Conditions for two linearly independent solutions of the ODE representation of a diffusion to be global...
$begingroup$
Let $W_t$ be a Brownian Motion, $X_t$ and $M(t)$ be two stochastic processes:
$$X_t=adt+bdW_t$$
$$M(t)=e^{-lambda t}Phi(X_t)$$
where $Phi$ is a deterministic function determined from the requirement that $M_t$ is a martingale and $X_t$ assumed to have values in the interval $[L,U]$.
By applying Ito's lemma and fixing the zero drift, we get
$$lambdaPhi(X_t)=aPhi^prime (X_t)+frac{b^2}{2}Phi^{primeprime}(X_t)$$
which can be represented by the ODE $lambdaPhi(x)=aPhi^prime(x)+frac{b^2}{2}Phi^{primeprime}(x)$, which has two linearly independent solutions, $phi_1(x)$ and $phi_2(x)$, yielding two local martingales: $M_{1,2}(t)=e^{-lambda t}phi_{1,2}(X_t)$.
What else do I need in order to get that $M_{1,2}(t)$ are global martingales?
Suppose that $X_t$ does not have any singular point. Does this suffice to get that $M_{1,2}(t)$ are bounded?
probability-theory stochastic-processes stochastic-calculus
asked Dec 15 '18 at 20:41
FunnyBuzerFunnyBuzer
24919
$endgroup$
add a comment |
$begingroup$
Let $W_t$ be a Brownian Motion, $X_t$ and $M(t)$ be two stochastic processes:
$$X_t=adt+bdW_t$$
$$M(t)=e^{-lambda t}Phi(X_t)$$
where $Phi$ is a deterministic function determined from the requirement that $M_t$ is a martingale and $X_t$ assumed to have values in the interval $[L,U]$.
By applying Ito's lemma and fixing the zero drift, we get
$$lambdaPhi(X_t)=aPhi^prime (X_t)+frac{b^2}{2}Phi^{primeprime}(X_t)$$
which can be represented by the ODE $lambdaPhi(x)=aPhi^prime(x)+frac{b^2}{2}Phi^{primeprime}(x)$, which has two linearly independent solutions, $phi_1(x)$ and $phi_2(x)$, yielding two local martingales: $M_{1,2}(t)=e^{-lambda t}phi_{1,2}(X_t)$.
What else do I need in order to get that $M_{1,2}(t)$ are global martingales?
Suppose that $X_t$ does not have any singular point. Does this suffice to get that $M_{1,2}(t)$ are bounded?
probability-theory stochastic-processes stochastic-calculus
asked Dec 15 '18 at 20:41
FunnyBuzerFunnyBuzer
24919
$endgroup$
$begingroup$
If $X$ is bounded as you assume, then $M$, being a bounded local martingale, is a true martingale. But unless $b=0$, how can $X$ be bounded?
$endgroup$
– AddSup
Dec 16 '18 at 6:33
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@AddSup that's why I was thinking of a condition on X. This is a hitting probability exercise where the two bounds are $L$ and $U$, respectively. Absence of singular point should imply that the function is smooth.
$endgroup$
– FunnyBuzer
Dec 17 '18 at 8:50
add a comment |
$begingroup$
Let $W_t$ be a Brownian Motion, $X_t$ and $M(t)$ be two stochastic processes:
$$X_t=adt+bdW_t$$
$$M(t)=e^{-lambda t}Phi(X_t)$$
where $Phi$ is a deterministic function determined from the requirement that $M_t$ is a martingale and $X_t$ assumed to have values in the interval $[L,U]$.
By applying Ito's lemma and fixing the zero drift, we get
$$lambdaPhi(X_t)=aPhi^prime (X_t)+frac{b^2}{2}Phi^{primeprime}(X_t)$$
which can be represented by the ODE $lambdaPhi(x)=aPhi^prime(x)+frac{b^2}{2}Phi^{primeprime}(x)$, which has two linearly independent solutions, $phi_1(x)$ and $phi_2(x)$, yielding two local martingales: $M_{1,2}(t)=e^{-lambda t}phi_{1,2}(X_t)$.
What else do I need in order to get that $M_{1,2}(t)$ are global martingales?
