Markov property for unbounded function
$begingroup$
Let $(X_t)$ be a Markov process with respect to a filtration $mathcal{F}_t$. Assume that $P(X_t>0 , forall tgeq 0) = 1 $.
Denote $E_x$ the expectation under the measure where $X_0=x$.
Is it true that
$$
E_x[X_{t+s}^{-1} e^{z X_{t+s}} rvert mathcal{F_t}] = E_{X_t}[X_{s}^{-1} e^{z X_{s}} ]?
$$
for $zin {z in mathbb{C} : Re(z) <0 } $ given both expressions exist.
My guess is yes, since $f_n(x) = x^{-1} e^{z x } 1_{{x>1/n}}$ is bounded and I can do
$$
E_x[ f_n(X_{t+s} ) rvert mathcal{F_t}] = E_{X_t}[ f_n(X_s) ]
$$
for all $n$ and take the limit. Does that work?
probability-theory stochastic-processes conditional-expectation markov-process
asked Dec 17 '18 at 14:29
HenrikHenrik
87411227
$endgroup$
add a comment |
$begingroup$
Let $(X_t)$ be a Markov process with respect to a filtration $mathcal{F}_t$. Assume that $P(X_t>0 , forall tgeq 0) = 1 $.
Denote $E_x$ the expectation under the measure where $X_0=x$.
Is it true that
$$
E_x[X_{t+s}^{-1} e^{z X_{t+s}} rvert mathcal{F_t}] = E_{X_t}[X_{s}^{-1} e^{z X_{s}} ]?
$$
for $zin {z in mathbb{C} : Re(z) <0 } $ given both expressions exist.
My guess is yes, since $f_n(x) = x^{-1} e^{z x } 1_{{x>1/n}}$ is bounded and I can do
$$
E_x[ f_n(X_{t+s} ) rvert mathcal{F_t}] = E_{X_t}[ f_n(X_s) ]
$$
for all $n$ and take the limit. Does that work?
probability-theory stochastic-processes conditional-expectation markov-process
asked Dec 17 '18 at 14:29
HenrikHenrik
87411227
$endgroup$
add a comment |
$begingroup$
Let $(X_t)$ be a Markov process with respect to a filtration $mathcal{F}_t$. Assume that $P(X_t>0 , forall tgeq 0) = 1 $.
Denote $E_x$ the expectation under the measure where $X_0=x$.
Is it true that
$$
E_x[X_{t+s}^{-1} e^{z X_{t+s}} rvert mathcal{F_t}] = E_{X_t}[X_{s}^{-1} e^{z X_{s}} ]?
$$
for $zin {z in mathbb{C} : Re(z) <0 } $ given both expressions exist.
My guess is yes, since $f_n(x) = x^{-1} e^{z x } 1_{{x>1/n}}$ is bounded and I can do
$$
E_x[ f_n(X_{t+s} ) rvert mathcal{F_t}] = E_{X_t}[ f_n(X_s) ]
$$
for all $n$ and take the limit. Does that work?
probability-theory stochastic-processes conditional-expectation markov-process
asked Dec 17 '18 at 14:29
HenrikHenrik
87411227
$endgroup$
Let $(X_t)$ be a Markov process with respect to a filtration $mathcal{F}_t$. Assume that $P(X_t>0 , forall tgeq 0) = 1 $.
Denote $E_x$ the expectation under the measure where $X_0=x$.
Is it true that
$$
E_x[X_{t+s}^{-1} e^{z X_{t+s}} rvert mathcal{F_t}] = E_{X_t}[X_{s}^{-1} e^{z X_{s}} ]?
$$
for $zin {z in mathbb{C} : Re(z) <0 } $ given both expressions exist.
My guess is yes, since $f_n(x) = x^{-1} e^{z x } 1_{{x>1/n}}$ is bounded and I can do
$$
E_x[ f_n(X_{t+s} ) rvert mathcal{F_t}] = E_{X_t}[ f_n(X_s) ]
$$
for all $n$ and take the limit. Does that work?
probability-theory stochastic-processes conditional-expectation markov-process
probability-theory stochastic-processes conditional-expectation markov-process
asked Dec 17 '18 at 14:29
HenrikHenrik
87411227
asked Dec 17 '18 at 14:29
HenrikHenrik
87411227
edited Dec 17 '18 at 14:47
Henrik
asked Dec 17 '18 at 14:29
HenrikHenrik
87411227
asked Dec 17 '18 at 14:29
HenrikHenrik
87411227
asked Dec 17 '18 at 14:29
HenrikHenrik
87411227
87411227
add a comment |
add a comment |
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