Why won't Bjork ever show that the integrability condition is satisfied?

A major technique employed throughout Bjork's "Arbitrage theory in Continuous Time" is that when taking the expectation of a stochastic integral, the result is 0.

This is based on a result presented in chapter 4, which states that if the integrand, say $\psi$, satisfies that $\int_0^t E \psi^2 ds < \infty$, then the expectation of $\int_0^t \psi dW_s$ is zero.

However, in subsequent arguments, regardless of how complex $\psi$ may have been defined in that setting, the author never demonstrates that this non-trivial condition is actually satisfied, which causes trouble for me in doing exercises, as I do not know whether I too can just assume that it is satisfied, or whether I have to go through some very tedious calculations in order to show that it is.

So, is it normal to not check it? Is there an easy way to see that it is satisfied?

• Usually, to rule out some type of doubling strategy, it is assumed ex ante that $\psi$ is a “well-behaved” process. As Maleki said, it is not trivial at all to demonstrate. – fni Jan 8 '17 at 9:55
• @fnic: What are some of the conditions qualifying as "well behaved" when the process in question is the solution of an SPDE? – Hans Jan 11 '17 at 20:57

Here is one sufficient condition for square integrability of $\psi(t,\omega)$ which may be very useful. Given stochastic process $X(t,\omega)$ satisfying SPDE $$dX = \mu(t,X)dt+\sigma(t,X)dW$$ where $(\mu(t,x),\sigma(t,x))$ is Liptschitz continuous, I think $\displaystyle\int_t^T \mathbf E[\sigma(s,X)^2]ds<\infty$. Note here that the functions depend on the spatial variable $x$ rather than the more general sample $\omega$. I will write out the proof when I have time.

We should check the martingale properties.

Let $$(\Omega ,\mathcal{F},\{\mathcal{F}\}_{t\ge 0},\mathbb{P})$$ be a filtered probability space. We define the class of functions ,$$\mathcal {V} =\mathcal {V}(t,T)$$, as follow $$\psi(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$$ such that

• $$(t,\omega)\to \psi(t,\omega)$$ is $$\mathcal{B}\times\mathcal{F}$$ where $$\mathcal{B}$$ denotes the Borel algebra on $$[0,\infty)$$.
• $$\psi(t,\omega)$$ is $$\mathcal{F}_t$$ adapted.
• $$\mathbb{E}\left[\int_{t}^{T}\psi^2(s,\omega)ds\right]<\infty$$

In this case, we have $$\mathbb{E}\left[\int_{t}^{T}\psi(s,\omega)dW_s\right]=0$$

Remark

If $$M_t$$ be an an arbitrary martingale with respect to $$\{\mathcal{F}\}_{t\ge 0}$$ and $$\psi(.,\omega)$$ be bounded then $$\int_{t}^{T}\psi(s,\omega)dM_s$$ is a martingale,and $$\mathbb{E}\left[\int_{t}^{T}\psi(s,\omega)dM_s\right]=0$$

Counter Example

The Constant elasticity of variance model ,CEV, describes a process which evolves according to the following stochastic differential equation: $$dS_t=\mu S_t dt+\sigma S_t^{\gamma} dW_t\tag 1$$ where The constant parameters $$\mu\,,\sigma$$ and $$\gamma$$ satisfy the conditions: $$\mu\in\mathbb{R}$$, $$\sigma\ge 0$$ and $$\gamma\ge 0$$.

The parameter $$\gamma$$ controls the relationship between volatility and price, and is the central feature of the model. When $$\gamma <1$$ we see the so-called leverage effect, commonly observed in equity markets, where the volatility of a stock increases as its price falls. Conversely, in commodity markets, we often observe $$\gamma>1$$ so-called inverse leverage effect, whereby the volatility of the price of a commodity tends to increase as its price increases.

I use the standard technique, I can integrate, take expectations, differentiate with respect to time and solve by ODE techniques !! . Now, I write the equation $$(1)$$ in integral form $$S_t=S_0+\mu\int_{0}^{t}S_u du+\sigma\int_{0}^{t}S_u^\gamma dW_u\tag 2$$

It is known that the expectation of a stochastic integral is zero, thus

$$\mathbb{E}[S_t]=S_0+\mu\int_{0}^{t}\mathbb{E} [S_u] du\tag 3$$

This can be differentiated to obtain the ordinary differential equation $$\frac{d\mathbb{E}[S_t]}{dt}=\mu \mathbb{E}[S_t]\tag 4$$ which has the unique solution $$\mathbb{E}[S_t]=S_0e^{\mu t}$$

Indeed this procedure is so wrong . For $$\gamma> 1$$, $$\mathbb{E}[S_t]. Indeed, if $$\gamma> 1$$ then the local martingale property holds and $$\int_{0}^{t}S_u^\gamma dW_u$$ is not a proper martingale, and has strictly negative expectation at all positive times. The reason that the martingale property fails here for $$\gamma>1$$ is that the coefficient $$\sigma S_t^\gamma$$ of $$dW_t$$ grows too fast in $$\ S_t$$.

• I do not think you are answering the OP's question. He is asking whether the finiteness of the expected square integral should be checked. Your answer nowhere addresses the issue. Usually $\phi(t,\omega)$ is not bounded especially over $\omega$. Your example gives just such a case. If by $\phi(\cdot,\omega)$ you mean fixing $\omega$ and varying $\cdot$, then your condition $\phi(\cdot,\omega)$ is bounded only mean it is bounded for each path. But it says nothing about whether it is bounded over all $(t,\omega)$. It is NOT a sufficient condition for the square integral boundedness. – Hans Jan 11 '17 at 20:37