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When the following expected value is finite?
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Let us consider the stochastic process $(X_t)_{tgeq0}$ that can be described by the following SDE:
$$
dX_t = alpha(X_t, t) dt + sigma(X_t, t) dB_t
$$
Now I consider the following expected value:
$$
mathbb{E}[e^{-int_t^T X_s ds}].
$$
My question is:
What conditions should be fulfilled by the functions $alpha(x,t)$ and $sigma(x,t)$ to make the above expectation finite?
Intuitively, I think that boundedness of $sigma(x, t)$ for negative $x$ is enough ensure finiteness of the above expected value, however I do not know how to prove it formally or even strengthen (or weaken) this condition.
stochastic-processes sde
asked Nov 16 at 9:31
MathMen
1167
add a comment |
up vote
1
down vote
favorite
Let us consider the stochastic process $(X_t)_{tgeq0}$ that can be described by the following SDE:
$$
dX_t = alpha(X_t, t) dt + sigma(X_t, t) dB_t
$$
Now I consider the following expected value:
$$
mathbb{E}[e^{-int_t^T X_s ds}].
$$
My question is:
What conditions should be fulfilled by the functions $alpha(x,t)$ and $sigma(x,t)$ to make the above expectation finite?
Intuitively, I think that boundedness of $sigma(x, t)$ for negative $x$ is enough ensure finiteness of the above expected value, however I do not know how to prove it formally or even strengthen (or weaken) this condition.
stochastic-processes sde
asked Nov 16 at 9:31
MathMen
1167
1. What exactly do you mean by a "bounded expectation"? Do you mean "finite expectation"? Or do you want to some boundedness w.r.t $T$..? 2. Do you have a specific application in mind? I would rather doubt that there are conditions which apply in this generality.
– saz
Nov 16 at 9:59
I mean finite expectation.I do not have any specific application in mind, but I remember that I read somewhere that if the process can take negative values and moreover $sigma(x, t)$ is not bounded for negative x, then this expectation I mentioned in the main post is not finite.
– MathMen
Nov 16 at 10:10
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let us consider the stochastic process $(X_t)_{tgeq0}$ that can be described by the following SDE:
$$
dX_t = alpha(X_t, t) dt + sigma(X_t, t) dB_t
$$
Now I consider the following expected value:
$$
mathbb{E}[e^{-int_t^T X_s ds}].
$$
My question is:
What conditions should be fulfilled by the functions $alpha(x,t)$ and $sigma(x,t)$ to make the above expectation finite?
Intuitively, I think that boundedness of $sigma(x, t)$ for negative $x$ is enough ensure finiteness of the above expected value, however I do not know how to prove it formally or even strengthen (or weaken) this condition.
stochastic-processes sde
asked Nov 16 at 9:31
MathMen
1167
Let us consider the stochastic process $(X_t)_{tgeq0}$ that can be described by the following SDE:
$$
dX_t = alpha(X_t, t) dt + sigma(X_t, t) dB_t
$$
Now I consider the following expected value:
$$
mathbb{E}[e^{-int_t^T X_s ds}].
$$
My question is:
What conditions should be fulfilled by the functions $alpha(x,t)$ and $sigma(x,t)$ to make the above expectation finite?
Intuitively, I think that boundedness of $sigma(x, t)$ for negative $x$ is enough ensure finiteness of the above expected value, however I do not know how to prove it formally or even strengthen (or weaken) this condition.
stochastic-processes sde
stochastic-processes sde
asked Nov 16 at 9:31
MathMen
1167
asked Nov 16 at 9:31
MathMen
1167
edited yesterday
asked Nov 16 at 9:31
MathMen
1167
asked Nov 16 at 9:31
MathMen
1167
asked Nov 16 at 9:31
MathMen
1167
1167
1. What exactly do you mean by a "bounded expectation"? Do you mean "finite expectation"? Or do you want to some boundedness w.r.t $T$..? 2. Do you have a specific application in mind? I would rather doubt that there are conditions which apply in this generality.
– saz
Nov 16 at 9:59
I mean finite expectation.I do not have any specific application in mind, but I remember that I read somewhere that if the process can take negative values and moreover $sigma(x, t)$ is not bounded for negative x, then this expectation I mentioned in the main post is not finite.
– MathMen
Nov 16 at 10:10
add a comment |
1. What exactly do you mean by a "bounded expectation"? Do you mean "finite expectation"? Or do you want to some boundedness w.r.t $T$..? 2. Do you have a specific application in mind? I would rather doubt that there are conditions which apply in this generality.
– saz
Nov 16 at 9:59
I mean finite expectation.I do not have any specific application in mind, but I remember that I read somewhere that if the process can take negative values and moreover $sigma(x, t)$ is not bounded for negative x, then this expectation I mentioned in the main post is not finite.
– MathMen
Nov 16 at 10:10
1. What exactly do you mean by a "bounded expectation"? Do you mean "finite expectation"? Or do you want to some boundedness w.r.t $T$..? 2. Do you have a specific application in mind? I would rather doubt that there are conditions which apply in this generality.
– saz
Nov 16 at 9:59
1. What exactly do you mean by a "bounded expectation"? Do you mean "finite expectation"? Or do you want to some boundedness w.r.t $T$..? 2. Do you have a specific application in mind? I would rather doubt that there are conditions which apply in this generality.
– saz
Nov 16 at 9:59
I mean finite expectation.I do not have any specific application in mind, but I remember that I read somewhere that if the process can take negative values and moreover $sigma(x, t)$ is not bounded for negative x, then this expectation I mentioned in the main post is not finite.
– MathMen
Nov 16 at 10:10
I mean finite expectation.I do not have any specific application in mind, but I remember that I read somewhere that if the process can take negative values and moreover $sigma(x, t)$ is not bounded for negative x, then this expectation I mentioned in the main post is not finite.
– MathMen
Nov 16 at 10:10
add a comment |
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1. What exactly do you mean by a "bounded expectation"? Do you mean "finite expectation"? Or do you want to some boundedness w.r.t $T$..? 2. Do you have a specific application in mind? I would rather doubt that there are conditions which apply in this generality.
– saz
Nov 16 at 9:59
I mean finite expectation.I do not have any specific application in mind, but I remember that I read somewhere that if the process can take negative values and moreover $sigma(x, t)$ is not bounded for negative x, then this expectation I mentioned in the main post is not finite.
– MathMen
Nov 16 at 10:10