10

First, let's check if these models are abritrage free. The first fundamental theorem of asset pricing says that if there exists an equivalent probability measure under which $\frac{S_t}{\beta_t} = e^{-t}S_t$ is a martingale, then the market is arbitrage free, so we will check whether such an equivalent martingale measure exists. This is where we will use ...


9

Consider the Heston (1993) model under the real world measure ($\mathbb{P}$) \begin{align*} \mathrm{d}S_t&=\mu^\mathbb{P} S_t\mathrm{d}t+\sqrt{v_t}S_t\mathrm{d}B_{S,t}^\mathbb{P}, \\ \mathrm{d}v_t&=\kappa^\mathbb{P} (\bar{v}^\mathbb{P}-v_t)\mathrm{d}t+\sigma^\mathbb{P}\sqrt{v_t}\mathrm{d}B_{v,t}^\mathbb{P}, \end{align*} where $\mathrm{d}B_{S,t}^\...


4

Let $Y_t= e^{B_t}$ and $Z_t = B_{t}-t / 2$. Then, \begin{align*} dX_t &= Z_t dY_t + Y_t dZ_t + d\langle Y, Z\rangle_t\\ &=(B_{t}-t / 2)e^{B_t}\big( dB_t + 1/2\,dt \big) + e^{B_t}\big(dB_t -1/2\, dt\big) + e^{B_t} dt\\ &=e^{B_t}(B_t-t / 2+1)dB_t + e^{B_t}(B_t/2-t / 4 -1/2+1)dt\\ &=e^{B_t}(B_t-t / 2+1)d\big(B_t+1/2t\big). \end{align*} We ...


4

I don't think that you can apply Girsanov's theorem in this way, noting that I don't understand your (short) argument in the comments. This is how I would proceed, which makes sense mathematically, but the economic interpretation is then a bit strange. Let us write the SDE's a bit different, noting that it still preservers the same correlation structure \...


3

I assume all three models are stated under the money-market measure: then there is no arbitrage if the discounted pay-off is a martingale under the money-market Numeraire. Therefore to show no arbitrage for all three models, we would want to show that: $$\mathbb{E}\left[\frac{S_t}{\beta_t}|\mathcal{F_0}\right]=\frac{S_0}{\beta_0}$$ Model a: $$\frac{S_0}{\...


2

(Now that I saw Gordon's solution, I can finish my attempt; I had noticed $dB_t +1/2dt$ immediately from product rule for $V_t$, zero quadratic covariation between $t/2$ and $e^{B_t}$, but hours later :) I was still perplexed by $U_t$.) $$ X_t = U_t - V_t $$ $$ V_t = e^{B_t}t/2$$ $$ U_t = e^{B_t}B_t $$ $$ dV_t = \boxed{1/2e^{B_t} dt} + 1/2V_t(dB_t + 1/2dt) $...


1

Your attempt is correct as far as the second equation is concerned. Nonetheless I will include this in my answer: You know that $$ d\left(\frac{P(t,T)}{B(t)}\right)=k(t,T)\,dW^*_t\quad\quad\quad\quad\quad\text{(1)} $$ (where I fixed your notation). Applying Ito's lemma to the LHS of this relation gives $$ \frac{dP}{B}-P\frac{dB}{B^2}=\frac{P}{B}\left(\frac{...


1

It sounds to me like you understand everything apart from how Girsanov's theorem defines the EMM. Girsanov's theorem tells us that if $B_t$ is standard Brownian motion under $P$, then for any adapted process $\gamma_t$ (satisfying certain conditions) the process $\hat{B}_t$ defined by: \begin{equation} d\hat{B}_t = \gamma_t dt +dB_t \end{equation} is ...


1

Intuition The stochastic discount factor (SDF) really has two jobs to do: it needs to incorporate the time value of money (discount) and take the riskiness of cash flows into account (stochastic). Thus, it can make sense to split the SDF into its two components, $$M_t=e^{-rt}A_t,$$ where $A_t$ does the risk compensation. Girsanov's theorem is concerned with $...


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