---
# System prepended metadata

title: Report

---

# Report

## Introduction

In this project, we are going to price an ugly option.



## Methodology

### Assumptions & Model Specification

The option payoff at maturity looks like
$$
N \cdot \max\left[0,\left(k-\frac{S(T)}{S(0)}\right).\left(\frac{L(T,T,T+\Delta)}{L(0,T,T+\Delta)}-k'\right)\right]
$$
~~<u>Assumption 1</u>: The short rates for both areas are independent from the index and exchange rate.~~

<u>Assumption 2</u>: The dividend paid by the component stocks of the index is frequent enough to be regarded in aggregation as a continuously paying fractional dividend by the index itself. Mathematically, the dividend  process denominated in foreign currency is $dD(t) = qZ(t)dt$.

<u>Assumption 3</u>: The short rate can be described by the Vasicek model. The Vasicek model allows negative risk free rate. It is suitable for Europe, which adopted negative interest on excess reserves from 2014 and endured for a significantly long time.

<u>Assumption 4</u>: The index and exchange rate can be described by the Geometric Brownian Motion.

<u>Assumption 5</u>: Volatilities and correlations remain constant (No regime change during current time to maturity and the extra 3 months).



Suppose $X(t)$, $Z(t)$, $r(t)$, and $r_f(t)$ denote the process of USD/EUR exchange rate, EURO STOXX 50 index in EUR, USA short rate, and EUR short rate respectively. We can model the processes under the historical measure $P$ as:

$$
\begin{aligned}
dZ(t) / Z(t) &= (\mu_Z(t) - q) dt + \sigma_Z dW^{(1)}(t) \\
dX(t) / X(t) &= \mu_X(t) dt + \sigma_X\left(\rho_X dW^{(1)}(t) + \sqrt{1 - \rho_X^2} dW^{(2)}(t)\right) \\
dr(t) &= a (b - r(t)) dt + \sigma dW^{(3)}(t) \\
dr_f(t) &= a_f (b_f - r_f(t)) dt + \sigma_f \left(\rho_f dW^{(3)}(t) + \sqrt{1 - \rho_f^2} dW^{(4)}(t)\right) \\
\end{aligned}
$$

where $W^{(i)}(t), i=1, \dots, 4$ are independent Brownian motions.

Obviously, we can express $S(t) = Z(t) X(t)$. Besides, under certain fairly relaxed technical conditions, such as the square integrable condition, we can express the the price at time $t$ of a US zero-coupon bond maturing at time $T$ with a payoff of $1$ as
$$
p(t, T) = \mathrm{E}^Q\left[\exp\left.\left(-\int_t^Tr(s)ds\right)\right|\mathcal{F}_t\right]
$$
The instantaneous forward rate can be calculated as
$$
f(t, T) = -\frac{\partial}{\partial T} \ln p(t, T)
$$
The forward rate $L(t, T, T + \Delta)$ can be derived as
$$
\begin{aligned}
\exp({-\Delta L(t, T, T + \Delta)}) &= \exp\left(-\int_T^{T + \Delta}f(t, s)ds\right) \\
\Rightarrow L(t, T, T + \Delta) &= \frac{1}{\Delta} \int_T^{T + \Delta}f(t, s)ds = \frac{\ln p(t, T) - \ln p(t, T + \Delta)}{\Delta}
\end{aligned}
$$
Under the Vasicek model, we have
$$
p(t, T) = e^{A(t,T)-B(t,T)r(t)}
$$
where
$$
B(t,T)=\frac{1-e^{-a(T-t)}}a \\
A(t,T)=\left(b-\frac{\sigma^2}{2a^2}\right)[B(t,T)-(T-t)]-\frac{\sigma^2}{4a}B^2(t,T)
$$
Therefore, the forward rate is
$$
L(t, T, T + \Delta) = \frac{\ln p(t, T) - \ln p(t, T + \Delta)}{\Delta} = \frac{A(t, T) - A(t, T + \Delta) - (B(t, T) - B(t, T + \Delta))r(t)}{\Delta}
$$

### Short Rate Inference

Given that we only have the term structure with certain maturities, we cannot obtain short rate data from data directly. Therefore, it is necessary to infer the short rate from the term structure panel data, which is obtained by the weighted least square.

