Mixing
Stochastic differential equation with respect to general stochastic process
$begingroup$
I am a beginner with stochastic processes and I want to clarify some ideas that I just learnt. Say I have a stochastic process:
$$X_t = X_0+ int_0^t sigma_s dW_s $$
where $W_s$ is a standard Brownian Motion. The Ito/Stochastic Differential Equation for the above can be written as
$$dX_t = sigma_t dW_t $$
But what if we have an integral with respect to a stochastic process that is not a standard Brownian Motion, e.g., say we have
$$V_t = V_0 + int_0^t sigma_s dI_s $$
where $I_s$ is some general stochastic process. Can I write the SDE as
$$dV_t = sigma_t dI_t $$?
stochastic-processes stochastic-integrals
asked Jan 11 at 13:23
user40333user40333
455517
$endgroup$
add a comment |
$begingroup$
I am a beginner with stochastic processes and I want to clarify some ideas that I just learnt. Say I have a stochastic process:
$$X_t = X_0+ int_0^t sigma_s dW_s $$
where $W_s$ is a standard Brownian Motion. The Ito/Stochastic Differential Equation for the above can be written as
$$dX_t = sigma_t dW_t $$
But what if we have an integral with respect to a stochastic process that is not a standard Brownian Motion, e.g., say we have
$$V_t = V_0 + int_0^t sigma_s dI_s $$
where $I_s$ is some general stochastic process. Can I write the SDE as
$$dV_t = sigma_t dI_t $$?
stochastic-processes stochastic-integrals
asked Jan 11 at 13:23
user40333user40333
455517
$endgroup$
$begingroup$
First you want to make sure that the integral with respect to your stochastic process exists
$endgroup$
– Gâteau-Gallois
Jan 11 at 13:52
$begingroup$
Ok, and under the assumption that it does exist, am I correct in my thinking?
$endgroup$
– user40333
Jan 11 at 14:07
add a comment |
$begingroup$
I am a beginner with stochastic processes and I want to clarify some ideas that I just learnt. Say I have a stochastic process:
$$X_t = X_0+ int_0^t sigma_s dW_s $$
where $W_s$ is a standard Brownian Motion. The Ito/Stochastic Differential Equation for the above can be written as
$$dX_t = sigma_t dW_t $$
But what if we have an integral with respect to a stochastic process that is not a standard Brownian Motion, e.g., say we have
$$V_t = V_0 + int_0^t sigma_s dI_s $$
where $I_s$ is some general stochastic process. Can I write the SDE as
$$dV_t = sigma_t dI_t $$?
stochastic-processes stochastic-integrals
asked Jan 11 at 13:23
user40333user40333
455517
$endgroup$
I am a beginner with stochastic processes and I want to clarify some ideas that I just learnt. Say I have a stochastic process:
$$X_t = X_0+ int_0^t sigma_s dW_s $$
where $W_s$ is a standard Brownian Motion. The Ito/Stochastic Differential Equation for the above can be written as
$$dX_t = sigma_t dW_t $$
But what if we have an integral with respect to a stochastic process that is not a standard Brownian Motion, e.g., say we have
$$V_t = V_0 + int_0^t sigma_s dI_s $$
where $I_s$ is some general stochastic process. Can I write the SDE as
$$dV_t = sigma_t dI_t $$?
stochastic-processes stochastic-integrals
stochastic-processes stochastic-integrals
asked Jan 11 at 13:23
user40333user40333
455517
asked Jan 11 at 13:23
user40333user40333
455517
asked Jan 11 at 13:23
user40333user40333
455517
asked Jan 11 at 13:23
user40333user40333
455517
asked Jan 11 at 13:23
user40333user40333
455517
455517
$begingroup$
First you want to make sure that the integral with respect to your stochastic process exists
$endgroup$
– Gâteau-Gallois
Jan 11 at 13:52
$begingroup$
Ok, and under the assumption that it does exist, am I correct in my thinking?
