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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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 $v_t-v_0= \int_{0}^{t} (a-b e^{a v_s}) ds+ \sigma W_t$.

  • $\begingroup$ 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

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