How Risk Premium Affects Calls in the Schwartz One-Factor Model

# Interpretation of Risk Premium for Schwartz one-factor model


# Interpretation of Risk Premium for Schwartz one-factor model












I have to deal with this one-factor model:

\begin{equation*} \begin{cases} dS_t = \alpha \bigl(\mu - \log(S_t) \bigr)S_t \, dt + \sigma S_{t} \, dW_t \, , t \geq 0,\\ S|_{t=0} = S_0 > 0, \end{cases} \end{equation*}

which gives me the following PDE for an European Call option:

\begin{equation*} \begin{cases} \frac{\partial V}{\partial t} + \Bigl [ \alpha \Bigl(\mu - \frac{\lambda}{\alpha} - \log (S) \Bigr) S \Bigr ] \frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} - rV = 0, \\ V(S,T) = (S -K)^+, \end{cases} \end{equation*}

where the parameter $\lambda$ represents the risk premium. Solving numerically the PDE, if I increase $\lambda$ (usually I take positive values) then the price of the option decreases. Is it possible? And what is the economic interpretation of this phenomenon? Thanks in advance.

## Answer by M. Jeunesse (score 1, accepted)

https://quant.stackexchange.com/a/28093

I assume $\alpha>0$.

Let $V^\lambda$ be the solution of : \begin{equation*} \begin{cases} \frac{\partial V^\lambda}{\partial t} + \Bigl [ \alpha \Bigl(\mu - \frac{\lambda}{\alpha} - \log (S) \Bigr) S \Bigr ] \frac{\partial V^\lambda}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V^\lambda}{\partial S^2} - rV^\lambda = 0, \\ V^\lambda(S,T) = (S -K)^+, \end{cases} \end{equation*}

then you want to prove :

$$\lambda<\lambda' \Rightarrow V^\lambda \geq V^{\lambda'}$$

- Use Feyman Kac to prove that, under $\mathbb{P}$, by denoting : $$dX^\lambda_t = \alpha(\mu-\frac{\lambda}{\alpha}-X^\lambda_t)dt-\frac{\sigma^2}{2}dt + \sigma dW_t$$

$$V^\lambda(t,S) = \mathbb{E}\left[ \left.e^{-r(T-t)}\left(e^{X^\lambda_{T}}-K\right)^+ \right| X^\lambda_t = \ln S \right]$$

- Prove using Ito's lemma that: $$X^{\lambda_1}_T - X^{\lambda_2}_T = e^{-\alpha(T-t)}(X^{\lambda_1}_t-X^{\lambda_2}_t)+(1-e^{-\alpha(T-t)})(\frac{\lambda_2}{\alpha}-\frac{\lambda_1}{\alpha})$$

- Using that $x\to (e^{x}-K)^+$ is increasing, prove that:

$$ \lambda_1 < \lambda_2 \Rightarrow V^{\lambda_1}(t,S)\geq V^{\lambda_2}(t,S)$$

### Economic intuition

$\lambda \uparrow \Rightarrow \mu-\frac{\lambda}{\alpha} \downarrow \Rightarrow$ drift of asset is lower, so lower up-trend, and since call option is increasing with the price of the asset, you get lower values.