robertness
2/21/2017 - 5:37 AM

Markov process derivation

Markov process derivation

\begin{align}
\frac{d}{dt}E_{\pi}(\rho(Y)) &= E_{\pi}(vX \rho(Y^o)) - E_{\pi}(\alpha \rho(Y)) \nonumber \\ 
&= vX\rho(N) - (vX + \alpha) E_{\pi}\rho(Y) \nonumber \\
&= vX\rho(N) - (vX + \alpha) \rho(N_y) \theta_y  
\end{align}

Setting $\frac{d}{dt}E_{\pi}(\rho(Y))$ to 0 provides the steady state expectation. 
\begin{align}
0 &= vX\rho(N_y) - (vX + \alpha) \rho(N_y) \theta_y  \nonumber \\
\theta_y &= \frac{vX}{vX + \alpha}  \\
&= \frac{\beta X}{1 + \beta X} = g(\beta X) 
\end{align}