為什麼比約克永遠不會證明可積性條件得到滿足？
Bjork 的“連續時間套利理論”中採用的一項主要技術是，當對隨機積分進行期望時，結果為 0。
這是基於第 4 章中提出的結果，該結果指出，如果被積函式，說 $ \psi $ , 滿足 $ \int_0^t E \psi^2 ds < \infty $ , 那麼期望 $ \int_0^t \psi dW_s $ 為零。
然而，在隨後的爭論中，無論多麼複雜 $ \psi $ 可能是在那種環境下定義的，作者從來沒有證明這個不平凡的條件實際上是滿足的，這給我做練習帶來了麻煩，因為我不知道我是否也可以假設它是滿足的，或者我是否我必須經過一些非常繁瑣的計算才能證明它是。
那麼，不檢查正常嗎？有沒有一種簡單的方法可以看出它是否滿意？
這是平方可積的一個充分條件 $ \psi(t,\omega) $ 這可能非常有用。給定隨機過程 $ X(t,\omega) $ 滿足 SPDE
$$ dX = \mu(t,X)dt+\sigma(t,X)dW $$ 在哪裡 $ (\mu(t,x),\sigma(t,x)) $ 是 Liptschitz 連續的，我認為 $ \displaystyle\int_t^T \mathbf E[\sigma(s,X)^2]ds<\infty $ . 請注意，函式取決於空間變數 $ x $ 而不是更一般的樣本 $ \omega $ . 有時間我會寫出證明。
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 $ , $$ \mathbb{E}[S_t]<S_0e^{\mu t}\tag 5 $$. 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 $ .