# Using black scholes to model a clawback in private equity

I am new to Black Scholes, and trying to use it to model a clawback in private equity. Essentially, a clawback gives the "limited" partners in the deal the option to pull some funds away from the "general" partner. I am essentially trying to value a put option, to assess the likelihood of the put option being in the money (i.e. $\mathcal{N} \left( -d_2 \right)$).

I believe the option in this case would rely on: Underlying value (i.e. "S"): net asset value of the investments + the distributions earned by the "limited" partners Exercise ("K"): the capital contributions (with a preferred growth rate) of the "limited" partners

My concern is primarily with the portion of "S", and not as much with "K". "S" has 2 separate components: the net asset value which has a volatility, and the distributions which does not have a volatility. I'm essentially trying to figure out how to model using Black Scholes when my "S" has a component with and without volatility. Has anyone had to deal with this before?

Thank you!

Update to question: thanks LocalVolatility! (I am now logged in using my own account).
Two clarifications:
1) I do not understand what the "+" after the brackets mean in your answer below? Thought maybe it would help that I understood it moving forward. Example: $\left( \hat{K} - X_T \right)^+$
2) Also, there will be times when the distribution $Y$ is larger than $K e^{\gamma T}$; so $\hat{K}$ will be less than 0. I think this is OK. $X$ (in this case the net present value, which follows a Brownian motion) is never less than 0. So, when $\hat{K}$ is less than 0 because $Y$ is larger than $K e^{\gamma T}$, the payoff $\left( \hat{K} - X_T \right)^+$ will end up being negative. We can just interpret this payoff as being 0. Further, the actual calculation of $\mathcal{N} \left( -d_2 \right)$ would just lead to an error. So, we can also interpret $\mathcal{N} \left( -d_2 \right)$ as being 0. This is because the likelihood of being in the money in this case is just 0. Does this align with your thinking?

• Note that the exercise probability of the put is $\mathcal{N} \left( -d_2 \right)$. Let me try to paraphrase your question. We can set $S_T = X_T + Y$ where $X$ follows a geometric Brownian motion and $Y$ is a constant. Then you want to price a European put with a strike price $K_T = K_0 e^{\gamma T}$ where $\gamma$ is a growth rate? If yes, then your payoff is $V_T = \left( K_T - S_T \right)^+ = \left( K_0 e^{\gamma T} - X_T - Y \right)^+$. But this is just a standard put on the asset $X$ with strike $K_0 e^{\gamma T} - Y$. I guess I misunderstood sth. here. Mar 14 '17 at 0:11
• $(a - b)^+$ is a common way to denote $\max \{ a - b, 0 \}$, which is just the put payoff in that case. Regarding $Y \geq K e^{\gamma T}$. You are right that the put is then worthless and the exercise probability is zero. However, you couldn't substitute the numbers in the equation as you'd end up trying to take the log of a negative number. Mar 14 '17 at 20:57
• @BobJansen - done! Mar 16 '17 at 14:23

This answer is an extended version of my comment on your question.

Here is how I understand you setup: The underlying value $S$ is given by the sum of two components: $S_t = X_t + Y$. Here $X$ is the net asset value following a geometric Brownian motion

\begin{equation} \mathrm{d}X_t = \mu X_t \mathrm{d}t + \sigma X_t \mathrm{d}W_t. \end{equation}

with drift $\mu$ and volatility $\sigma$. $Y$ is the value of the distributions which is fixed.

You want to price a European put option with maturity in $T$ and strike price $K e^{\gamma T}$ where $\gamma$ is the growth rate of the strike. The terminal payoff of this option is

\begin{eqnarray} V_T & = & \left( K e^{\gamma T} - S_T \right)^+\\ & = & \left( K e^{\gamma T} - X_T - Y \right)^+\\ & = & \left( \hat{K} - X_T \right)^+, \end{eqnarray}

where $\hat{K} = K e^{\gamma T} - Y$. I.e. the value of this option is the value of European plain vanilla put option on the asset $X$. The exercise probability is $\mathcal{N} \left( -d_- \right)$ as usual, where

\begin{equation} d_- = \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{X_0}{\hat{K}} \right) + \left( \mu - \frac{1}{2} \sigma^2 \right) T \right). \end{equation}