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I want to find the covariance of the following SDE:

$$x_{t} = x_{0}e^{-\alpha t} + \rho \int_{0}^{t} e^{-\alpha(t-s)}dW_{s}$$

To start, I find the mean, it is simply:

$$ E[x_{t}] = E[x_{0}] e^{-\alpha t} $$

Then I need to compute:

$$ Cov(x_{t}, x_{s}) = E \left[ (x_{t} - E[x_{t}]) (x_{s} - E[x_{s}]) \right] $$

Expanding this out, gives me:

$$ Cov(x_{t}, x_{s}) = E\left[ (x_{0}e^{-\alpha t} + \rho \int_{0}^{t} e^{-\alpha(t-k)}dW_{k} - E[x_{0}] e^{-\alpha t}) (x_{0}e^{-\alpha s} + \rho \int_{0}^{s} e^{-\alpha(s-q)}dW_{q} - E[x_{0}] e^{-\alpha s}) \right] $$

$$ Cov(x_{t}, x_{s}) = E\left[ x_{0}^{2} e^{-\alpha(t+s) } + x_{0} \rho\int_{0}^{s} e^{-\alpha (s-q)} dW_{q} - x_{0}E[x_{0}] e^{-\alpha(t+s)} + a_{0}\rho e^{-\alpha s} \int_{0}^{t} e^{-\alpha(t-k)} dW_{k} + \rho^{2} \int_{0}^{t} \int_{0}^{s} e^{-\alpha (t-k)} e^{-\alpha (s-q)} dW_{k} dW_{q} - \rho E[x_{0}] e^{-\alpha s} \int_{0}^{t} e^{-\alpha(t-k)} dW_{k} - x_{0}E[x_{0}] e^{-\alpha (t+s)} - \rho E[x_{0}] e^{-\alpha t} \int_{0}^{s} e^{-\alpha (s-q)} dW_{q} + E[x_{0}]^{2} e^{-\alpha(t+s)} \right]$$

Since $E\left[\rho \int_{0}^{t} e^{-\alpha(t-k)}dW_{k} \right] = 0$, we have:

$$ Cov(x_{t}, x_{s}) = \left(E[x_{0}^{2}] - E[x_{0}]^{2}\right)e^{-\alpha (t+s)} + \rho^{2} E\left[\int_{0}^{t} \int_{0}^{s} e^{-\alpha (t-k)} e^{-\alpha (s-q)} dW_{k} dW_{q} \right]$$

This is where I get stuck.

Question: How can I compute $E\left[\int_{0}^{t} \int_{0}^{s} e^{-\alpha (t-k)} e^{-\alpha (s-q)} dW_{k} dW_{q} \right]$?

It doesn't seem to follow the Wiener isometry rule.

Any ideas?

the_src_dude
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  • @GeorgeDewhirst So I need to rewrite it as: $E\left[ (\int_{0}^{t} e^{-\alpha(t-k)} dW_{k} ) ( \int_{0}^{s} e^{-\alpha(s-q)} dW_{q}) \right]$ then I can claim the integrals are independent, and so I rewrite it as the product of the expected values $E\left[ \int_{0}^{t} e^{-\alpha(t-k)} dW_{k} \right] \cdot E\left[\int_{0}^{s} e^{-\alpha(s-q)} dW_{q} \right] = 0$. Is that correct? – the_src_dude Dec 09 '19 at 16:32
  • I just found https://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process. The answer seems to be there actually. However it is different from your answer and different from mine, and I don't understand how they finalize the answer anyway. Hmm... – the_src_dude Dec 09 '19 at 16:38

2 Answers2

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Itô's isometry shows that

$$\mathbb{E} \left( \left[ \int_0^t f(s) \, dW_s \right]^2 \right) = \mathbb{E} \left( \int_0^t f(s)^2 \, ds \right).$$

Using the polarization identity $$x \cdot y = \frac{1}{4} ((x+y)^2-(x-y)^2),$$ this gives

$$\mathbb{E} \left( \int_0^t f(s) \, dW_s \int_0^t g(r) \, dW_r \right) = \mathbb{E} \left( \int_0^t f(s) g(s) \, ds \right). \tag{1}$$

$(1)$ allows you to compute the expectation of the product of stochastic integrals, which appears in the computation of the covariance.