$endgroup$
– user40333
Jan 11 at 14:07
add a comment |
$begingroup$
First you want to make sure that the integral with respect to your stochastic process exists
$endgroup$
– Gâteau-Gallois
Jan 11 at 13:52
$begingroup$
Ok, and under the assumption that it does exist, am I correct in my thinking?
$endgroup$
– user40333
Jan 11 at 14:07
$begingroup$
First you want to make sure that the integral with respect to your stochastic process exists
$endgroup$
– Gâteau-Gallois
Jan 11 at 13:52
$begingroup$
First you want to make sure that the integral with respect to your stochastic process exists
$endgroup$
– Gâteau-Gallois
Jan 11 at 13:52
$begingroup$
Ok, and under the assumption that it does exist, am I correct in my thinking?
$endgroup$
– user40333
Jan 11 at 14:07
$begingroup$
Ok, and under the assumption that it does exist, am I correct in my thinking?
$endgroup$
– user40333
Jan 11 at 14:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can if you want. Soon you will learn Ito's lemma, which states that
$$
{rm d}f(X_t)=f'(X_t),{rm d}X_t+frac{1}{2}f''(X_t),{rm d}left<Xright>_t
$$
for all Ito processes $X_t$ and twice continuously differentiable functions $f$, where $left<Xright>_t$ denotes the quadratic variation process of $X_t$. Here ${rm d}X_t$ appears explicitly, probably as you want. However, provided that $X_t$ is an Ito process, there must exist some predictable processes $mu_t$ and $sigma_t$, such that
$$
{rm d}X_t=mu_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $W_t$ denotes the Wiener process. Therefore, while ${rm d}X_t$ appears explicitly, you may re-express it so that only ${rm d}t$ and ${rm d}W_t$ remain.
Nevertheless, the above argument is only for Ito processes. For general stochastic processes, you may still define its Ito stochastic integral if it exists. However, it is not always funny to consider the "most general" case - beyond curiosity, it is more realistic to consider, at least at the very beginning, those that help to shape our world.
Among all stochastic processes, martingales have witnessed their wide applications in physics and finance. Later you will learn the martingale representation theorem, which states that every martingale is actually an Ito process without drift, i.e., for any given martingale $X_t$, there exists some predictable process $sigma_t$, such that
$$
{rm d}X_t=sigma_t,{rm d}W_t.
$$
See? Martingales are defined independent from Brownian motions. However, they can be represented by the standard Brownian motion. This beauty partly explains why we focus on Ito processes: they are, at least for the moment, general enough to model most of the cases in physics and finance.
Finally, if you are math-majored, you probably will also learn Levy's characterization of Brownian motion, which provides an equivalent definition for the Brownian motion. While this theorem itself might be abstract, one of its profound corollary goes that
$$
X_t=int_0^tsigma_s,{rm d}W_s
$$
has the same distribution as
$$
B_{int_0^tsigma_s^2,{rm d}s},
$$
where, in case of ambiguity, $B_t$ here denotes another Wiener process. This conclusion is extremely powerful. Combined with the martingale representation theorem, it tells that all martingales are no more than a time-changed Brownian motion. Concisely, every martingale observes a nature in the Brownian motion. Perhaps this is why we choose to pay particular attention to the Brownian motion: we find martingales are useful, and martingales are closely related to Brownian motions in the sense of both representation and nature.
That's it! Hope this explanation could be somewhat helpful for you.
answered Jan 11 at 15:54
hypernovahypernova
4,789414
$endgroup$
$begingroup$
This is super helpful, thank you so much!
$endgroup$
– user40333
Jan 12 at 2:43
$begingroup$
@user40333: My pleasure. Best wishes for your stochastic process journey!
