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Question about integral notation in a Markov process + how to evaluate said integral
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I'm reading Chapter 11 of Puterman's book on Markov Decision Processes (in particular, about continuous-time Markov processes). There's a lot of notation involved, but I've tried to distill the question. Puterman defines a function $Q(t,j|s,a)$, which, as a simple example, might equal
$$
Q(t,j|s,a)=frac{1}{4}(1-e^{-mu{t}})
$$
for some $mu>0$. The function $Q$ is a joint probability distribution in $tgeq0$ and $jin{S}$ for finite $S$ (in the example above, the product of the CDF of an exponential random variable with a constant). He then writes down the integral
$$
int_0^infty e^{-alpha{t}}Q(dt,j|s,a),
$$
and asserts that the value of this integral is $<1$. Puterman states "[w]e use $Q(dt,j|s,a)$ to represent a time-differential", but I don't know what this means in the context of integration.
Question 1 What kind of integral is this? Seems like Riemann-Stieltjes or Lebesgue, but I can't tell. I thought it might be strange notation for
$$
int_0^infty e^{-alpha{t}}Q(t,j|s,a)dt,
$$
but it seems that's not the case (as then the integral can easily be $geq1$).
Question 2 How do you evaluate such an integral? Is there e.g. a closed-form for the $Q$ defined above?
integration lebesgue-integral markov-process
asked Jan 14 at 22:27
David M.David M.
1,734418
$endgroup$
add a comment |
$begingroup$
I'm reading Chapter 11 of Puterman's book on Markov Decision Processes (in particular, about continuous-time Markov processes). There's a lot of notation involved, but I've tried to distill the question. Puterman defines a function $Q(t,j|s,a)$, which, as a simple example, might equal
$$
Q(t,j|s,a)=frac{1}{4}(1-e^{-mu{t}})
$$
for some $mu>0$. The function $Q$ is a joint probability distribution in $tgeq0$ and $jin{S}$ for finite $S$ (in the example above, the product of the CDF of an exponential random variable with a constant). He then writes down the integral
$$
int_0^infty e^{-alpha{t}}Q(dt,j|s,a),
$$
and asserts that the value of this integral is $<1$. Puterman states "[w]e use $Q(dt,j|s,a)$ to represent a time-differential", but I don't know what this means in the context of integration.
Question 1 What kind of integral is this? Seems like Riemann-Stieltjes or Lebesgue, but I can't tell. I thought it might be strange notation for
$$
int_0^infty e^{-alpha{t}}Q(t,j|s,a)dt,
$$
but it seems that's not the case (as then the integral can easily be $geq1$).
Question 2 How do you evaluate such an integral? Is there e.g. a closed-form for the $Q$ defined above?
integration lebesgue-integral markov-process
asked Jan 14 at 22:27
David M.David M.
1,734418
$endgroup$
$begingroup$
The integral would be $$ int_0^infty e^{-alpha t}frac14 mu e^{-mu t} mathsf dt = fracmu{4(alpha+mu)}. $$
$endgroup$
– Math1000
Jan 14 at 23:07
$begingroup$
@Math1000 Weird notation. Post as an answer (maybe with a small explanation?) and I’ll accept!
$endgroup$
– David M.
Jan 14 at 23:18
add a comment |
$begingroup$
I'm reading Chapter 11 of Puterman's book on Markov Decision Processes (in particular, about continuous-time Markov processes). There's a lot of notation involved, but I've tried to distill the question. Puterman defines a function $Q(t,j|s,a)$, which, as a simple example, might equal
$$
Q(t,j|s,a)=frac{1}{4}(1-e^{-mu{t}})
$$
for some $mu>0$. The function $Q$ is a joint probability distribution in $tgeq0$ and $jin{S}$ for finite $S$ (in the example above, the product of the CDF of an exponential random variable with a constant). He then writes down the integral
$$
int_0^infty e^{-alpha{t}}Q(dt,j|s,a),
$$
and asserts that the value of this integral is $<1$. Puterman states "[w]e use $Q(dt,j|s,a)$ to represent a time-differential", but I don't know what this means in the context of integration.
