Backward Kolmogorov equation to find probability
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From lecture notes in a course on SDE's. We are tasked with using the backward Kolmogorov equation to find.
$mathbb{P}^{X_t=x}left(X_Tgeq2 right)$
I am confused by the terminology here. We are looking for the probability that a process at time $t$ is equal to $x$, conditioned on the terminal value being equal to $2$. Do we then solve for the density in the backward equation and integrate over the time interval?
probability-theory stochastic-processes stochastic-calculus markov-process sde
asked Nov 29 '18 at 18:22
thaumoctopusthaumoctopus
9519
$endgroup$
add a comment |
$begingroup$
From lecture notes in a course on SDE's. We are tasked with using the backward Kolmogorov equation to find.
$mathbb{P}^{X_t=x}left(X_Tgeq2 right)$
I am confused by the terminology here. We are looking for the probability that a process at time $t$ is equal to $x$, conditioned on the terminal value being equal to $2$. Do we then solve for the density in the backward equation and integrate over the time interval?
probability-theory stochastic-processes stochastic-calculus markov-process sde
asked Nov 29 '18 at 18:22
thaumoctopusthaumoctopus
9519
$endgroup$
1
$begingroup$
$P^{X_t=x}(X_Tge 2)$ denotes the probability that $X_Tge 2$ given $X_t=x$ (not the probability that $X_t=x$ given $X_Tge 2$).
$endgroup$
– AddSup
Nov 30 '18 at 6:25
$begingroup$
Thanks for the clarification. I would then solve for the density in the backwards equation, which would give me a probability density $k(x,t)$, expressed as a function of the initial conditions. Would I then just integrate the relevant area in the probability space?
$endgroup$
– thaumoctopus
Dec 3 '18 at 20:52
add a comment |
$begingroup$
From lecture notes in a course on SDE's. We are tasked with using the backward Kolmogorov equation to find.
$mathbb{P}^{X_t=x}left(X_Tgeq2 right)$
I am confused by the terminology here. We are looking for the probability that a process at time $t$ is equal to $x$, conditioned on the terminal value being equal to $2$. Do we then solve for the density in the backward equation and integrate over the time interval?
probability-theory stochastic-processes stochastic-calculus markov-process sde
asked Nov 29 '18 at 18:22
thaumoctopusthaumoctopus
9519
$endgroup$
From lecture notes in a course on SDE's. We are tasked with using the backward Kolmogorov equation to find.
$mathbb{P}^{X_t=x}left(X_Tgeq2 right)$
I am confused by the terminology here. We are looking for the probability that a process at time $t$ is equal to $x$, conditioned on the terminal value being equal to $2$. Do we then solve for the density in the backward equation and integrate over the time interval?
probability-theory stochastic-processes stochastic-calculus markov-process sde
probability-theory stochastic-processes stochastic-calculus markov-process sde
asked Nov 29 '18 at 18:22
thaumoctopusthaumoctopus
9519
asked Nov 29 '18 at 18:22
thaumoctopusthaumoctopus
9519
asked Nov 29 '18 at 18:22
thaumoctopusthaumoctopus
9519
asked Nov 29 '18 at 18:22
thaumoctopusthaumoctopus
9519
asked Nov 29 '18 at 18:22
thaumoctopusthaumoctopus
9519
9519
1
$begingroup$
$P^{X_t=x}(X_Tge 2)$ denotes the probability that $X_Tge 2$ given $X_t=x$ (not the probability that $X_t=x$ given $X_Tge 2$).
$endgroup$
– AddSup
Nov 30 '18 at 6:25
$begingroup$
Thanks for the clarification. I would then solve for the density in the backwards equation, which would give me a probability density $k(x,t)$, expressed as a function of the initial conditions. Would I then just integrate the relevant area in the probability space?
$endgroup$
– thaumoctopus
Dec 3 '18 at 20:52
add a comment |
1
$begingroup$
$P^{X_t=x}(X_Tge 2)$ denotes the probability that $X_Tge 2$ given $X_t=x$ (not the probability that $X_t=x$ given $X_Tge 2$).
$endgroup$
– AddSup
Nov 30 '18 at 6:25
$begingroup$
Thanks for the clarification. I would then solve for the density in the backwards equation, which would give me a probability density $k(x,t)$, expressed as a function of the initial conditions. Would I then just integrate the relevant area in the probability space?
$endgroup$
– thaumoctopus
Dec 3 '18 at 20:52
1
1
$begingroup$
$P^{X_t=x}(X_Tge 2)$ denotes the probability that $X_Tge 2$ given $X_t=x$ (not the probability that $X_t=x$ given $X_Tge 2$).
$endgroup$
– AddSup
Nov 30 '18 at 6:25
$begingroup$
$P^{X_t=x}(X_Tge 2)$ denotes the probability that $X_Tge 2$ given $X_t=x$ (not the probability that $X_t=x$ given $X_Tge 2$).
$endgroup$
– AddSup
Nov 30 '18 at 6:25
$begingroup$
Thanks for the clarification. I would then solve for the density in the backwards equation, which would give me a probability density $k(x,t)$, expressed as a function of the initial conditions. Would I then just integrate the relevant area in the probability space?
$endgroup$
– thaumoctopus
Dec 3 '18 at 20:52
$begingroup$
Thanks for the clarification. I would then solve for the density in the backwards equation, which would give me a probability density $k(x,t)$, expressed as a function of the initial conditions. Would I then just integrate the relevant area in the probability space?
$endgroup$
– thaumoctopus
Dec 3 '18 at 20:52
add a comment |
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1
$begingroup$
$P^{X_t=x}(X_Tge 2)$ denotes the probability that $X_Tge 2$ given $X_t=x$ (not the probability that $X_t=x$ given $X_Tge 2$).
$endgroup$
– AddSup
Nov 30 '18 at 6:25
$begingroup$
Thanks for the clarification. I would then solve for the density in the backwards equation, which would give me a probability density $k(x,t)$, expressed as a function of the initial conditions. Would I then just integrate the relevant area in the probability space?
$endgroup$
– thaumoctopus
Dec 3 '18 at 20:52