For any time $t$, we regress the forward rate on the features constructed by the maturity The model is specified as
$$
f(t, t + T) = \sum_{i=0}^{4} \beta_0 T^i + \epsilon_t
$$
The loss function is
$$
L = \sum_{i=1}^N \left[\hat f(t, t + T_i) - f(t, t + T_i) \right]^2 \frac{1}{T_i}
$$
The short rate is obtained as
$$
r(t) = \hat f(t, t + 0) = \hat \beta_0
$$
The reason to assign loss weight as $\frac{1}{T_i}$ is that because we want to estimate the short rate, we would want to emphasize the term structure behavior when the maturity $T$ is closer to $0$.

### Risk-neutral Measure

Note that because the zero bonds are priced under the domestic risk-neutral measure, denoted as $Q$ measure, the par yield implied by the zero bonds is therefore under the $Q$ measure, which means the short rate models calibrated by par yield inherit parameters under the $Q$ measure.

Now, we are interested in finding the risk-neutral representation for $S(t)$.

Considering two bank accounts $B(t)$, $B_f(t)$ for USD, EUR, respectively. The value process of the bank account can be represented as
$$
B(t) = \exp\left(\int_t^Tr(s)ds\right) \\
B_f(t) = \exp\left(\int_t^Tr_f(s)ds\right) \\
$$
The value of the foreign money-market account denominated in domestic currency, is denoted as $N(t) := B_f(t) X(t)$, we have
$$
\begin{aligned}
dN(t) &= dB_f(t)X(t) + dX(t) B_f(t) + dX(t)dB_f(t) \\
&= r_f(t)B_f(t)X(t) dt + X(t)B_f(t)\left[\mu_X(t) dt + \sigma_X\left(\rho_X dW^{(1)}(t) + \sqrt{1 - \rho_X^2} dW^{(2)}(t)\right)\right] + 0 \\
&= N(t) \left[(\mu_X(t) + r_f(t)) dt + \sigma_X\left(\rho_X dW^{(1)}(t) + \sqrt{1 - \rho_X^2} dW^{(2)}(t)\right)\right]
\end{aligned}
$$
Under the $Q$ measure, we have $\mu_X(t) + r_f(t) = r(t)$ and therefore
$$
\mu_X(t) = r(t) - r_f(t)
$$
Next, we want to find the $Q$ measure representation for the quantoed index. First, we look at the foreign portfolio process $\Pi(t)$ of owning $a_t$ shares of the index. The portfolio value comes from 3 parts, namely the index value, dividend value, and the value from depositing the remaining capital after purchasing the index into the foreign bank account. Therefore, we have the process:
$$
\begin{aligned}
d\Pi(t) &= a_t(dZ(t) + qZ(t)dt) + r_f(t) (\Pi(t) - a_tZ(t))dt \\
&= r_f(t)\Pi(t)dt + a_t(\mu_Z(t) - r_f(t) )Z(t) dt + a_tZ(t)\sigma_Z dW^{(1)}(t)
\end{aligned}
$$
We now find the portfolio process for the quantoed portfolio process $\Pi'(t) = \Pi(t)X(t)$.

$$
\begin{aligned}
d\Pi'(t) &= d\Pi(t)X(t) + dX(t) \Pi(t) + dX(t)d\Pi(t) \\
&= X(t)\left[r_f(t)\Pi(t)dt + a_t(\mu_Z(t) - r_f(t) )Z(t) dt + a_tZ(t)\sigma_Z dW^{(1)}(t)\right] \\
& \quad  + X(t)\Pi(t)\left[\mu_X(t) dt + \sigma_X\left(\rho_X dW^{(1)}(t) + \sqrt{1 - \rho_X^2} dW^{(2)}(t)\right)\right] \\
& \quad + a_t\rho_X \sigma_X \sigma_Z X(t)Z(t) dt \\
&= \mu_\Pi(t) dt + \beta_\Pi^{(1)} dW^{(1)}(t) + \beta_\Pi^{(1)} dW^{(2)}(t)
\end{aligned}
$$

