Martingality Theorem: Solving expectation of a stochastic integral

Question

I am trying to prove that:

$$ \Bbb E\left[\int_s^t\sigma e^{-k(t-u)}\sqrt{V_u}dW_u\right] =0$$

Where: $$ dV_t=k~(\theta-V_t)~dt+\sigma\sqrt{V_t}dW_t $$

I have attempted to use Ito's formula on the SDE with $f(x)=\sqrt{x}$ but this doesn't seem to be a good way to go as the argument becomes circular.

Further, the theorem of 'martingality' states that if the integrand is progressively measureable and square integrable then the result follows, but I'm not able to show that $V_t$ satisfies these conditions.

Answer 1

Lets assume that we are in the regime in which $V_t > 0$ for every $t>0$. Consider the sequence of stopping times $$\tau_n = \inf \left\{ t: V_t > n \right\}, \quad n \in \mathbf{N},$$ then, as you mentioned before, if we can show that $\tau_n \to \infty$ as $n \to \infty$ in probability the result follows by standard localization: the expectation at time $t \wedge \tau_n$ would be zero (since the integrand is bounded and hence square integrable), and by taking the limit $n\to \infty$ the expectation will be zero for every $t>0$.

Fix $T>0$, and observe that Ito's product rule combined with some calculus, we can find an upper bound of $\mathbf{P} ( \tau_n < T ) = \mathbf{P}\left( \sup_{ t < T \wedge \tau_n } V_t \geq n \right)$ as a sum of two quantities $$ p_n = \mathbf{P}\left( \sup_{ t \leq T\wedge \tau_n } \left| \theta + \left( V_0 - \theta \right) e^{-kt} \right| > n/2\right), \text{ and } q_n = \mathbf{P}\left( \sigma \sup_{ t \leq T\wedge \tau_n } \left| \int_0^t e^{-k(t - s )}V_sdW_s \right| > n/2\right).$$ Obviously $p_n = 0$ for $n$ large enough, so we need to focus on $q_n$. From Chebychev's and Burkholder-Davis-Gundy Inequalities, it follows that $$ q_n \leq \frac{ 4 \sigma^2 }{n^2} \mathbf{E} \sup_{ t \leq T\wedge \tau_n } e^{-2t}\left| \int_0^{t\wedge \tau_n }e^{s}\sqrt{ V_s} dW(s)\right|^2 \leq\frac{ 4 \sigma^2 }{n^2} \mathbf{E}\int_0^{T\wedge \tau_n }e^{2s}V_sds, $$ which, using the definition of $\tau_n$, implies that $q_n \leq 4 \sigma^2e^{2T}n^{-1}.$ Hence, $\lim_{n \to \infty} \mathbf{P} ( \tau_n < T ) = 0$ and the result follows by localization as before.

Answer 2

Hi as mentioned in my comment the proof given does not allows us to conclude by localization. I have finally found a rigorous and elementary proof of the fact that a CIR process possesses moments of all orders which is necessary to get the result (order 1 is enough for our need) .

First two obersvations :

• by Yamada-Watanabe's theorem that the CIR's SDE has strong solution at all times. Indeed $\int_0^\epsilon 1/\rho^2(x)dx=\infty$ for $\rho(x)=\sigma.\sqrt{x}$.

• Feller's test for explosion can give us the result (nicely proved by Innombrabre in is post) that there is no explosion for CIR process under suitable condition on the parameters.

Now the proof is based on an adaptation of the argument in the proof of Proposition 2.3 in an article of Carr, Ekström, and Tysk which goes well beyond continuous processes but is almost applicable to our case.

Let's fix $T>0$ and let's start by localizing the process $V$ using $\tau_n$ as in the post of Innombrabre. Again in this case the stopped process $V^{\tau_n}$ is a bounded (continuous) semimartingale so that the local martingale part of it is a true martingale ( for the reason mentioned by Innombrabre in his post).

