Testing Arbitrage and Completeness with Girsanov’s Theorem

# Change of measure and Girsanov's Theorem: Do the following models admit arbitrage and are they complete?


# Change of measure and Girsanov's Theorem: Do the following models admit arbitrage and are they complete?












Let $S_{t}$ denote the price of stock, $\beta_{t}$ denote the savings account. For each model below state with reason whether it admits arbitrage and whether it is complete.

(a) $\beta_{t}=e^{t}, S_{t}=B_{t}+1$

(b) $\beta_{t}=e^{t}, S_{t}=e^{t+\int_{0}^{t} s d B_{s}}$

(c) $\beta_{t}=e^{t}, S_{t}=e^{t+\int_{0}^{t} B_{s} d s}$

Here's my approach to these questions:

$$\text { (a) } \frac{S_{t}}{\beta_{t}} =\frac{1+B_{t}}{e^{t}} =e^{-t}+e^{-t} B_{t}$$

$e^{-t}$ is determimistic, but not constant and hence this model admits arbitrage and is not complete. Is this right logic?

$$\text { (b) } \quad \frac{S_{t}}{\beta_{t}}=\frac{e^{t} e^{\int_{0}^{t} s d B_{s}}}{e^{t}} = e^{\int_{0}^{t} s d B_{s}} $$

How do I proceed from here? I'm unsure what to do.. same with (c) as well.

$$\text { (c) } \quad \frac{S_{t}}{\beta_{t}}=\frac{e^{t} e^{\int_{0}^{t} B_{s}ds}}{e^{t}} = e^{\int_{0}^{t} B_{s}ds} $$

How do I apply Girsanov's theorem to (b) and (c)? I'm not sure how to prove it for the last two, any help would be appreciated thank you

## Answer by user6247850 (score 10, accepted)

https://quant.stackexchange.com/a/66301

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 Girsanov's theorem, which states that if $Z_t = \exp\left(\int_0^t \theta_s dB_s - \frac 12 \int_0^t \theta_s^2 ds\right)$ and $d\tilde{\mathbb{P}} = Z_T d\mathbb{P}$, then $\tilde{B}_t = B_t - \int_0^t \theta_s ds$ is a Brownian motion under $\tilde{\mathbb{P}}$. We could also write this as $d\tilde{B}_t = dB_t - \theta_t dt$.

In a), we use Ito's lemma to compute $d(e^{-t}S_t) = e^{-t}(dS_t - S_tdt) = e^{-t}(dB_t - S_tdt)$. We want this to be a martingale under $\tilde{\mathbb P}$, so we want $d\tilde B_t = dB_t - S_tdt$. This suggests setting $\theta_t = S_t$ for all $t$ in Girsanov's theorem, i.e. define $Z_t = e^{\int_0^t S_sdB_s - \frac 12 \int_0^tS_s^2ds}$ and $d\tilde{\mathbb{P}} = Z_T d\mathbb{P}$. Then $d(e^{-t}S_t) = e^{-t}d\tilde B_t$ is a martingale under $\tilde{\mathbb{P}}$, so this model is arbitrage free.

In b), we compute \begin{align*}d(e^{-t}S_t) &= d(e^{\int_0^t s dB_s}) \\ &= e^{\int_0^t s dB_s}(tdB_t + \frac 12 t^2 dt) \\ &= te^{\int_0^t s dB_s}(dB_t + \frac 12 t dt).\end{align*} We again want to find a probability measure that makes this a martingale, so we want $d\tilde B_t = dB_t + \frac 12 t dt$. This suggests setting $\theta_t = -\frac 12 t$ in Girsanov's theorem, so define $Z_t := \exp\left(-\frac 12 \int_0^t s dB_s - \frac 18 \int_0^t s^2 ds\right)$ and $d\tilde{\mathbb{P}} := Z_T d\mathbb{P}$. Then $d(e^{-t}S_t) = te^{\int_0^t s dB_s}d\tilde B_t$ is a martingale under $\tilde{ \mathbb{P}},$ so this model is also arbitrage free.

In c), we compute \begin{align*}d(e^{-t}S_t) &= d(e^{\int_0^t B_sds}) \\ &= e^{\int_0^t B_sds}B_tdt.\end{align*} No matter how we change the measure, we cannot make this a martingale because there is no $dB_t$ term. Hence this model is not arbitrage free.

