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Markov Process, holding times are exponential
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im looking at the holding times for a markov process where c(.,.) are the transition rates. My question comes at the end of this working, can you explain when the sentance at the end means? I get that this quality (product to sum) holds for the exponential function but is it unique to the exponential function?
Let
$$tau_x:=inftgeq 0 hspace1.5mm : hspace1.5mm omega_t neq x $$
then if $sum_yin Omegac(x,y)>0$ it turns out that $tau_xsim Exp(sum_yin Omegac(x,y)) $. Proof:
Since our process is Markov it only cares about its current state and not how much time it has spent in its current state. i.e the holding time $tau$ has the 'loss of memory property $$P^x(tau_x>s+t hspace1.5mm | hspace1.5mm tau_x>s)=P^x(tau_x>s+t hspace1.5mm | hspace1.5mm omega_s=x)=P^x(tau_x>t).$$
Hence
$$P^x(tau_x>s+t)=P^x(tau_x>s+t|tau_x>s)P^x(tau_x>s)=P^x(tau_x>t)cdot P^x(tau_x>s)$$
????
Then this is the functional equation for the exponential and implies that $P^x(tau_x>t)=e^lambda t$
????
stochastic-processes exponential-function markov-chains
asked Jul 25 at 13:34
Monty
15212
add a comment |Â
up vote
1
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im looking at the holding times for a markov process where c(.,.) are the transition rates. My question comes at the end of this working, can you explain when the sentance at the end means? I get that this quality (product to sum) holds for the exponential function but is it unique to the exponential function?
Let
$$tau_x:=inftgeq 0 hspace1.5mm : hspace1.5mm omega_t neq x $$
then if $sum_yin Omegac(x,y)>0$ it turns out that $tau_xsim Exp(sum_yin Omegac(x,y)) $. Proof:
Since our process is Markov it only cares about its current state and not how much time it has spent in its current state. i.e the holding time $tau$ has the 'loss of memory property $$P^x(tau_x>s+t hspace1.5mm | hspace1.5mm tau_x>s)=P^x(tau_x>s+t hspace1.5mm | hspace1.5mm omega_s=x)=P^x(tau_x>t).$$
Hence
$$P^x(tau_x>s+t)=P^x(tau_x>s+t|tau_x>s)P^x(tau_x>s)=P^x(tau_x>t)cdot P^x(tau_x>s)$$
????
Then this is the functional equation for the exponential and implies that $P^x(tau_x>t)=e^lambda t$
????
stochastic-processes exponential-function markov-chains
asked Jul 25 at 13:34
Monty
15212
Let $g(t) = P^X(tau_x > t)$. The last inequality implies $g(t+s) = g(t)g(s)$ for all $s,tgeqslant 0$. This implies that $g$ is identically zero (which is impossible since $tau_x$ is finite a.s.) or $g(t) = e^lambda t$ for some real $lambda$.
– Math1000
Aug 1 at 13:48
inequality? you mean equality. why is it zero or e ? are there no other functions for which g(t+s)=g(t)g(s) holds?
– Monty
Aug 2 at 18:08
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
im looking at the holding times for a markov process where c(.,.) are the transition rates. My question comes at the end of this working, can you explain when the sentance at the end means? I get that this quality (product to sum) holds for the exponential function but is it unique to the exponential function?
Let
$$tau_x:=inftgeq 0 hspace1.5mm : hspace1.5mm omega_t neq x $$
then if $sum_yin Omegac(x,y)>0$ it turns out that $tau_xsim Exp(sum_yin Omegac(x,y)) $. Proof:
Since our process is Markov it only cares about its current state and not how much time it has spent in its current state. i.e the holding time $tau$ has the 'loss of memory property $$P^x(tau_x>s+t hspace1.5mm | hspace1.5mm tau_x>s)=P^x(tau_x>s+t hspace1.5mm | hspace1.5mm omega_s=x)=P^x(tau_x>t).$$
Hence
$$P^x(tau_x>s+t)=P^x(tau_x>s+t|tau_x>s)P^x(tau_x>s)=P^x(tau_x>t)cdot P^x(tau_x>s)$$
????
