Difference between weak ( or martingale ) and strong solutions to SDEs
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Hi Im fairly new to SDE theory and am struggling with the difference between a weak ( or martingale ) solution and a strong solution to an SDE :
$$ d(X_{t})=b(t,X_{t})dt + sigma(t,X_{t})dW_{t} $$
Are these two differences and what do they really mean in detail?
For a strong solution we are given an initial value, whereas for weak solutions only a probability law?
For strong solutions we know what probability space we are working in and have a Brownian Motion $W$ in that space. For a weak solution we can only say that there exists some probability space where the SDE holds (with a new brownian motion in the space).
As you can tell I am confused with this topic some clarifications would be amazing.
stochastic-processes stochastic-calculus brownian-motion sde
add a comment |
up vote
4
down vote
favorite
Hi Im fairly new to SDE theory and am struggling with the difference between a weak ( or martingale ) solution and a strong solution to an SDE :
$$ d(X_{t})=b(t,X_{t})dt + sigma(t,X_{t})dW_{t} $$
Are these two differences and what do they really mean in detail?
For a strong solution we are given an initial value, whereas for weak solutions only a probability law?
For strong solutions we know what probability space we are working in and have a Brownian Motion $W$ in that space. For a weak solution we can only say that there exists some probability space where the SDE holds (with a new brownian motion in the space).
As you can tell I am confused with this topic some clarifications would be amazing.
stochastic-processes stochastic-calculus brownian-motion sde
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Hi Im fairly new to SDE theory and am struggling with the difference between a weak ( or martingale ) solution and a strong solution to an SDE :
$$ d(X_{t})=b(t,X_{t})dt + sigma(t,X_{t})dW_{t} $$
Are these two differences and what do they really mean in detail?
For a strong solution we are given an initial value, whereas for weak solutions only a probability law?
For strong solutions we know what probability space we are working in and have a Brownian Motion $W$ in that space. For a weak solution we can only say that there exists some probability space where the SDE holds (with a new brownian motion in the space).
As you can tell I am confused with this topic some clarifications would be amazing.
stochastic-processes stochastic-calculus brownian-motion sde
Hi Im fairly new to SDE theory and am struggling with the difference between a weak ( or martingale ) solution and a strong solution to an SDE :
$$ d(X_{t})=b(t,X_{t})dt + sigma(t,X_{t})dW_{t} $$
Are these two differences and what do they really mean in detail?
For a strong solution we are given an initial value, whereas for weak solutions only a probability law?
For strong solutions we know what probability space we are working in and have a Brownian Motion $W$ in that space. For a weak solution we can only say that there exists some probability space where the SDE holds (with a new brownian motion in the space).
As you can tell I am confused with this topic some clarifications would be amazing.
stochastic-processes stochastic-calculus brownian-motion sde
stochastic-processes stochastic-calculus brownian-motion sde
edited Nov 9 at 7:52
saz
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76k755116
asked Nov 6 at 21:46
Monty
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The main difference between weak and strong solutions is indeed that for strong solutions we are given a Brownian motion on a given probability space whereas for weak solutions we are free to choose the Brownian motion.
Definition: Let $(B_t)_{t geq 0}$ be a Brownian motion with admissible filtration $(mathcal{F}_t)_{t geq 0}$. A progressively measurable process $(X_t,mathcal{F}_t)$ is a strong solution with initial condition $xi$ if $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds, qquad X_0 =xi tag{1}$$ holds almost surely for all $t geq 0$.
Definition: A stochastic process $(X_t,mathcal{F}_t)$ on some probability space $(Omega,mathcal{F},mathbb{P})$ is called a weak solution with initial distribution $mu$ if there exists a Brownian motion $(B_t)_{t geq 0}$ on $(Omega,mathcal{F},mathbb{P})$ such that $(mathcal{F}_t)_{t geq 0}$ is an admissible filtration, $mathbb{P}(X_0 in cdot) = mu(cdot)$ and $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds$$ holds almost surely for all $t geq 0$.
As a consequence of these definitions, we have to consider different notions of uniqueness. For strong solutions we are typically looking for pathwise unique solutions, i.e. if $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$ are strong solutions to $(1)$ with the same initial condition, then pathwise uniqueness means $$mathbb{P} left( sup_{t geq 0} |X_t^{(1)}-X_t^{(2)}|=0 right)=1.$$ As the following simple example shows it doesn't make sense to talk about pathwise uniqueness of weak solutions.