Suppose that $X_t$ does not have any singular point. Does this suffice to get that $M_{1,2}(t)$ are bounded?
probability-theory stochastic-processes stochastic-calculus
asked Dec 15 '18 at 20:41
FunnyBuzerFunnyBuzer
24919
$endgroup$
Let $W_t$ be a Brownian Motion, $X_t$ and $M(t)$ be two stochastic processes:
$$X_t=adt+bdW_t$$
$$M(t)=e^{-lambda t}Phi(X_t)$$
where $Phi$ is a deterministic function determined from the requirement that $M_t$ is a martingale and $X_t$ assumed to have values in the interval $[L,U]$.
By applying Ito's lemma and fixing the zero drift, we get
$$lambdaPhi(X_t)=aPhi^prime (X_t)+frac{b^2}{2}Phi^{primeprime}(X_t)$$
which can be represented by the ODE $lambdaPhi(x)=aPhi^prime(x)+frac{b^2}{2}Phi^{primeprime}(x)$, which has two linearly independent solutions, $phi_1(x)$ and $phi_2(x)$, yielding two local martingales: $M_{1,2}(t)=e^{-lambda t}phi_{1,2}(X_t)$.
What else do I need in order to get that $M_{1,2}(t)$ are global martingales?
Suppose that $X_t$ does not have any singular point. Does this suffice to get that $M_{1,2}(t)$ are bounded?
probability-theory stochastic-processes stochastic-calculus
probability-theory stochastic-processes stochastic-calculus
asked Dec 15 '18 at 20:41
FunnyBuzerFunnyBuzer
24919
asked Dec 15 '18 at 20:41
FunnyBuzerFunnyBuzer
24919
asked Dec 15 '18 at 20:41
FunnyBuzerFunnyBuzer
24919
asked Dec 15 '18 at 20:41
FunnyBuzerFunnyBuzer
24919
asked Dec 15 '18 at 20:41
FunnyBuzerFunnyBuzer
24919
24919
$begingroup$
If $X$ is bounded as you assume, then $M$, being a bounded local martingale, is a true martingale. But unless $b=0$, how can $X$ be bounded?
$endgroup$
– AddSup
Dec 16 '18 at 6:33
$begingroup$
@AddSup that's why I was thinking of a condition on X. This is a hitting probability exercise where the two bounds are $L$ and $U$, respectively. Absence of singular point should imply that the function is smooth.
$endgroup$
– FunnyBuzer
Dec 17 '18 at 8:50
add a comment |
$begingroup$
If $X$ is bounded as you assume, then $M$, being a bounded local martingale, is a true martingale. But unless $b=0$, how can $X$ be bounded?
$endgroup$
– AddSup
Dec 16 '18 at 6:33
$begingroup$
@AddSup that's why I was thinking of a condition on X. This is a hitting probability exercise where the two bounds are $L$ and $U$, respectively. Absence of singular point should imply that the function is smooth.
$endgroup$
– FunnyBuzer
Dec 17 '18 at 8:50
$begingroup$
If $X$ is bounded as you assume, then $M$, being a bounded local martingale, is a true martingale. But unless $b=0$, how can $X$ be bounded?
$endgroup$
– AddSup
Dec 16 '18 at 6:33
$begingroup$
If $X$ is bounded as you assume, then $M$, being a bounded local martingale, is a true martingale. But unless $b=0$, how can $X$ be bounded?
$endgroup$
– AddSup
Dec 16 '18 at 6:33
$begingroup$
@AddSup that's why I was thinking of a condition on X. This is a hitting probability exercise where the two bounds are $L$ and $U$, respectively. Absence of singular point should imply that the function is smooth.
$endgroup$
– FunnyBuzer
Dec 17 '18 at 8:50
$begingroup$
@AddSup that's why I was thinking of a condition on X. This is a hitting probability exercise where the two bounds are $L$ and $U$, respectively. Absence of singular point should imply that the function is smooth.
$endgroup$
– FunnyBuzer
Dec 17 '18 at 8:50
add a comment |
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$begingroup$
If $X$ is bounded as you assume, then $M$, being a bounded local martingale, is a true martingale. But unless $b=0$, how can $X$ be bounded?
$endgroup$
– AddSup
Dec 16 '18 at 6:33
$begingroup$
@AddSup that's why I was thinking of a condition on X. This is a hitting probability exercise where the two bounds are $L$ and $U$, respectively. Absence of singular point should imply that the function is smooth.
$endgroup$
– FunnyBuzer
Dec 17 '18 at 8:50