Modeling a Private Equity Clawback as a Put Option

# Using black scholes to model a clawback in private equity


# 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?

## Answer by LocalVolatility (score 1)

https://quant.stackexchange.com/a/33021

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}