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I'm currently looking at the CIR model and trying to calculate the expectation of $r_t$. I know that by Fubini's theorem, $\mathbb{E}\left(\int_0^t\kappa(\theta-r_s)\,\textrm{d}s\right)=\kappa\theta t-\int_0^t\mathbb{E}(r_s)\,\textrm{d}s$ and we can create an ODE from it, but as for the second term I know I can't use Fubini's theorem, so what can I do? I know one way is to multiply the SDE by $\textrm{e}^{\kappa t}$ but I haven't seen how it helps with the square root term, and Shreve claims it is an Ito integral but I don't see how $\sqrt{r_t}\in\mathcal{L}^2_{\mathcal{F}}$. All help is appreciated, thanks!

DMcMor
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user107224
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    I guess by $\mathcal L^2_{\mathcal F}$ you mean the space of ${\mathcal F_t }$-adapted processes $X$ such that $E \int_0^T |X_u|^2 du < \infty.$ Since the function $f(x)=\sqrt{x}$ has linear growth, $\sqrt{r_t} \in \mathcal L^2_{\mathcal F}.$ – UBM Dec 02 '20 at 22:49
  • @UBM this feels incredibly silly to ask, but how does linear growth imply square integrability? surely there is a potential for divergence? – user107224 Dec 03 '20 at 12:17
  • It's not silly at all, it requires a proof. You can see a proof in the book from Xuerong Mao for example (page 51). Most likely you have it as well in Oksendal. I could type it here but it may be quicker for you to check it in the book. – UBM Dec 03 '20 at 12:46
  • @UBM I’ve just taken a look — it looks like I have to satisfy both Lipschitz and linear growth conditions? This is quite confusing, is it alright if you could describe it as an answer? – user107224 Dec 03 '20 at 16:54
  • The square root function doesn't satisfy the global Lipschitz condition. But linear growth is enough for what you need. I'll write an answer later or tomorrow. – UBM Dec 03 '20 at 17:23

1 Answers1

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We need the following Lemma.

Lemma. Consider the SDE $$dX_t = f(X_t)dt + g(X_t)dB_t, \quad X_0 = x_0 \in \mathbb R^{+}. \tag{1}$$

with solution $\{x_t; 0 \leq t \leq T\}.$ Assume that the linear growth condition holds, i.e.

There exists a constant $K \in \mathbb R$ such that for all $x \in \mathbb R$ $$|f(x)|^2 \bigvee |g(x)|^2 \leq K(1+x^2).$$ Then there exists a constant $C$ such that $$E[\sup_{0 \leq t \leq T} X_t^2] \leq C.$$ (For a proof of this lemma see for example lemma 3.2 - page 51 in Xuerong Mao's book)

Now going to your problem: your SDE is of the SDE (1) type with $f(x)= \kappa(\theta -x)$ and $g(x) = \sigma \sqrt{x}.$ We can see that clearly the linear growth condition holds, so the lemma applies and there exists $C>0$ such that $E[\sup_{0 \leq t \leq T} r_t^2] \leq C.$ This implies that $E\int_0^T (\sigma \sqrt{r_t})^2dt < \infty.$ Thus the process $I_t:= \int_0^t (\sigma \sqrt{r_t})dB_t, 0 \leq t \leq T$ is a martingale and has constant expectation. Therefore, $E[I_T]=E[I_0]=0.$

UBM
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  • I hope you don’t mind that I’ve returned to this old question but I have another query — in the comment you’ve mentioned it does not satisfy the Lipschitz condition, but why do we only need the linear growth? Also, I read that the CIR model has a unique non-negative solution, doesn’t this require BOTH Lipschitz AND linear growth to be satisfied? Thanks! – user107224 Dec 16 '20 at 15:07
  • question 1: We only need the linear growth condition because that's enough to prove what we want. As I said before you can see a proof on Xuerong Mao's book (page 51). Do you have the book? If you don't have it and you want to see a full proof, you may want to ask a separate question and I'll try to answer it. Also, other people can answer it or comment on it. Question 2: you are right, the CIR model has unique solution but remember that the Lipschitz and linear growth conditions are sufficient conditions for a unique solution to exists but they are not necessary. – UBM Dec 16 '20 at 16:02
  • The function $f(x)=\sqrt{x}$ is not Lipschitz. See for example: https://math.stackexchange.com/questions/667346/sqrtx-isnt-lipschitz-function – UBM Dec 16 '20 at 16:09
  • ah yes I read the chapter, still trying to digest it! I’m probably combining many concepts at once and confusing myself. May I just check that the two conditions together are sufficient, or either one is sufficient (I read somewhere else on MSE that (global??) Lipschitz implies linear growth but I am still trying to understand that too)? – user107224 Dec 16 '20 at 16:12
  • Everytime I said the Lipschitz condition before I meant the global Lipschitz condition. Since it wasn't relevant for your original question, I didn't want to mention it to no create confusion. Yes, as you said the global Lipschitz implies the linear growth condition. Also, the function $f(x)=\sqrt{x}$ doesn't satisfy the global linear condition but it satisfies the local Lipschitz condition. Local Lipschitz and linear growth is enough to prove existence and uniquess (Theorem 3.4 - page 56 in Mao's book). – UBM Dec 16 '20 at 18:52
  • But once we know there exist a unique solution, we only need the linear growth condition to answer your original question. – UBM Dec 16 '20 at 18:53