Example 1: Let $(W_t^{(1)})_{t geq 0}$ and $(W_t^{(2)})_{t geq 0}$ be two Brownian motions (possibly defined on different probability spaces), then both $X_t^{(1)} := W_t^{(1)}$ and $X_t^{(2)} := W_t^{(2)}$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 = 0$$ Why? According to the definition we are free choose the driving Brownian motion, so we can set $B_t^{(1)} := W_t^{(1)}$ and $B_t^{(2)} := W_t^{(2)}$, respectively, and then $$dX_t^{(i)} = dB_t^{(i)} quad text{for $i=1,2$}.$$
What do we learn from this? Since weak solutions might be defined on different probability spaces, there is no (immediate) way to compute probabilities of the form $mathbb{P}(X_t^{(1)}=X_t^{(2)})$ for two weak solutions $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$, and therefore we cannot even attempt to talk about pathwise uniqueness. For the same reason, it doesn't make sense to talk about pointwise initial conditions $xi$ for weak solutions (... for this we would need to fix some probability space on which $xi$ lives...); instead we only prescribe the initial distribution of $X_0$.
The next example shows that we cannot expect to have pathwise uniqueness even if the weak solutions are defined on the same probability space.
Example 2: Let $(W_t)_{t geq 0}$ be a Brownian motion. It follows from Example 1 that $X_t^{(1)} := W_t$ and $X_t^{(2)} := -W_t$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 =0.$$ Clearly, $mathbb{P}(X_t^{(1)} = X_t^{(2)}) = mathbb{P}(W_t=0)=0$.
The "good" notion of uniqueness for weak solutions is weak uniqueness, i.e. uniqueness in distribution (= the solutions have the same finite-dimensional distributions).
Typically it is much easier to prove the existence (and/or uniqueness of) a weak solution the the existence (and/or uniqueness) of a strong solution.
Example 3: The SDE $$dX_t = - text{sgn},(X_t) , dB_t, qquad X_0 = 0 tag{2}$$ has a weak solution but no strong solution.
Let's prove that the SDE has a weak solution. Let $(X_t,mathcal{F}_t)_{t geq 0}$ be some Brownian motion and define $$W_t := -int_0^t text{sgn} , (X_s) , dX_s.$$ It follows from Lévy's characterization that $(W_t,mathcal{F}_t)$ is also a Brownian motion. Since $$dW_t = - text{sgn} , (X_t) , dX_t$$ implies $$dX_t = - text{sgn} , (X_t) , dW_t$$ this means that $(X_t)_{t geq 0}$ is a weak solution to $(2)$. For a proof that a strong solution does not exist see e.g. Example 19.16 in the book by Schilling & Partzsch on Brownian motion.
Let me finally mention that weak solutions are closely related to martingale problems; in this answer I tried to give some insights on the connection between the two notions.
Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way.
– Monty
3 hours ago
@Monty Thank you; I fixed it.
– saz
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The main difference between weak and strong solutions is indeed that for strong solutions we are given a Brownian motion on a given probability space whereas for weak solutions we are free to choose the Brownian motion.
Definition: Let $(B_t)_{t geq 0}$ be a Brownian motion with admissible filtration $(mathcal{F}_t)_{t geq 0}$. A progressively measurable process $(X_t,mathcal{F}_t)$ is a strong solution with initial condition $xi$ if $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds, qquad X_0 =xi tag{1}$$ holds almost surely for all $t geq 0$.
Definition: A stochastic process $(X_t,mathcal{F}_t)$ on some probability space $(Omega,mathcal{F},mathbb{P})$ is called a weak solution with initial distribution $mu$ if there exists a Brownian motion $(B_t)_{t geq 0}$ on $(Omega,mathcal{F},mathbb{P})$ such that $(mathcal{F}_t)_{t geq 0}$ is an admissible filtration, $mathbb{P}(X_0 in cdot) = mu(cdot)$ and $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds$$ holds almost surely for all $t geq 0$.
As a consequence of these definitions, we have to consider different notions of uniqueness. For strong solutions we are typically looking for pathwise unique solutions, i.e. if $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$ are strong solutions to $(1)$ with the same initial condition, then pathwise uniqueness means $$mathbb{P} left( sup_{t geq 0} |X_t^{(1)}-X_t^{(2)}|=0 right)=1.$$ As the following simple example shows it doesn't make sense to talk about pathwise uniqueness of weak solutions.
