The Black Scholes model does not assume that all assets have rate of return equal to the interest rate. The risk neutral measure is a computational trick. It is not a statement about the expected rate of return of risky assets. $\endgroup$
    – user9403
    Mar 11, 2017 at 23:57

I'll define expected return as $\mathbb{E}\left[\frac{H(T)-V(0)}{V(0)T}\right]$ where $V(0)$ is the current Black Scholes price and $H(T)$ is the payoff function. The expectation is under the real world measure.

Since $V(0)$ is known, the only stochastic element in the expectation is the expected value of the payoff. This is $$S_0 e^{\alpha T} N\left(\hat{d_1}\right)-KN\left(\hat{d_2}\right)$$ where $\hat{d_1}=\frac{log\left(\frac{S_0}{K}\right)+(\alpha+\sigma^2 /2 )T }{\sigma \sqrt{T}}$ and $\hat{d_2}=\hat{d_1}-\sigma \sqrt{T}$. Putting the equations together, the expected return is

$$\frac{S_0 \left(e^{\alpha T}N\left(\hat{d_1}\right)-N\left(d_1\right)\right)-K\left(N\left(\hat{d_2}\right)-e^{-rT} N\left(d_2\right)\right)}{\left(S_0 N\left(d_1 \right)-Ke^{-rT}N\left(d_2\right)\right)T}$$

At first glance this would seem to confirm that the expected value declines since the term multiplying $S_0$ in the numerator will be close to zero while the term in the denominator moves quite strongly with $S_0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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