I am trying to understand the solution to the following exercise, however it is kind of poorly written. Can someone please explain it to me?

For $V = (V_t)$ the solution to the Ornstein-Uhlenbeck SDE (OU) $dV = −βV dt + cdB$

(i) By using the Ito isometry, or otherwise, show that $V_t$ has distribution $N(0, σ^2 (1 − e^{ −2βt})/(2β))$.

So the solution starts like this: $V_t$ has mean $v_0e^ {−βt}$, as $E[e^{ βu}dBu = ∫ t_ 0 e^ {βu}E[dBu] = 0$.

Can someone explain how we derived the mean of $V_t$? Was I required to solve the SDE to obtain that?

  • $\begingroup$ "Can someone explain how we derived the mean of $V_t$? Was I required to solve the SDE to obtain that?" No this is not required. The direct approach: $$V_t=-\beta\int_0^tV_sds+cB_t\implies E(V_t)=-\beta E\left(\int_0^tV_sds\right)+cE(B_t)=-\beta\int_0^tE(V_s)ds.$$ $\endgroup$ – Did May 9 '15 at 19:35
  • $\begingroup$ can you explain how I will find $\int_0^t E(V_s)ds$? I mean $E(V_s)$ is what im looking for isnt it? I dont see it... $\endgroup$ – vounoo May 9 '15 at 20:06
  • $\begingroup$ Thus $v(t)=E(V_t)$ solves the integral equation $$v(t)=-\beta\int_0^tv(s)ds.$$ Surely you can transform this into a differential equation. $\endgroup$ – Did May 9 '15 at 21:32
  • $\begingroup$ Indeed. Thank you. That was useful. $\endgroup$ – vounoo May 9 '15 at 22:53

$$dV = -\beta Vdt +\sigma dB, \space V(0)=0$$ Treat this like an integrating factor ODE with initial condition $V(0) =0$.

Define $Y=e^{\beta t}V$, using Ito:

1)$$dY = \beta e^{\beta t}Vdt + e^{\beta t}dV = \beta e^{\beta t}Vdt - \beta e^{\beta t}Vdt + e^{\beta t}cdB = e^{\beta t}cdB$$ 2) $$\Rightarrow Y(t) = Y(0) + \sigma \int_{0}^te^{\beta s}dB \Rightarrow E[Y(t)]= E[Y(0)] +\sigma \int_0^te^{\beta s}E[dB] = E[Y(0)]=0$$

Where the third to last equality follows from Fubini

3) $$Var[Y(t)] = \sigma^2Var[\int_0^te^{\beta s}dB] = \sigma^2E[(\int_0^te^{\beta s}dB)^2]= \sigma^2E[\int_0^te^{2\beta s}dt] = \sigma^2\int_0^te^{2\beta s}dt$$

where the second to last equality follows from ito isometry and the last again by Fubini

4) $$Var[Y(t)] = \frac {\sigma^2}{2\beta}(e^{2\beta t}-1)$$

Now, $V(t) = e^{-\beta t}Y(t)$


$$\Rightarrow E[V(t)]=e^{-\beta t}E[Y(t)] = e^{-\beta t}*0 = 0$$ and

6) $$\Rightarrow Var[V(t)] = Var[e^{-\beta t}Y(t)] = e^{-2\beta t}Var[Y(t)] = e^{-2\beta t}\frac {\sigma^2}{2\beta}(e^{2\beta t}-1)= \frac {\sigma^2}{2\beta}(1-e^{-2\beta t})$$

The main tricks used in 2) and 3) were passing the expectation under the integral (justified by Fubini's theorem) and Ito isometry

If you want more practice, solve:

$$\frac {dS}{S}= k(\theta-lnS)dt + \sigma dW$$

Use an absolutely similar argument. This is a common model for energy and agricultural commodities.

  • $\begingroup$ Very helpful thank you $\endgroup$ – vounoo May 9 '15 at 0:29

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