Now for fixed integer $k>0$ let's define the function $h:\mathbb{R^+}\to \mathbb{R^+}$ such that is it twice continuously differentiable and has the following properties :

• $h(x)=1$ on $[0,1/2]$
• $h(x)\geq x^k$ on $(1/2,2)$
• $h(x)= x^k$ on $[2,+\infty[$

This function has the following very interesting properties for us as there exists a constant $C$ such that the following inequality holds $\forall x\geq0$ :

$h''(x).\frac{\sigma^2.x}{2}+h'(x).k.(\theta-x)\leq C.h(x)$

For the moment let's just admit this fact which we shall show later.

Now let's apply Itô lemma to the stopped process $V^{\tau_n}$ using $h$ we get :

$h(V^{\tau_n}_t)=h(V_0)+\int_0^{t\wedge \tau_n}h''(V_s).\frac{\sigma^2.V_s}{2}+h'(V_s).k.(\theta-V_s)ds + M_{t}$ where $M_t$ is a true martingale term .

Taking expectation gives :

$f(t)=h(V_0)+E[\int_0^{t\wedge \tau_n}h''(V_s).\frac{\sigma^2.V_s}{2}+h'(V_s).k.(\theta-V_s)ds ]$

Now using our property on $h$ :

$f(t)\leq h(V_0)+E[\int_0^{t\wedge \tau_n}C.h(V_s)ds]\leq h(V_0)+E[\int_0^{t}C.h(V_s)ds]=f(0)+ \int_0^{t}C.f(s)ds $.(please observe that Fubini is licit since everything is positive)

Applying Gronwall's lemma now gives us :

$f(t)\leq h(V_0).e^{C.t}$

Now observe that $\forall n>0$, $E[(V^{\tau_n}_t)^k]\leq f(t)=E[h(V^{\tau_n}_t)]\leq h(V_0).e^{C.t}$

This allows us to conclude that $E[V_t^k]\leq h(V_0).e^{C.t}$ and hence has finite moments of all orders. In our case the bound obtained with $k=1$ is enough to show that $E[<\int_0^t\sigma e^{-k(t-u)}\sqrt{V_u}dW_u>]=<+\infty$. Let's take a look at the quadratic variation term :

$E[<\int_0^t\sigma e^{-k(t-u)}\sqrt{V_u}dW_u>]=E[\int_0^t\sigma^2 e^{-2k(t-u)}V_u.du]=\int_0^t\sigma^2 e^{-2k(t-u)}E[V_u].du$

$\leq \int_0^t\sigma^2 e^{-2k(t-u)}h(V_0).e^{C.u}.du<\infty$ (here again Fubini is ok)

This shows that $\int_0^t\sigma e^{-k(t-u)}\sqrt{V_u}dW_u$ is a true martingale started 0, and then it has 0 expectation.

Best regards

Edit : Proof of the inequality : $h''(x).\frac{\sigma^2.x}{2}+h'(x).k.(\theta-x)\leq C.h(x)$

First for any $k\leq 1$ and $x\in[0;1/2]$, $C\geq 0$ is enough as $h'(x)=h''(x)=0$, if we have $x\in[2,\infty)$, then the inequality after a few simplification turn into:

$\frac{1}{x}.(k.(k-1)\sigma^2/2+k.\kappa.\theta)-k\kappa\leq C$

For this to be true it is enough that $C>1/2.(k.(k-1)\sigma^2/2+|k.\kappa.\theta|)+|k.\kappa|$.

Now the last interval has $h$ is continuous it attains its minimum on the interval $(1/2;2)$ say $A>0$, the continuity argument tells us that $h''(x).\frac{\sigma^2.x}{2}+h'(x).k.(\theta-x)$ attains it maximum say $B$ so that taking $C>|B/A|$ is enough which concludes the proof.

NB: I still need to prove that absence of explosion is not enough to get the result.