where

$$
\begin{aligned}
\mu_\Pi(t) &= r_f(t) \Pi'(t) + a_tX(t)Z(t)(\mu_Z(t) - r_f(t) + \rho_X \sigma_X \sigma_Z) + \Pi'(t) \mu_X(t) \\
\beta_\Pi^{(1)} &= a_t X(t)Z(t) \sigma_Z + \Pi'(t) \sigma_X \rho_X \\
\beta_\Pi^{(2)} &= \Pi'(t) \sigma_X \sqrt{1 - \rho_X^2}
\end{aligned}
$$
Under the $Q$ measure, we desire $\mu_\Pi(t) = r(t)\Pi'(t)$. Recalling that $\mu_X(t) = r(t) - r_f(t)$ under the domestic risk-neutral measure, we have
$$
r(t) \Pi'(t) + a_tX(t)Z(t)(\mu_Z(t) - r_f(t) + \rho_X \sigma_X \sigma_Z) = r(t)\Pi'(t) \\
\Rightarrow \mu_Z(t) = r_f(t) - \rho_X \sigma_X \sigma_Z
$$
In conclusion, the processes under the $Q$ measure can be represented as
$$
\begin{aligned}
dZ(t) / Z(t) &= (r_f(t) - \rho_X \sigma_X \sigma_Z - q) dt + \sigma_Z d \tilde W^{(1)}(t) \\
dX(t) / X(t) &= (r(t) - r_f(t)) dt + \sigma_X\left(\rho_X d\tilde W^{(1)}(t) + \sqrt{1 - \rho_X^2} d\tilde W^{(2)}(t)\right) \\
dr(t) &= a (b - r(t)) dt + \sigma d\tilde W^{(3)}(t) \\
dr_f(t) &= a_f (b_f - r_f(t)) dt + \sigma_f \left(\rho_f d\tilde W^{(3)}(t) + \sqrt{1 - \rho_f^2} d\tilde W^{(4)}(t)\right) \\
\end{aligned}
$$
where $\tilde W^{(i)}(t), i=1, \dots, 4$ are independent Brownian motions under the $Q$ measure.

The process $S(t)$ under the $Q$ measure can be expressed as
$$
\begin{aligned}
d S(t) &= d(Z(t)X(t)) \\
&= dZ(t)X(t) + dX(t)Z(t) + dX(t)dZ(t) \\
&= S(t)\left[(r_f(t) - \rho_X \sigma_X \sigma_Z - q) dt + \sigma_Z d \tilde W^{(1)}(t)\right] \\
& \quad + S(t)\left[ (r(t) - r_f(t)) dt + \sigma_X\left(\rho_X d\tilde W^{(1)}(t) + \sqrt{1 - \rho_X^2} d\tilde W^{(2)}(t)\right) \right] \\
& \quad + S(t) \sigma_Z \sigma_X \rho_X dt \\
&= \mu_S(t) dt + \beta_S^{(1)} d\tilde W^{(1)}(t) + \beta_S^{(2)} d\tilde W^{(2)}(t)
\end{aligned}
$$
where
$$
\begin{aligned}
\mu_S(t) &= S(t)\left( r(t) - q \right) \\
\beta_S^{(1)} &= S(t) (\sigma_Z + \sigma_X \rho_X) \\
\beta_S^{(2)} &= S(t) \sigma_X \sqrt{1 - \rho_X^2}
\end{aligned}
$$

