Value iteration method of optimal stopping
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In Lawler's introdcution to Stochastic process p89~93, the value iteration method is given for homogeneous Markov chain: $u_1(x)$ equal to the payoff function $f(x)$ if $x$ is an absorbing state and otherwise equal the maximum value of $f(x)$. Then by the recursive relation:
$$u_{k+1}(x)=max {boldsymbol{P}u_k(x),,f(x) } $$
where $boldsymbol{P}$ is transition matrix.
Let $u(z)= lim_{n to infty}u_n(z) $ then we can get the optimal strategy $T^*$ by compare $u(x)$ and f(x): stop when $u(x)=f(x)$ and continue when $u(x) >f(x)$.
My problem is how $u(z)le v(z)$ is deduced, where $v(z)$ is the value function
defined by $$v(x) =max_T mathbb{E}[f(X_T)|X_0=x]$$.
$T$ is a stopping time random variable. Lawler just say that it is because $v(z)$ is the largest expected value over all choice of stopping sets. But it is not obvious how do we have $u(x) = mathbb{E}[f(X_{T*})|X_0=x]$
stochastic-processes markov-chains optimal-control
asked Dec 14 '18 at 0:33
RikeijinRikeijin
999
$endgroup$
add a comment |
$begingroup$
In Lawler's introdcution to Stochastic process p89~93, the value iteration method is given for homogeneous Markov chain: $u_1(x)$ equal to the payoff function $f(x)$ if $x$ is an absorbing state and otherwise equal the maximum value of $f(x)$. Then by the recursive relation:
$$u_{k+1}(x)=max {boldsymbol{P}u_k(x),,f(x) } $$
where $boldsymbol{P}$ is transition matrix.
Let $u(z)= lim_{n to infty}u_n(z) $ then we can get the optimal strategy $T^*$ by compare $u(x)$ and f(x): stop when $u(x)=f(x)$ and continue when $u(x) >f(x)$.
My problem is how $u(z)le v(z)$ is deduced, where $v(z)$ is the value function
defined by $$v(x) =max_T mathbb{E}[f(X_T)|X_0=x]$$.
$T$ is a stopping time random variable. Lawler just say that it is because $v(z)$ is the largest expected value over all choice of stopping sets. But it is not obvious how do we have $u(x) = mathbb{E}[f(X_{T*})|X_0=x]$
stochastic-processes markov-chains optimal-control
asked Dec 14 '18 at 0:33
RikeijinRikeijin
999
$endgroup$
add a comment |
$begingroup$
In Lawler's introdcution to Stochastic process p89~93, the value iteration method is given for homogeneous Markov chain: $u_1(x)$ equal to the payoff function $f(x)$ if $x$ is an absorbing state and otherwise equal the maximum value of $f(x)$. Then by the recursive relation:
$$u_{k+1}(x)=max {boldsymbol{P}u_k(x),,f(x) } $$
where $boldsymbol{P}$ is transition matrix.
Let $u(z)= lim_{n to infty}u_n(z) $ then we can get the optimal strategy $T^*$ by compare $u(x)$ and f(x): stop when $u(x)=f(x)$ and continue when $u(x) >f(x)$.
My problem is how $u(z)le v(z)$ is deduced, where $v(z)$ is the value function
defined by $$v(x) =max_T mathbb{E}[f(X_T)|X_0=x]$$.
$T$ is a stopping time random variable. Lawler just say that it is because $v(z)$ is the largest expected value over all choice of stopping sets. But it is not obvious how do we have $u(x) = mathbb{E}[f(X_{T*})|X_0=x]$
stochastic-processes markov-chains optimal-control
asked Dec 14 '18 at 0:33
RikeijinRikeijin
999
$endgroup$
In Lawler's introdcution to Stochastic process p89~93, the value iteration method is given for homogeneous Markov chain: $u_1(x)$ equal to the payoff function $f(x)$ if $x$ is an absorbing state and otherwise equal the maximum value of $f(x)$. Then by the recursive relation:
$$u_{k+1}(x)=max {boldsymbol{P}u_k(x),,f(x) } $$
where $boldsymbol{P}$ is transition matrix.
Let $u(z)= lim_{n to infty}u_n(z) $ then we can get the optimal strategy $T^*$ by compare $u(x)$ and f(x): stop when $u(x)=f(x)$ and continue when $u(x) >f(x)$.
My problem is how $u(z)le v(z)$ is deduced, where $v(z)$ is the value function
defined by $$v(x) =max_T mathbb{E}[f(X_T)|X_0=x]$$.
$T$ is a stopping time random variable. Lawler just say that it is because $v(z)$ is the largest expected value over all choice of stopping sets. But it is not obvious how do we have $u(x) = mathbb{E}[f(X_{T*})|X_0=x]$
stochastic-processes markov-chains optimal-control
stochastic-processes markov-chains optimal-control
asked Dec 14 '18 at 0:33
RikeijinRikeijin
999
asked Dec 14 '18 at 0:33
RikeijinRikeijin
999
edited Dec 14 '18 at 0:50
Rikeijin
asked Dec 14 '18 at 0:33
RikeijinRikeijin
999
asked Dec 14 '18 at 0:33
RikeijinRikeijin
999
asked Dec 14 '18 at 0:33
RikeijinRikeijin
999
999
add a comment |
add a comment |
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