$endgroup$
– hypernova
Jan 12 at 2:55
add a comment |
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$begingroup$
You can if you want. Soon you will learn Ito's lemma, which states that
$$
{rm d}f(X_t)=f'(X_t),{rm d}X_t+frac{1}{2}f''(X_t),{rm d}left<Xright>_t
$$
for all Ito processes $X_t$ and twice continuously differentiable functions $f$, where $left<Xright>_t$ denotes the quadratic variation process of $X_t$. Here ${rm d}X_t$ appears explicitly, probably as you want. However, provided that $X_t$ is an Ito process, there must exist some predictable processes $mu_t$ and $sigma_t$, such that
$$
{rm d}X_t=mu_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $W_t$ denotes the Wiener process. Therefore, while ${rm d}X_t$ appears explicitly, you may re-express it so that only ${rm d}t$ and ${rm d}W_t$ remain.
Nevertheless, the above argument is only for Ito processes. For general stochastic processes, you may still define its Ito stochastic integral if it exists. However, it is not always funny to consider the "most general" case - beyond curiosity, it is more realistic to consider, at least at the very beginning, those that help to shape our world.
Among all stochastic processes, martingales have witnessed their wide applications in physics and finance. Later you will learn the martingale representation theorem, which states that every martingale is actually an Ito process without drift, i.e., for any given martingale $X_t$, there exists some predictable process $sigma_t$, such that
$$
{rm d}X_t=sigma_t,{rm d}W_t.
$$
See? Martingales are defined independent from Brownian motions. However, they can be represented by the standard Brownian motion. This beauty partly explains why we focus on Ito processes: they are, at least for the moment, general enough to model most of the cases in physics and finance.
Finally, if you are math-majored, you probably will also learn Levy's characterization of Brownian motion, which provides an equivalent definition for the Brownian motion. While this theorem itself might be abstract, one of its profound corollary goes that
$$
X_t=int_0^tsigma_s,{rm d}W_s
$$
has the same distribution as
$$
B_{int_0^tsigma_s^2,{rm d}s},
$$
where, in case of ambiguity, $B_t$ here denotes another Wiener process. This conclusion is extremely powerful. Combined with the martingale representation theorem, it tells that all martingales are no more than a time-changed Brownian motion. Concisely, every martingale observes a nature in the Brownian motion. Perhaps this is why we choose to pay particular attention to the Brownian motion: we find martingales are useful, and martingales are closely related to Brownian motions in the sense of both representation and nature.
That's it! Hope this explanation could be somewhat helpful for you.
answered Jan 11 at 15:54
hypernovahypernova
4,789414
$endgroup$
$begingroup$
This is super helpful, thank you so much!
$endgroup$
– user40333
Jan 12 at 2:43
$begingroup$
@user40333: My pleasure. Best wishes for your stochastic process journey!
$endgroup$
– hypernova
Jan 12 at 2:55
add a comment |
$begingroup$
You can if you want. Soon you will learn Ito's lemma, which states that
$$
{rm d}f(X_t)=f'(X_t),{rm d}X_t+frac{1}{2}f''(X_t),{rm d}left<Xright>_t
$$
for all Ito processes $X_t$ and twice continuously differentiable functions $f$, where $left<Xright>_t$ denotes the quadratic variation process of $X_t$. Here ${rm d}X_t$ appears explicitly, probably as you want. However, provided that $X_t$ is an Ito process, there must exist some predictable processes $mu_t$ and $sigma_t$, such that
$$
{rm d}X_t=mu_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $W_t$ denotes the Wiener process. Therefore, while ${rm d}X_t$ appears explicitly, you may re-express it so that only ${rm d}t$ and ${rm d}W_t$ remain.
Nevertheless, the above argument is only for Ito processes. For general stochastic processes, you may still define its Ito stochastic integral if it exists. However, it is not always funny to consider the "most general" case - beyond curiosity, it is more realistic to consider, at least at the very beginning, those that help to shape our world.
Among all stochastic processes, martingales have witnessed their wide applications in physics and finance. Later you will learn the martingale representation theorem, which states that every martingale is actually an Ito process without drift, i.e., for any given martingale $X_t$, there exists some predictable process $sigma_t$, such that
$$
{rm d}X_t=sigma_t,{rm d}W_t.
$$
See? Martingales are defined independent from Brownian motions. However, they can be represented by the standard Brownian motion. This beauty partly explains why we focus on Ito processes: they are, at least for the moment, general enough to model most of the cases in physics and finance.