Example 1: Let $(W_t^{(1)})_{t geq 0}$ and $(W_t^{(2)})_{t geq 0}$ be two Brownian motions (possibly defined on different probability spaces), then both $X_t^{(1)} := W_t^{(1)}$ and $X_t^{(2)} := W_t^{(2)}$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 = 0$$ Why? According to the definition we are free choose the driving Brownian motion, so we can set $B_t^{(1)} := W_t^{(1)}$ and $B_t^{(2)} := W_t^{(2)}$, respectively, and then $$dX_t^{(i)} = dB_t^{(i)} quad text{for $i=1,2$}.$$
What do we learn from this? Since weak solutions might be defined on different probability spaces, there is no (immediate) way to compute probabilities of the form $mathbb{P}(X_t^{(1)}=X_t^{(2)})$ for two weak solutions $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$, and therefore we cannot even attempt to talk about pathwise uniqueness. For the same reason, it doesn't make sense to talk about pointwise initial conditions $xi$ for weak solutions (... for this we would need to fix some probability space on which $xi$ lives...); instead we only prescribe the initial distribution of $X_0$.
The next example shows that we cannot expect to have pathwise uniqueness even if the weak solutions are defined on the same probability space.
Example 2: Let $(W_t)_{t geq 0}$ be a Brownian motion. It follows from Example 1 that $X_t^{(1)} := W_t$ and $X_t^{(2)} := -W_t$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 =0.$$ Clearly, $mathbb{P}(X_t^{(1)} = X_t^{(2)}) = mathbb{P}(W_t=0)=0$.
The "good" notion of uniqueness for weak solutions is weak uniqueness, i.e. uniqueness in distribution (= the solutions have the same finite-dimensional distributions).
Typically it is much easier to prove the existence (and/or uniqueness of) a weak solution the the existence (and/or uniqueness) of a strong solution.
Example 3: The SDE $$dX_t = - text{sgn},(X_t) , dB_t, qquad X_0 = 0 tag{2}$$ has a weak solution but no strong solution.
Let's prove that the SDE has a weak solution. Let $(X_t,mathcal{F}_t)_{t geq 0}$ be some Brownian motion and define $$W_t := -int_0^t text{sgn} , (X_s) , dX_s.$$ It follows from Lévy's characterization that $(W_t,mathcal{F}_t)$ is also a Brownian motion. Since $$dW_t = - text{sgn} , (X_t) , dX_t$$ implies $$dX_t = - text{sgn} , (X_t) , dW_t$$ this means that $(X_t)_{t geq 0}$ is a weak solution to $(2)$. For a proof that a strong solution does not exist see e.g. Example 19.16 in the book by Schilling & Partzsch on Brownian motion.
Let me finally mention that weak solutions are closely related to martingale problems; in this answer I tried to give some insights on the connection between the two notions.
Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way.
– Monty
3 hours ago
@Monty Thank you; I fixed it.
– saz
3 hours ago
add a comment |
up vote
3
down vote
accepted
The main difference between weak and strong solutions is indeed that for strong solutions we are given a Brownian motion on a given probability space whereas for weak solutions we are free to choose the Brownian motion.
Definition: Let $(B_t)_{t geq 0}$ be a Brownian motion with admissible filtration $(mathcal{F}_t)_{t geq 0}$. A progressively measurable process $(X_t,mathcal{F}_t)$ is a strong solution with initial condition $xi$ if $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds, qquad X_0 =xi tag{1}$$ holds almost surely for all $t geq 0$.
Definition: A stochastic process $(X_t,mathcal{F}_t)$ on some probability space $(Omega,mathcal{F},mathbb{P})$ is called a weak solution with initial distribution $mu$ if there exists a Brownian motion $(B_t)_{t geq 0}$ on $(Omega,mathcal{F},mathbb{P})$ such that $(mathcal{F}_t)_{t geq 0}$ is an admissible filtration, $mathbb{P}(X_0 in cdot) = mu(cdot)$ and $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds$$ holds almost surely for all $t geq 0$.
As a consequence of these definitions, we have to consider different notions of uniqueness. For strong solutions we are typically looking for pathwise unique solutions, i.e. if $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$ are strong solutions to $(1)$ with the same initial condition, then pathwise uniqueness means $$mathbb{P} left( sup_{t geq 0} |X_t^{(1)}-X_t^{(2)}|=0 right)=1.$$ As the following simple example shows it doesn't make sense to talk about pathwise uniqueness of weak solutions.
