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I have to solve the following SDE. $$ \mathrm dY_t= f(X_t) \mathrm dt, \tag{1} $$ where $X_t$ is a two-state Markov Process possesses states $a$ and $b$.

Moreover, I would like to solve $$ \mathrm dY_t= Z_t \mathrm dt, \tag{2} $$ where $Z_t$ is an rv conditioned on $X_t$, i.e., $Z_t \mid X_t \sim Bin(n,g(X_t))$.


After seeing Did's comment and Jay.H's answer, I tried to derive $Y_t$ with Riemann Sum. Let us introduce $$ Y_t := \lim_{\lambda \rightarrow 0} \sum_{t_i} f(X_{t_i}) \Delta t, $$ where $0=t_0<t_1<t_2<\cdots<t_n<t_{n+1}=t$, $\lambda=\max(t_{j+1}-t_j)$ for $j=0 \ldots n$. Assume that the Markov process has a stationary state $\pi$, and $\pi_a$, $\pi_b$ are probabilities of the state $a$ and $b$ respectively. Since the terms added are infinite, thus $$ \Pr \{ \lim_{\lambda \rightarrow 0} \sum_{t_i} f(X_{t_i}) \Delta t\ =\pi_a f(X_a) + \pi_b f(X_b)\} = 1, $$ namely $$ \Pr \{ Y_t =\pi_a f(X_a) + \pi_b f(X_b)\} = 1. $$ Therefore, we can define $Y_t:=\pi_a f(X_a) + \pi_b f(X_b)$ with probability 1.


It's very strange that $Y_t$ is a deterministic real number, what's wrong with the above derivation?

Any suggestions? Thanks in advance!

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    $\begingroup$ How does one define $dX_t$ when $X$ is a (presumably, real-valued) two-state process? $\endgroup$
    – Did
    Jan 6, 2016 at 9:27
  • $\begingroup$ Sorry for the typo, it's $\mathrm dt$ instead of $\mathrm dX_t$.@Did $\endgroup$
    – robit
    Jan 6, 2016 at 9:51
  • $\begingroup$ What exactly do you mean by "solve"? What kind of result do you expect? $\endgroup$
    – saz
    Jan 6, 2016 at 13:04
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    $\begingroup$ Not an SDE anymore, then. A (strong) solution of the (new) equation is simply $$Y_t(\omega)=Y_0(\omega)+\int_0^tf(X_s(\omega))\,ds=Y_0(\omega)+tf(a)+{}{}{}{}{}{}L_t^b(\omega)(f(b)-f(a)),$$ where $$L_t^b(\omega)=\int_0^t\mathbf 1_{X_s(\omega)=b}\,ds.$$ $\endgroup$
    – Did
    Jan 6, 2016 at 14:57
  • $\begingroup$ I would like to obtain an explicit formula of $Y_t$. @saz $\endgroup$
    – robit
    Jan 7, 2016 at 3:24

1 Answer 1

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Let $a$,$b$ be the two states of $X_t$, and let $T_{a,t}$ be the time that $X_s$ spend in state $a$, for $s\in [0\ t]$. The answer for (1) is:

$Y_t = f(a)T_{a,t} + f(b)(t-T_{a,t})$

For (2), I'm not sure the question is even well defined, it seems that each $Z_t$ has some "randomness" which is totally unrelated to each other, as a result, a sample path $Z_.$ may not be measurable in the usual sense (for example, Borel or Lebesgue)

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