Finally, if you are math-majored, you probably will also learn Levy's characterization of Brownian motion, which provides an equivalent definition for the Brownian motion. While this theorem itself might be abstract, one of its profound corollary goes that
$$
X_t=int_0^tsigma_s,{rm d}W_s
$$
has the same distribution as
$$
B_{int_0^tsigma_s^2,{rm d}s},
$$
where, in case of ambiguity, $B_t$ here denotes another Wiener process. This conclusion is extremely powerful. Combined with the martingale representation theorem, it tells that all martingales are no more than a time-changed Brownian motion. Concisely, every martingale observes a nature in the Brownian motion. Perhaps this is why we choose to pay particular attention to the Brownian motion: we find martingales are useful, and martingales are closely related to Brownian motions in the sense of both representation and nature.
That's it! Hope this explanation could be somewhat helpful for you.
answered Jan 11 at 15:54
hypernovahypernova
4,789414
$endgroup$
$begingroup$
This is super helpful, thank you so much!
$endgroup$
– user40333
Jan 12 at 2:43
$begingroup$
@user40333: My pleasure. Best wishes for your stochastic process journey!
$endgroup$
– hypernova
Jan 12 at 2:55
add a comment |
$begingroup$
You can if you want. Soon you will learn Ito's lemma, which states that
$$
{rm d}f(X_t)=f'(X_t),{rm d}X_t+frac{1}{2}f''(X_t),{rm d}left<Xright>_t
$$
for all Ito processes $X_t$ and twice continuously differentiable functions $f$, where $left<Xright>_t$ denotes the quadratic variation process of $X_t$. Here ${rm d}X_t$ appears explicitly, probably as you want. However, provided that $X_t$ is an Ito process, there must exist some predictable processes $mu_t$ and $sigma_t$, such that
$$
{rm d}X_t=mu_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $W_t$ denotes the Wiener process. Therefore, while ${rm d}X_t$ appears explicitly, you may re-express it so that only ${rm d}t$ and ${rm d}W_t$ remain.
Nevertheless, the above argument is only for Ito processes. For general stochastic processes, you may still define its Ito stochastic integral if it exists. However, it is not always funny to consider the "most general" case - beyond curiosity, it is more realistic to consider, at least at the very beginning, those that help to shape our world.
Among all stochastic processes, martingales have witnessed their wide applications in physics and finance. Later you will learn the martingale representation theorem, which states that every martingale is actually an Ito process without drift, i.e., for any given martingale $X_t$, there exists some predictable process $sigma_t$, such that
$$
{rm d}X_t=sigma_t,{rm d}W_t.
$$
See? Martingales are defined independent from Brownian motions. However, they can be represented by the standard Brownian motion. This beauty partly explains why we focus on Ito processes: they are, at least for the moment, general enough to model most of the cases in physics and finance.
Finally, if you are math-majored, you probably will also learn Levy's characterization of Brownian motion, which provides an equivalent definition for the Brownian motion. While this theorem itself might be abstract, one of its profound corollary goes that
$$
X_t=int_0^tsigma_s,{rm d}W_s
$$
has the same distribution as
$$
B_{int_0^tsigma_s^2,{rm d}s},
$$
where, in case of ambiguity, $B_t$ here denotes another Wiener process. This conclusion is extremely powerful. Combined with the martingale representation theorem, it tells that all martingales are no more than a time-changed Brownian motion. Concisely, every martingale observes a nature in the Brownian motion. Perhaps this is why we choose to pay particular attention to the Brownian motion: we find martingales are useful, and martingales are closely related to Brownian motions in the sense of both representation and nature.