Proof of (1): $$\begin{align*} & \mathbb{E} \left[ \left( \int_0^t f(s) \, dW_s \right) \cdot \left( \int_0^t g(r) \, dW_r \right) \right]\\ &= \frac{1}{4} \left( \mathbb{E} \left[ \left( \int_0^t f(s)\, dW_s + \int_0^t g(r) \, dW_r \right)^2 \right] - \mathbb{E} \left[ \left( \int_0^t f(s) \, dW_s - \int_0^t g(r) \, dW_r \right)^2 \right] \right) \\ &= \frac{1}{4} \left( \mathbb{E} \left[ \left( \int_0^t f(s)+g(s) dW_s \right)^2 \right] - \mathbb{E} \left[ \left( \int_0^t f(s)-g(s) \, dW_s \right)^2 \right] \right) \\ &= \frac{1}{4} \left( \mathbb{E} \left( \int_0^t (f(s)+g(s))^2 \, ds \right) - \mathbb{E} \left( \int_0^t (f(s)-g(s))^2 \, ds \right)\right) \\ &= \mathbb{E} \left[ \left( \int_0^t f(s) \cdot g(s) \, ds \right) \right] \end{align*}$$

saz
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  • RIght, I get that idea, but we have $dW_{k} dW_{q}$ and not $dW_{s} dW_{s}$. Ie $k \neq q$. – the_src_dude Dec 09 '19 at 17:45
  • @the_src_dude It doesn't matter how you call the variables of integration... anyway, I edited my answer, so hopefully this fixes your concern – saz Dec 09 '19 at 18:20
  • Even if the limits of the integrals are different? For instance, one ends at t, and the other s. They are not necessarily equal. – the_src_dude Dec 09 '19 at 19:43
  • @the_src_dude Write the upper limit using indicator functios, i.e. use (1) for the functions $1_{[0,u]}(s) f (s) $ and $1_{[0,v]}(r) g (r)$. – saz Dec 09 '19 at 20:27
  • That is an interesting idea and I will have to figure that out. ATM I have no idea how to do that. – the_src_dude Dec 09 '19 at 20:35
  • @the_src_dude What do you mean...? I already told you what to do. Take $u,v<t$ and define $\tilde{f}(s) := 1_{[0,u]}(s) f(s)$ and $\tilde{g}(r) := 1_{[0,v]}(r) g(r)$, then \begin{align} \mathbb{E} \left( \int_0^u f(s) , dW_s \int_0^v g(r) , dW_r \right) &= \mathbb{E} \left( \int_0^t \tilde{f}(s) , dW_s \int_0^t \tilde{g}(r) , dW_r \right) \ &\stackrel{(1)}{=} \mathbb{E}\left( \int_0^t \tilde{f}(s) \tilde{g}(s) , ds \right) \ &= \mathbb{E} \left( \int_0^{\min{u,v}} f(s) g(s) , ds \right). \end{align} – saz Dec 09 '19 at 21:52
  • @saz: It seems a good general technique but I don't understand how you get (1) from the polarization identity and Ito's Isometry. Could you explain please? (I can start a new question if necessary) – UBM Dec 10 '19 at 12:53
  • @UBM See my edited answer. – saz Dec 10 '19 at 13:10
  • @saz: Thank you! – UBM Dec 10 '19 at 13:21
  • @saz Thank you for the explanation. It is always better to be more verbose and explain things clearly on QA sites like these. I awarded you the answer because, even though it took some prodding, you answered the question very well. – the_src_dude Dec 10 '19 at 13:49
  • @the_src_dude Prodding.... You see, I prefer OPs which try to solve problems on their own; I do not want people to copy my solution and hand them in as a solution for their homework problem. From my point of view, you can learn mathematics only by making your own hands dirty... if you haven't tried hard enough on your own, you don't learn anything from the (complete) solution. If you had searched for "Itô isometry" and "polarization", you would have found a very similar question here on SE which contains a proof of (1). – saz Dec 10 '19 at 15:31
  • (... and I do not care much about reputation.. though it is good to know that there is somebody who appreciates the answer.) – saz Dec 10 '19 at 15:32
  • @saz I apologize for not being as mathematically gifted as you are. I will try to solve problems that are closer to my level moving forward. – the_src_dude Dec 10 '19 at 15:50
  • @the_src_dude Sure, I was just saying that I prefer OPs who try to solve the problems on their own (... and you certainly did this). – saz Dec 10 '19 at 16:13
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If $s \leq t,$ \begin{align*} E\left[\int_{0}^{t} \int_{0}^{s} e^{-\alpha (t-k)} e^{-\alpha (s-q)} dW_{k} dW_{q} \right] &= E\left[\int_{0}^{t}e^{-\alpha (s-q)}dW_{q} \int_{0}^{s} e^{-\alpha (t-k)} dW_{k} \right] \\ &=E\left[\left(\int_{0}^{s}e^{-\alpha (s-q)}dW_{q} +\int_{s}^{t}e^{-\alpha (s-q)}dW_{q} \right) \int_{0}^{s} e^{-\alpha (t-k)} dW_{k} \right] \\ &=e^{-\alpha(t-s)}E\left[\left(\int_{0}^{s} e^{-\alpha (s-k)} dW_{k} \right)^2 \right]+ e^{-\alpha(t-s)}E\left[\int_{s}^{t}e^{-\alpha (s-q)}dW_{q}\right] \\ &=e^{-\alpha(t-s)} E \int_{0}^{s} e^{- 2 \alpha (s-k)} dk \\ &= \frac{1}{2 \alpha}(e^{- \alpha(t-s)}-e^{- \alpha (t+s)}). \end{align*} Thus, if $s > t,$ we get $\frac{1}{2 \alpha}(e^{- \alpha(s-t)}-e^{- \alpha (t+s)}).$ Therefore, the general result is $\frac{1}{2 \alpha}(e^{- \alpha |s-t|}-e^{- \alpha (t+s)}).$