Example 1: Let $(W_t^{(1)})_{t geq 0}$ and $(W_t^{(2)})_{t geq 0}$ be two Brownian motions (possibly defined on different probability spaces), then both $X_t^{(1)} := W_t^{(1)}$ and $X_t^{(2)} := W_t^{(2)}$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 = 0$$ Why? According to the definition we are free choose the driving Brownian motion, so we can set $B_t^{(1)} := W_t^{(1)}$ and $B_t^{(2)} := W_t^{(2)}$, respectively, and then $$dX_t^{(i)} = dB_t^{(i)} quad text{for $i=1,2$}.$$
What do we learn from this? Since weak solutions might be defined on different probability spaces, there is no (immediate) way to compute probabilities of the form $mathbb{P}(X_t^{(1)}=X_t^{(2)})$ for two weak solutions $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$, and therefore we cannot even attempt to talk about pathwise uniqueness. For the same reason, it doesn't make sense to talk about pointwise initial conditions $xi$ for weak solutions (... for this we would need to fix some probability space on which $xi$ lives...); instead we only prescribe the initial distribution of $X_0$.
The next example shows that we cannot expect to have pathwise uniqueness even if the weak solutions are defined on the same probability space.
Example 2: Let $(W_t)_{t geq 0}$ be a Brownian motion. It follows from Example 1 that $X_t^{(1)} := W_t$ and $X_t^{(2)} := -W_t$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 =0.$$ Clearly, $mathbb{P}(X_t^{(1)} = X_t^{(2)}) = mathbb{P}(W_t=0)=0$.
The "good" notion of uniqueness for weak solutions is weak uniqueness, i.e. uniqueness in distribution (= the solutions have the same finite-dimensional distributions).
Typically it is much easier to prove the existence (and/or uniqueness of) a weak solution the the existence (and/or uniqueness) of a strong solution.
Example 3: The SDE $$dX_t = - text{sgn},(X_t) , dB_t, qquad X_0 = 0 tag{2}$$ has a weak solution but no strong solution.
Let's prove that the SDE has a weak solution. Let $(X_t,mathcal{F}_t)_{t geq 0}$ be some Brownian motion and define $$W_t := -int_0^t text{sgn} , (X_s) , dX_s.$$ It follows from Lévy's characterization that $(W_t,mathcal{F}_t)$ is also a Brownian motion. Since $$dW_t = - text{sgn} , (X_t) , dX_t$$ implies $$dX_t = - text{sgn} , (X_t) , dW_t$$ this means that $(X_t)_{t geq 0}$ is a weak solution to $(2)$. For a proof that a strong solution does not exist see e.g. Example 19.16 in the book by Schilling & Partzsch on Brownian motion.
Let me finally mention that weak solutions are closely related to martingale problems; in this answer I tried to give some insights on the connection between the two notions.
Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way.
– Monty
3 hours ago
@Monty Thank you; I fixed it.
– saz
3 hours ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The main difference between weak and strong solutions is indeed that for strong solutions we are given a Brownian motion on a given probability space whereas for weak solutions we are free to choose the Brownian motion.
Definition: Let $(B_t)_{t geq 0}$ be a Brownian motion with admissible filtration $(mathcal{F}_t)_{t geq 0}$. A progressively measurable process $(X_t,mathcal{F}_t)$ is a strong solution with initial condition $xi$ if $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds, qquad X_0 =xi tag{1}$$ holds almost surely for all $t geq 0$.
Definition: A stochastic process $(X_t,mathcal{F}_t)$ on some probability space $(Omega,mathcal{F},mathbb{P})$ is called a weak solution with initial distribution $mu$ if there exists a Brownian motion $(B_t)_{t geq 0}$ on $(Omega,mathcal{F},mathbb{P})$ such that $(mathcal{F}_t)_{t geq 0}$ is an admissible filtration, $mathbb{P}(X_0 in cdot) = mu(cdot)$ and $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds$$ holds almost surely for all $t geq 0$.
As a consequence of these definitions, we have to consider different notions of uniqueness. For strong solutions we are typically looking for pathwise unique solutions, i.e. if $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$ are strong solutions to $(1)$ with the same initial condition, then pathwise uniqueness means $$mathbb{P} left( sup_{t geq 0} |X_t^{(1)}-X_t^{(2)}|=0 right)=1.$$ As the following simple example shows it doesn't make sense to talk about pathwise uniqueness of weak solutions.