Question 1 What kind of integral is this? Seems like Riemann-Stieltjes or Lebesgue, but I can't tell. I thought it might be strange notation for
$$
int_0^infty e^{-alpha{t}}Q(t,j|s,a)dt,
$$
but it seems that's not the case (as then the integral can easily be $geq1$).
Question 2 How do you evaluate such an integral? Is there e.g. a closed-form for the $Q$ defined above?
integration lebesgue-integral markov-process
asked Jan 14 at 22:27
David M.David M.
1,734418
$endgroup$
I'm reading Chapter 11 of Puterman's book on Markov Decision Processes (in particular, about continuous-time Markov processes). There's a lot of notation involved, but I've tried to distill the question. Puterman defines a function $Q(t,j|s,a)$, which, as a simple example, might equal
$$
Q(t,j|s,a)=frac{1}{4}(1-e^{-mu{t}})
$$
for some $mu>0$. The function $Q$ is a joint probability distribution in $tgeq0$ and $jin{S}$ for finite $S$ (in the example above, the product of the CDF of an exponential random variable with a constant). He then writes down the integral
$$
int_0^infty e^{-alpha{t}}Q(dt,j|s,a),
$$
and asserts that the value of this integral is $<1$. Puterman states "[w]e use $Q(dt,j|s,a)$ to represent a time-differential", but I don't know what this means in the context of integration.
Question 1 What kind of integral is this? Seems like Riemann-Stieltjes or Lebesgue, but I can't tell. I thought it might be strange notation for
$$
int_0^infty e^{-alpha{t}}Q(t,j|s,a)dt,
$$
but it seems that's not the case (as then the integral can easily be $geq1$).
Question 2 How do you evaluate such an integral? Is there e.g. a closed-form for the $Q$ defined above?
integration lebesgue-integral markov-process
integration lebesgue-integral markov-process
asked Jan 14 at 22:27
David M.David M.
1,734418
asked Jan 14 at 22:27
David M.David M.
1,734418
edited Jan 14 at 22:32
David M.
asked Jan 14 at 22:27
David M.David M.
1,734418
asked Jan 14 at 22:27
David M.David M.
1,734418
asked Jan 14 at 22:27
David M.David M.
1,734418
1,734418
$begingroup$
The integral would be $$ int_0^infty e^{-alpha t}frac14 mu e^{-mu t} mathsf dt = fracmu{4(alpha+mu)}. $$
$endgroup$
– Math1000
Jan 14 at 23:07
$begingroup$
@Math1000 Weird notation. Post as an answer (maybe with a small explanation?) and I’ll accept!
$endgroup$
– David M.
Jan 14 at 23:18
add a comment |
$begingroup$
The integral would be $$ int_0^infty e^{-alpha t}frac14 mu e^{-mu t} mathsf dt = fracmu{4(alpha+mu)}. $$
$endgroup$
– Math1000
Jan 14 at 23:07
$begingroup$
@Math1000 Weird notation. Post as an answer (maybe with a small explanation?) and I’ll accept!
$endgroup$
– David M.
Jan 14 at 23:18
$begingroup$
The integral would be $$ int_0^infty e^{-alpha t}frac14 mu e^{-mu t} mathsf dt = fracmu{4(alpha+mu)}. $$
$endgroup$
– Math1000
Jan 14 at 23:07
$begingroup$
The integral would be $$ int_0^infty e^{-alpha t}frac14 mu e^{-mu t} mathsf dt = fracmu{4(alpha+mu)}. $$
$endgroup$
– Math1000
Jan 14 at 23:07
$begingroup$
@Math1000 Weird notation. Post as an answer (maybe with a small explanation?) and I’ll accept!
$endgroup$
– David M.
Jan 14 at 23:18
$begingroup$
@Math1000 Weird notation. Post as an answer (maybe with a small explanation?) and I’ll accept!
$endgroup$
– David M.
Jan 14 at 23:18
add a comment |
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$begingroup$
The integral would be $$ int_0^infty e^{-alpha t}frac14 mu e^{-mu t} mathsf dt = fracmu{4(alpha+mu)}. $$
$endgroup$
– Math1000
Jan 14 at 23:07
$begingroup$
@Math1000 Weird notation. Post as an answer (maybe with a small explanation?) and I’ll accept!
$endgroup$
– David M.
Jan 14 at 23:18