That's it! Hope this explanation could be somewhat helpful for you.
answered Jan 11 at 15:54
hypernovahypernova
4,789414
$endgroup$
You can if you want. Soon you will learn Ito's lemma, which states that
$$
{rm d}f(X_t)=f'(X_t),{rm d}X_t+frac{1}{2}f''(X_t),{rm d}left<Xright>_t
$$
for all Ito processes $X_t$ and twice continuously differentiable functions $f$, where $left<Xright>_t$ denotes the quadratic variation process of $X_t$. Here ${rm d}X_t$ appears explicitly, probably as you want. However, provided that $X_t$ is an Ito process, there must exist some predictable processes $mu_t$ and $sigma_t$, such that
$$
{rm d}X_t=mu_t,{rm d}t+sigma_t,{rm d}W_t,
$$
where $W_t$ denotes the Wiener process. Therefore, while ${rm d}X_t$ appears explicitly, you may re-express it so that only ${rm d}t$ and ${rm d}W_t$ remain.
Nevertheless, the above argument is only for Ito processes. For general stochastic processes, you may still define its Ito stochastic integral if it exists. However, it is not always funny to consider the "most general" case - beyond curiosity, it is more realistic to consider, at least at the very beginning, those that help to shape our world.
Among all stochastic processes, martingales have witnessed their wide applications in physics and finance. Later you will learn the martingale representation theorem, which states that every martingale is actually an Ito process without drift, i.e., for any given martingale $X_t$, there exists some predictable process $sigma_t$, such that
$$
{rm d}X_t=sigma_t,{rm d}W_t.
$$
See? Martingales are defined independent from Brownian motions. However, they can be represented by the standard Brownian motion. This beauty partly explains why we focus on Ito processes: they are, at least for the moment, general enough to model most of the cases in physics and finance.
Finally, if you are math-majored, you probably will also learn Levy's characterization of Brownian motion, which provides an equivalent definition for the Brownian motion. While this theorem itself might be abstract, one of its profound corollary goes that
$$
X_t=int_0^tsigma_s,{rm d}W_s
$$
has the same distribution as
$$
B_{int_0^tsigma_s^2,{rm d}s},
$$
where, in case of ambiguity, $B_t$ here denotes another Wiener process. This conclusion is extremely powerful. Combined with the martingale representation theorem, it tells that all martingales are no more than a time-changed Brownian motion. Concisely, every martingale observes a nature in the Brownian motion. Perhaps this is why we choose to pay particular attention to the Brownian motion: we find martingales are useful, and martingales are closely related to Brownian motions in the sense of both representation and nature.
That's it! Hope this explanation could be somewhat helpful for you.
answered Jan 11 at 15:54
hypernovahypernova
4,789414
answered Jan 11 at 15:54
hypernovahypernova
4,789414
answered Jan 11 at 15:54
hypernovahypernova
4,789414
answered Jan 11 at 15:54
hypernovahypernova
4,789414
4,789414
$begingroup$
This is super helpful, thank you so much!
$endgroup$
– user40333
Jan 12 at 2:43
$begingroup$
@user40333: My pleasure. Best wishes for your stochastic process journey!
$endgroup$
– hypernova
Jan 12 at 2:55
add a comment |
$begingroup$
This is super helpful, thank you so much!
$endgroup$
– user40333
Jan 12 at 2:43
$begingroup$
@user40333: My pleasure. Best wishes for your stochastic process journey!
$endgroup$
– hypernova
Jan 12 at 2:55
$begingroup$
This is super helpful, thank you so much!
$endgroup$
– user40333
Jan 12 at 2:43
$begingroup$
This is super helpful, thank you so much!
$endgroup$
– user40333
Jan 12 at 2:43
$begingroup$
@user40333: My pleasure. Best wishes for your stochastic process journey!
$endgroup$
– hypernova
Jan 12 at 2:55
$begingroup$
@user40333: My pleasure. Best wishes for your stochastic process journey!
$endgroup$
– hypernova
Jan 12 at 2:55
add a comment |
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$begingroup$
First you want to make sure that the integral with respect to your stochastic process exists
$endgroup$
– Gâteau-Gallois
Jan 11 at 13:52
$begingroup$
Ok, and under the assumption that it does exist, am I correct in my thinking?
$endgroup$
– user40333
Jan 11 at 14:07