### Model Calibration

#### Vasicek Model

The closed form solution for $r_f(T) | F_t$ can be derived
$$
\begin{aligned}
dr_f(s) &= a_f (b_f - r_f(s)) ds + \sigma_f \left(\rho_f d\tilde W^{(3)}(s) + \sqrt{1 - \rho_f^2} d\tilde W^{(4)}(s)\right) \\
dr_f(s) + a_fr_f(s) dt &= a_fb_f ds + \sigma_f \left(\rho_f d\tilde W^{(3)}(s) + \sqrt{1 - \rho_f^2} d\tilde W^{(4)}(s)\right) \\
e^{a_fs} dr_f(s) + e^{a_fs} a_fr_f(s) dt &= e^{a_fs} \left[ a_fb_f ds + \sigma_f \left(\rho_f d\tilde W^{(3)}(s) + \sqrt{1 - \rho_f^2} d\tilde W^{(4)}(s)\right) \right] \\
\int_t^T d(e^{a_fs}r_f(s)) &= b_f (e^{a_fT} - e^{a_ft}) + \sigma_f \int_t^T e^{a_fs} \left(\rho_f d\tilde W^{(3)}(s) + \sqrt{1 - \rho_f^2} d\tilde W^{(4)}(s)\right) \\
r_f(T) &= r_f(t)e^{-a_f(T - t)} + b_f\left(1 - e^{-a_f(T - t)}\right) + \sigma_f \int_t^T e^{-a_f(T - s)} \left(\rho_f d\tilde W^{(3)}(s) + \sqrt{1 - \rho_f^2} d\tilde W^{(4)}(s)\right)
\end{aligned}
$$
Similarly the analytical solution for $r(T) | F_t$ is
$$
r(T) = r(t)e^{-a(T - t)} + b\left(1 - e^{-a(T - t)}\right) + \sigma \int_t^T e^{-a(T - s)} d\tilde W^{(3)}(s)
$$
We can calculate the statistics of the two processes:
$$
\begin{aligned}
E\left(r(T) | F_t\right) &= r(t)e^{-a(T - t)} + b\left(1 - e^{-a(T - t)}\right) \\
E\left(r_f(T) | F_t\right) &= r_f(t)e^{-a_f(T - t)} + b_f\left(1 - e^{-a_f(T - t)}\right) \\
Var\left(r(T) | F_t\right) &= \sigma^2 \int_t^T e^{-2a(T - s)} ds = \frac{\sigma^2}{2a}\left(1 - e^{-2a(T - t)}\right) \\
Var\left(r_f(T) | F_t\right) &= \sigma_f^2 \int_t^T \left(\rho_f^2 + 1 - \rho_f^2\right) e^{-2a_f(T - s)} ds  = \frac{\sigma_f^2}{2a_f}\left(1 - e^{-2a_f(T - t)}\right) \\
Cov\left(r(T), r_f(T) | F_t\right) &= \sigma\sigma_f \rho_f \int_t^T e^{-(a + a_f)(T - s)} ds = \frac{\sigma\sigma_f \rho_f}{a + a_f}\left(1 - e^{-(a + a_f)(T - t)}\right)
\end{aligned}
$$
We can write the conditional distributions as

$$
\begin{aligned}
(r(T) | F_t) &= Z^{(3)} \sim N\left(r(t)e^{-a(T - t)} + b\left(1 - e^{-a(T - t)}\right), \frac{\sigma^2}{2a}\left(1 - e^{-2a(T - t)}\right)\right) \\
(r_f(T) | F_t) &= Z^{(4)} \sim N\left(r_f(t)e^{-a_f(T - t)} + b_f\left(1 - e^{-a_f(T - t)}\right), \frac{\sigma_f^2}{2a_f}\left(1 - e^{-2a_f(T - t)}\right)\right) \\
Cov(Z^{(3)}, Z^{(4)}) &= \frac{\sigma\sigma_f \rho_f}{a + a_f}\left(1 - e^{-(a + a_f)(T - t)}\right)
\end{aligned}
$$


#### Geometric Brownian Motion

The processes for the index and exchange rate have the closed form solutions
$$
\begin{aligned}
Z(T) &= Z(t) \exp \left\{ \int_t^T \left(r_f(s) - \rho_X \sigma_X \sigma_Z - q -\frac{\sigma_Z^2}{2}\right) ds + \sigma_Z \sqrt{T - t} Z^{(1)} \right\} \\
X(T) &= X(t) \exp \left\{ \int_t^T \left(r(s) - r_f(s) -\frac{\sigma_X^2}{2}\right) ds + \sigma_X \sqrt{T - t} \left(\rho_X Z^{(1)} + \sqrt{1 - \rho_X^2} Z^{(2)}\right) \right\} \\
\end{aligned}
$$
Obviously, the combined data follows multivariate normal distribution
$$
\boldsymbol R(t, T) = \left(\begin{matrix}{\ln\frac{Z(T)}{Z(t)}}\\{\ln\frac{X(T)}{X(t)}}\end{matrix}\right) \sim N\left( \boldsymbol{\mu}_R, \boldsymbol \Sigma_R \right)
$$
where
$$
\begin{aligned}
\boldsymbol{\mu}_R &= \begin{bmatrix}{\left(r_f(t) - \rho_X \sigma_X \sigma_Z - q -\frac{\sigma_Z^2}{2}\right) (T - t)}\\{\left(r(t) - r_f(t) -\frac{\sigma_X^2}{2}\right) (T - t)} \end{bmatrix} \\
\boldsymbol{\Sigma}_R &= 
\begin{bmatrix}
\sigma_Z^2 (T - t) & \rho_X \sigma_X \sigma_X (T - t) \\
\rho_X \sigma_X \sigma_X (T - t) & \sigma_X^2 (T - t)
\end{bmatrix}
\end{aligned}
$$


### Monte Carlo Simulation

After model calibration, we have obtained the estimations for all parameters in the four processes. We can adopt Monte Carlo Simulation to price the option.

