2
$\begingroup$

I would like to ask if there is a way to find the expected value and the variance of the following process $$ dv_t=(a-be^{\alpha v_t})dt+\sigma dW_t, \quad v_t=v_0 $$ where $a\in (-\infty,+\infty), b>0, \sigma>0, \alpha>0$.

Thank you for your time. I truly appreciate it.

Added (after the question of The Bridge)

Let $V_t=e^{2v_t}$, then applying Ito's lemma we can show that

$$ V_t=\frac{V_0\exp(2at +2\sigma W_t)}{\left(1+\alpha b V_0^{\alpha/2}\int_0^t exp(\alpha(as+\sigma W_s))ds\right)^{\alpha/2}} $$ From here, we can see that the above SDE has a unique solution.

$\endgroup$
  • $\begingroup$ First question should is there (or under which assumptions) a solution to this SDE. Could you provide any insights on this ? Best regards $\endgroup$ – The Bridge May 4 '16 at 9:53
  • 1
    $\begingroup$ Thank you for your question. I added a paragraph into it. $\endgroup$ – Duy Nguyen May 4 '16 at 15:25
  • $\begingroup$ You can use this reference vixra.org/abs/1706.0425 to solve this pb. $\endgroup$ – Guest 111 Oct 18 '17 at 9:51
-1
$\begingroup$

First, integrate your SDE. Then you need to use the fact that $v_t$ has Gaussian distribution with mean $m_t$ and variance $n_t$, and apply the expectation to your equation to get the differential equation of the first moment. We know the expected value of $e^{\alpha v_t}$ because $v_t$ is Gaussian. Hence you get the differential equation of the first moment. You also need to derive the differential equation of the variance...

The integral of the SDE is $v_t-v_0= \int_{0}^{t} (a-b e^{a v_s}) ds+ \sigma W_t$.

$\endgroup$
  • $\begingroup$ And how exactly do you compute $\int \dfrac {1} {a - b \mathrm e ^{\alpha v}} \ \mathrm d v$? Also, please do not post two answers to the same question, unless really necessary. Rather, edit your already given answer and append new material to it. $\endgroup$ – Alex M. Oct 19 '17 at 13:41
  • $\begingroup$ $v_t$ would be Gaussian if $b=0$, but that's excluded ($b>0$). $\endgroup$ – Jean Duchon Oct 19 '17 at 13:49
  • $\begingroup$ $v_t$ is Gaussian for any b real. $\endgroup$ – Guest 111 Oct 19 '17 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.