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Having trouble finding expected value and variance of a stochastic process defined by SDE:

$dX_{t} = a X_{t} dt + b dB_{t}$

$X_0 = x$, $a$ and $b$ are constant values, $B_t$~$N(0,t)$

Thank you for any help or pointers. I will now post my progresses:

1) Considering an auxiliary process $Z_t=X_te^{-at}$

2) Using Ito's formula to derive $dZ_t$ omitting quadratic covariance term due to it being between a stochastic and a determinist process (=0):

$dZ_t = e^{-at} dX_t+X_t(-ae^{-at}) dt$

3) Plugging in $dX_t$:

$dZ_t=e^{-at}(aX_tdt+bdB_t)-aX_te^{-at}dt$

$dZ_t=aX_te^{-at}dt+be^{-at}dB_t-aX_te^{-at}dt$

$dZ_t=be^{-at}dB_t$

4) Applying the integral on both sides ( change of variable from t to s inside the integral)

$\int_{0}^{t}dZ_s = b \int_{0}^{t}e^{-as}dB_s $

$Z_t-Z_0 = b \int_{0}^{t}e^{-as}dB_s $

5) Plugging back in $Z_t=X_te^{-at}$ and isolating $X_t$

$X_te^{-at}-X_0=b\int_{0}^{t}e^{-as}dB_s$

$X_t=X_0e^{at} + be^{at}\int_{0}^{t}e^{-as}dB_s$

6) Applying expectations

$\mathbb{E}(X_t)=xe^{at} + be^{at}\,\mathbb{E}(\int_{0}^{t}e^{-as}dB_s)$

I end up with this expression, now I know that I should use the Doléans exponential to prove that the stochastic integral is 0. But how can I proceed further to recover both expected value and variance?

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  • $\begingroup$ So what happens if you plug in $dX_t$? (And I suppose $(B_t)_{t \geq 0}$ is a Brownian motion. If so, the statement "$B_t \sim N(0,1)$" is not correct.) $\endgroup$ – saz Jan 11 '15 at 17:01
  • $\begingroup$ You plugged in $dX_t$, so there should be no "$dX_t$" in the last 3 lines. (Note that the expression $dB_t dX_t$ doesn't even make sense...) Consequently, you end up with $$dZ_t = b dB_t.$$ Solve it! $\endgroup$ – saz Jan 11 '15 at 17:07
  • $\begingroup$ yes, thank you I'll fix the mistakes you pointed out and try to solve it, kinda new to Latex and stoch calculus, please bear with me $\endgroup$ – Clemente Cortile Jan 11 '15 at 17:09
  • $\begingroup$ You are welcome. (There is a typo in my previous comment, it should read $dZ_t = b e^{-a t} \, dB_t$.) $\endgroup$ – saz Jan 11 '15 at 17:24
  • $\begingroup$ So far your calculations are correct. What does this tell you about $X_t$? $\endgroup$ – saz Jan 11 '15 at 17:43
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Hint: The stochastic integral

$$M_t := \int_0^t e^{-as} \, dB_s$$

is a martingale. Hence, $\mathbb{E}M_t = \mathbb{E}M_0=0$. In order to calculate the variance of $X_t$ use Itô's isometry.


Alternative approach: Since stochastic integrals are martingales, we have

$$\mathbb{E}X_t- \mathbb{E}X_0 = \int_0^t a \cdot \mathbb{E}X_s \, ds,$$

i.e. $m(t) := \mathbb{E}X_t$ solves the ordinary differential equation (ODE)

$$m'(t) = a \cdot m(t) \qquad m(0) = \mathbb{E}X_0.$$

The (unique) solution is $m(t) =e^{at} \mathbb{E}X_0$. Similarly, using Itô's formla, one can show that

$$\mathbb{E}(X_t^2)-\mathbb{E}(X_0^2) = \int_0^t \left( 2a \mathbb{E}(X_s^2) + b^2 \right) \, ds.$$

Consequently, $\sigma(t) := \mathbb{E}(X_t^2)$ solves

$$\sigma'(t) = 2a \sigma(t)+b, \qquad \sigma(0) = \mathbb{E}(X_0^2).$$

Solving this (linear) ODE yields $\mathbb{E}(X_t^2)$.


Remark: The process $(X_t)_{t \geq 0}$ is called Ornstein-Uhlenbeck process.

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  • $\begingroup$ How is the expected value of a process defined? Does it depend on $t$? $\endgroup$ – user415535 Sep 13 '17 at 10:03
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    $\begingroup$ @user21312 Yes, it does depend on $t$; the expected value is the mapping $m(t) := \mathbb{E}(X_t)$. $\endgroup$ – saz Sep 13 '17 at 12:51

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