Now we want to check if these models are complete. Typically the definition of complete requires that the model be arbitrage free, so we can rule out c) immediately. The second fundamental theorem of asset pricing says that an arbitrage free model is complete if and only if the equivalent martingale measure is unique. In a) and b) we saw that there was only a single choice of $\theta_t$ to make $e^{-t}S_t$ a martingale, so both of these models are also complete. A good rule of thumb is that models are complete when there are the same number of risky assets as sources of uncertainty (i.e. Brownian motions).

EDIT: I should probably close a small gap in my answer to c). The first fundamental theorem of asset pricing doesn't have a (simple) converse, so the fact that there isn't an equivalent martingale measure does not imply there is an arbitrage. Instead, we can explicitly construct an arbitrage strategy. If we consider the wealth $X_t$ of an investor who holds $\Delta_t$ shares of stock at time $t$, then their wealth dynamics are $$dX_t = \Delta_t dS_t + (X_t - \Delta_t S_t)dt = (\Delta_t S_t (1+B_t) + (X_t-\Delta_t S_t))dt = (X_t + \Delta_t S_t B_t)dt.$$ Setting $\Delta_t = \operatorname{sgn}(B_t)$ then gives $dX_t = (X_t + S_t |B_t|)dt$. Since the drift is non-negative and is strictly positive whenever $B_t \ne 0$, we conclude this is an arbitrage because starting with $X_0 = 0$ we end up with $X_T \ge 0$ a.s. and $\mathbb{P}(X_T > 0) > 0$.

## Answer by Jan Stuller (score 3)

https://quant.stackexchange.com/a/66297

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:

$$\frac{S_0}{\beta_0}=\frac{1+B_0}{e^0}=1$$

$$\mathbb{E}\left[\frac{S_t}{\beta_t}|\mathcal{F_0}\right]=\mathbb{E}[e^{-t}+e^{-t}B_t|\mathcal{F_0}]=e^{-t}+e^{-t}\mathbb{E}[B_t|B_0]=e^{-t}\neq1$$

Therefore this model admits arbitrage as you've rightly pointed out.

Model b:

$$\frac{S_0}{\beta_0}=1$$

$$\mathbb{E}\left[\frac{S_t}{\beta_t}|\mathcal{F_0}\right]=\mathbb{E}\left[\frac{e^{t+\int_0^thdB_h}}{e^t}|\mathcal{F_0}\right]=\mathbb{E}\left[e^{\int_0^thdB_h}|\mathcal{F_0}\right]$$

Now by Ito formula, $\int_{h=0}^{h=t}hdB_h=tB_t-\int_{h=0}^{h=t}B_hdh$, and this quantity is Normally distributed with expectation of zero and variance of $\frac{1}{3}t^3$ (by Ito Isometry, see end of post*). So we know that $e^{\int_0^thdB_h}$ is log-normally distributed, and we can write:

$$\mathbb{E}\left[e^{\int_0^hhdW_h}\right]=e^{0+0.5*\frac{1}{3}t^3}=e^{\frac{1}{6}t^3}\neq1$$

So again, this model is not arbitrage-free.

Model c:

$$\frac{S_0}{\beta_0}=1$$

By Ito formula, $\int_{h=0}^{h=t}B_hdh=tB_t-\int_{h=0}^{h=t}hdB_h$. This has expectation of zero again and variance is again $\frac{1}{3}t^3$ (see here), so that:

$$\mathbb{E}\left[e^{\int_{h=0}^{h=t}B_hdh}\right]=e^{\frac{1}{6}t^3}\neq1$$

So this model is not arbitrage-free.

*Ito Isometry states that, for any adapted Stochastic process $X_t$:

$$\mathbb{E}\left[\left(\int_{h=0}^{h=t}X_hdB_h\right)^2\right]=\mathbb{E}\left[\int_{h=0}^{h=t}X_h^2dh\right]$$

So we have ($X_t=t$):

$$\mathbb{E}\left[\left(\int_{h=0}^{h=t}hdB_h\right)^2\right]=\mathbb{E}\left[\int_{h=0}^{h=t}h^2dh\right]=\left[\frac{1}{3}h^3\right]_{h=0}^{h=t}=\frac{1}{3}t^3$$