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I have the following two SDEs

$$dN_1=(2a-1)pN_1dt+\alpha_1 N_1dW_1$$

$$dN_2=(2pN_1-\mu N_2)dt+\alpha_2 N_2dW_2$$

where $W$ is the standard Brownian motion/Wiener process. This isn't homework, I'm just curious. I can solve the first one but the second one is in terms of $N_1$ and $N_2$ so I don't know how to go about it. I'm new to SDEs so any help is appreciated!

$$N_1(t)=N_1(0)exp\left\{((2a-1)p-\frac{1}{2}\alpha_1^2)t+\alpha_1 W_1\right\}$$

Herr Schrödinger
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Carol
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  • @ Carol : I suggest you use Girsanov theorem to find a probability measure under wich the second equation take the same for mas the first one and then using the Radon Nikodym process express the solution into the original probability space (of course you have to check under which conditions you can use Girsanov with respct to $N_1$). Best regards – TheBridge Dec 04 '14 at 08:19
  • It's a 2d Geometric Brownian Motion, with triangular coefficient matrices: have you tried looking for that on the web? – SBF Dec 04 '14 at 10:23
  • Thanks for the replies @TheBridge and Ilya. I am a second-year undergraduate management student so my math background is weak. I will look into your suggestions. – Carol Dec 04 '14 at 10:56
  • Solve the first equation (it's linear, just gbm), plug into the second and solve it with variation of constants method. – zhoraster Jul 18 '18 at 15:23
  • @zhoraster Would you provide more hints on how to treat the stochastic process of $N_1$ in the second equation? Thanks a lot! – huighlh Aug 03 '18 at 01:59
  • Use the variation of constants technique. Assume $N_2(t) = C(t) E(t)$ with $E(t) = \exp\left{ \big(2p-\frac12\alpha_2^2\big)t + \alpha_2W_2(t)\right}$. Then write $dN_2(t) = C(t) dE(t) + E(t) dC(t) + d[E,C]_t$ and equate it to the right-hand side of the second equation. (Seems that $C$ will be absolutely continuous, so you will get $[E,C]_t = 0$.) – zhoraster Aug 03 '18 at 06:32
  • @zhoraster what motivates using that expression for E(t)? – dleal Jan 24 '20 at 17:16
  • @dleal, actually, it should be $E(t) = \exp\left{ -\big(\mu+\frac12\alpha_2^2\big)t + \alpha_2W_2(t)\right}$. This is a solution to $dE(t) = E(t)(-\mu dt + \alpha_2 dW_2(t))$. – zhoraster Jan 27 '20 at 08:34

1 Answers1

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As mentioned in the comments the SDE

$$dN_1=(2a-1)pN_1dt+\alpha_1 N_1dW_1$$

is decoupled and so one can solve it

$$N_{1,t}=N_{1,0}\exp \left(\left(\mu -{\frac {\sigma ^{2}}{2}}\right)t+\sigma W_{1,t}\right).$$

for $\mu:=(2a-1)p$ and $\sigma:=\alpha_1$. Then since $W_{1}$ is independent of $W_{2}$ we can condition by it and so $N_{1,t}$ is simply a continuous process.

Then for the second equation,

$$dN_2=(2pN_1-\mu N_2)dt+\alpha_2 N_2dW_2$$

we simply use the Solution to General Linear SDE

\begin{align} dX_t = \big( a(t) X_t + b(t) \big) dt + \big( g(t) X_t \big) dB_t. \end{align}

\begin{align*} N_{2,t} =& \exp\left( \int_0^t\left( a(s) - \frac{1}{2}g^2(s) \right) \mathrm{d}s + \int_0^t g(s)\mathrm{d}B_s\right) \\ &\cdot \left(N_{2,0}+ \int_0^{t} b(s)\exp\left( \int_0^s\left( \frac{1}{2}g^2(r) - a(r)\right) \mathrm{d}r - \int_0^s g(r)\mathrm{d}B_r\right)\mathrm{d}s\right). \end{align*}

Thomas Kojar
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