Then this is the functional equation for the exponential and implies that $P^x(tau_x>t)=e^lambda t$
????
stochastic-processes exponential-function markov-chains
asked Jul 25 at 13:34
Monty
15212
im looking at the holding times for a markov process where c(.,.) are the transition rates. My question comes at the end of this working, can you explain when the sentance at the end means? I get that this quality (product to sum) holds for the exponential function but is it unique to the exponential function?
Let
$$tau_x:=inftgeq 0 hspace1.5mm : hspace1.5mm omega_t neq x $$
then if $sum_yin Omegac(x,y)>0$ it turns out that $tau_xsim Exp(sum_yin Omegac(x,y)) $. Proof:
Since our process is Markov it only cares about its current state and not how much time it has spent in its current state. i.e the holding time $tau$ has the 'loss of memory property $$P^x(tau_x>s+t hspace1.5mm | hspace1.5mm tau_x>s)=P^x(tau_x>s+t hspace1.5mm | hspace1.5mm omega_s=x)=P^x(tau_x>t).$$
Hence
$$P^x(tau_x>s+t)=P^x(tau_x>s+t|tau_x>s)P^x(tau_x>s)=P^x(tau_x>t)cdot P^x(tau_x>s)$$
????
Then this is the functional equation for the exponential and implies that $P^x(tau_x>t)=e^lambda t$
????
stochastic-processes exponential-function markov-chains
asked Jul 25 at 13:34
Monty
15212
asked Jul 25 at 13:34
Monty
15212
asked Jul 25 at 13:34
Monty
15212
asked Jul 25 at 13:34
Monty
15212
15212
Let $g(t) = P^X(tau_x > t)$. The last inequality implies $g(t+s) = g(t)g(s)$ for all $s,tgeqslant 0$. This implies that $g$ is identically zero (which is impossible since $tau_x$ is finite a.s.) or $g(t) = e^lambda t$ for some real $lambda$.
– Math1000
Aug 1 at 13:48
inequality? you mean equality. why is it zero or e ? are there no other functions for which g(t+s)=g(t)g(s) holds?
– Monty
Aug 2 at 18:08
add a comment |Â
Let $g(t) = P^X(tau_x > t)$. The last inequality implies $g(t+s) = g(t)g(s)$ for all $s,tgeqslant 0$. This implies that $g$ is identically zero (which is impossible since $tau_x$ is finite a.s.) or $g(t) = e^lambda t$ for some real $lambda$.
– Math1000
Aug 1 at 13:48
inequality? you mean equality. why is it zero or e ? are there no other functions for which g(t+s)=g(t)g(s) holds?
– Monty
Aug 2 at 18:08
Let $g(t) = P^X(tau_x > t)$. The last inequality implies $g(t+s) = g(t)g(s)$ for all $s,tgeqslant 0$. This implies that $g$ is identically zero (which is impossible since $tau_x$ is finite a.s.) or $g(t) = e^lambda t$ for some real $lambda$.
– Math1000
Aug 1 at 13:48
Let $g(t) = P^X(tau_x > t)$. The last inequality implies $g(t+s) = g(t)g(s)$ for all $s,tgeqslant 0$. This implies that $g$ is identically zero (which is impossible since $tau_x$ is finite a.s.) or $g(t) = e^lambda t$ for some real $lambda$.
– Math1000
Aug 1 at 13:48
inequality? you mean equality. why is it zero or e ? are there no other functions for which g(t+s)=g(t)g(s) holds?
– Monty
Aug 2 at 18:08
inequality? you mean equality. why is it zero or e ? are there no other functions for which g(t+s)=g(t)g(s) holds?
– Monty
Aug 2 at 18:08
add a comment |Â
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Let $g(t) = P^X(tau_x > t)$. The last inequality implies $g(t+s) = g(t)g(s)$ for all $s,tgeqslant 0$. This implies that $g$ is identically zero (which is impossible since $tau_x$ is finite a.s.) or $g(t) = e^lambda t$ for some real $lambda$.
– Math1000
Aug 1 at 13:48
inequality? you mean equality. why is it zero or e ? are there no other functions for which g(t+s)=g(t)g(s) holds?
– Monty
Aug 2 at 18:08