(The fourth equality follows from Ito's Isometry and the fact that $E\int_{s}^{t}e^{-\alpha (s-q)}dW_{q} = 0$)

UBM
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  • So this assumes that $t > s$, but how can we simultaneously handle the case where $s <= t$? – the_src_dude Dec 09 '19 at 17:46
  • I'm not sure if I understand your question. The case $s=t$ gives a second integral $\int_s^s ... = 0.$ So in this solution we are assuming $ s \leq t.$ – UBM Dec 09 '19 at 17:57
  • Yes but what about when $s > t$ then? There should be a way to put both cases together... – the_src_dude Dec 09 '19 at 18:02
  • We can assume without loss of generality that $s \leq t.$ If not, you would have to use $min(s,t)$ in every place and make the notation unnecessarily difficult. – UBM Dec 09 '19 at 18:15
  • In which places would you use $min(s, t)$? We know that $min(s, t) = \frac{s+t - |s-t|}{2}$ so maybe that helps? – the_src_dude Dec 09 '19 at 19:45
  • In fact you don't need to use $min(s,t)$ for anything. We have two cases: a)$s \leq t$ and b)$s > t.$ For case a) we have what I wrote. For case b) we get $\frac{1}{2 \alpha}(e^{- \alpha(s-t)}-e^{- \alpha(s+t)})$ so the general results would be $\frac{1}{2 \alpha}(e^{- \alpha|s-t|}- e^{- \alpha(s+t)}).$ – UBM Dec 09 '19 at 19:56
  • That was what I was hoping you would write haha. If you could edit your answer, I will accept it as the answer. – the_src_dude Dec 09 '19 at 20:35