For simplicity, we can rewrite the process representation as
$$
\begin{aligned}
dZ(t) &= Z(t) (r_f(t) - \rho_X \sigma_X \sigma_Z - q) dt + \sigma_Z Z(t) d B^{(1)}(t) \\
dX(t) &= X(t) (r(t) - r_f(t)) dt + \sigma_X X(t) d B^{(2)}(t) \\
dr(t) &= a (b - r(t)) dt + \sigma dB^{(3)}(t) \\
dr_f(t) &= a_f (b_f - r_f(t)) dt + \sigma_f dB^{(4)}(t) \\
\end{aligned}
$$
We have $dB^{(1)}(t)dB^{(2)}(t) = \rho_X dt$ and $dB^{(3)}(t)dB^{(4)}(t) = \rho_f dt$.

In this project, we adopt Milstein method to approximate the processes. The method states that for the process $dX_t = a(X_t, t) dt + b(X_t, t) dB_t$, we can discretize it as
$$
\hat X_{t_{j + 1}} = \hat X_{t_{j}} + a(\hat X_{t_{j}}, t) \Delta t + b(\hat X_{t_{j}}, t_j) \sqrt{\Delta t} Z_j + \frac12 b(\hat X_{t_{j}}, t_j) b_x (\hat X_{t_{j}}, t_j) \Delta t (Z_j^2 - 1)
$$
Hence, the 4 processes are discretized as
$$
\begin{aligned}
\hat Z_{t_{j + 1}} &= \hat Z_{t_{j}} + (r_{f, t_j} - \rho_X \sigma_X \sigma_Z - q) \Delta t + \sigma_Z \hat Z_{t_{j}} \sqrt{\Delta t} Z_j^{(1)} + \frac{\sigma_Z^2}{2} \hat Z_{t_{j}} \Delta t ((Z_j^{(1)})^2 - 1) \\
\hat X_{t_{j + 1}} &= \hat X_{t_{j}} + (r_{f, t_j} - r_{t_j}) \Delta t + \sigma_X \hat X_{t_{j}} \sqrt{\Delta t} Z_j^{(2)} + \frac{\sigma_X^2}{2} \hat X_{t_{j}} \Delta t ((Z_j^{(2)})^2 - 1) \\
\hat r_{t_{j + 1}} &= \hat r_{t_{j}} + a(b - \hat r_{t_{j}}) \Delta t + \sigma \sqrt{\Delta t} Z_j^{(3)} \\
\hat r_{f, t_{j + 1}} &= \hat r_{f, t_{j}} + a_f(b_f - \hat r_{f, t_{j}}) \Delta t + \sigma_f \sqrt{\Delta t} Z_j^{(4)} \\
\end{aligned}
$$
For each step, we sample the normal variables $Z^{(i)}_j, i=1, \dots, 4$ from the multivariate normal distribution $N(0, \boldsymbol \Sigma_Z)$ where
$$
\boldsymbol \Sigma_Z = \begin{bmatrix}
1 & \rho_X & 0 & 0 \\
\rho_X & 1 & 0 & 0 \\
0 & 0 & 1 & \rho_f \\
0 & 0 & \rho_f & 1 \\
\end{bmatrix}
$$
Suppose we sampled $M$ paths for each of the process and each path contains $R$ steps. The option price is given by
$$
p = \frac{N}{M} \sum_{i=1}^{M} \left\{ \exp\left[ -\sum_{j=1}^{R} \hat r_{t_j} \Delta t \right] \cdot \max\left[0,\left(k - \frac{\hat Z_T \hat X_T}{Z_t X_t}\right).\left(\frac{B(T, T + \Delta)\hat r_T - A(T, T + \Delta)}{A(t, T) - A(t, T + \Delta) - (B(t, T) - B(t, T + \Delta))r_t}-k'\right)\right] \right\}
$$


## Data



## Empirical Results



## Conclusion