Example 1: Let $(W_t^{(1)})_{t geq 0}$ and $(W_t^{(2)})_{t geq 0}$ be two Brownian motions (possibly defined on different probability spaces), then both $X_t^{(1)} := W_t^{(1)}$ and $X_t^{(2)} := W_t^{(2)}$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 = 0$$ Why? According to the definition we are free choose the driving Brownian motion, so we can set $B_t^{(1)} := W_t^{(1)}$ and $B_t^{(2)} := W_t^{(2)}$, respectively, and then $$dX_t^{(i)} = dB_t^{(i)} quad text{for $i=1,2$}.$$
What do we learn from this? Since weak solutions might be defined on different probability spaces, there is no (immediate) way to compute probabilities of the form $mathbb{P}(X_t^{(1)}=X_t^{(2)})$ for two weak solutions $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$, and therefore we cannot even attempt to talk about pathwise uniqueness. For the same reason, it doesn't make sense to talk about pointwise initial conditions $xi$ for weak solutions (... for this we would need to fix some probability space on which $xi$ lives...); instead we only prescribe the initial distribution of $X_0$.
The next example shows that we cannot expect to have pathwise uniqueness even if the weak solutions are defined on the same probability space.
Example 2: Let $(W_t)_{t geq 0}$ be a Brownian motion. It follows from Example 1 that $X_t^{(1)} := W_t$ and $X_t^{(2)} := -W_t$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 =0.$$ Clearly, $mathbb{P}(X_t^{(1)} = X_t^{(2)}) = mathbb{P}(W_t=0)=0$.
The "good" notion of uniqueness for weak solutions is weak uniqueness, i.e. uniqueness in distribution (= the solutions have the same finite-dimensional distributions).
Typically it is much easier to prove the existence (and/or uniqueness of) a weak solution the the existence (and/or uniqueness) of a strong solution.
Example 3: The SDE $$dX_t = - text{sgn},(X_t) , dB_t, qquad X_0 = 0 tag{2}$$ has a weak solution but no strong solution.
Let's prove that the SDE has a weak solution. Let $(X_t,mathcal{F}_t)_{t geq 0}$ be some Brownian motion and define $$W_t := -int_0^t text{sgn} , (X_s) , dX_s.$$ It follows from Lévy's characterization that $(W_t,mathcal{F}_t)$ is also a Brownian motion. Since $$dW_t = - text{sgn} , (X_t) , dX_t$$ implies $$dX_t = - text{sgn} , (X_t) , dW_t$$ this means that $(X_t)_{t geq 0}$ is a weak solution to $(2)$. For a proof that a strong solution does not exist see e.g. Example 19.16 in the book by Schilling & Partzsch on Brownian motion.
Let me finally mention that weak solutions are closely related to martingale problems; in this answer I tried to give some insights on the connection between the two notions.
The main difference between weak and strong solutions is indeed that for strong solutions we are given a Brownian motion on a given probability space whereas for weak solutions we are free to choose the Brownian motion.
Definition: Let $(B_t)_{t geq 0}$ be a Brownian motion with admissible filtration $(mathcal{F}_t)_{t geq 0}$. A progressively measurable process $(X_t,mathcal{F}_t)$ is a strong solution with initial condition $xi$ if $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds, qquad X_0 =xi tag{1}$$ holds almost surely for all $t geq 0$.
Definition: A stochastic process $(X_t,mathcal{F}_t)$ on some probability space $(Omega,mathcal{F},mathbb{P})$ is called a weak solution with initial distribution $mu$ if there exists a Brownian motion $(B_t)_{t geq 0}$ on $(Omega,mathcal{F},mathbb{P})$ such that $(mathcal{F}_t)_{t geq 0}$ is an admissible filtration, $mathbb{P}(X_0 in cdot) = mu(cdot)$ and $$X_t-X_0 = int_0^t sigma(s,X_s) , dB_s + int_0^t b(s,X_s) , ds$$ holds almost surely for all $t geq 0$.
As a consequence of these definitions, we have to consider different notions of uniqueness. For strong solutions we are typically looking for pathwise unique solutions, i.e. if $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$ are strong solutions to $(1)$ with the same initial condition, then pathwise uniqueness means $$mathbb{P} left( sup_{t geq 0} |X_t^{(1)}-X_t^{(2)}|=0 right)=1.$$ As the following simple example shows it doesn't make sense to talk about pathwise uniqueness of weak solutions.
Example 1: Let $(W_t^{(1)})_{t geq 0}$ and $(W_t^{(2)})_{t geq 0}$ be two Brownian motions (possibly defined on different probability spaces), then both $X_t^{(1)} := W_t^{(1)}$ and $X_t^{(2)} := W_t^{(2)}$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 = 0$$ Why? According to the definition we are free choose the driving Brownian motion, so we can set $B_t^{(1)} := W_t^{(1)}$ and $B_t^{(2)} := W_t^{(2)}$, respectively, and then $$dX_t^{(i)} = dB_t^{(i)} quad text{for $i=1,2$}.$$
What do we learn from this? Since weak solutions might be defined on different probability spaces, there is no (immediate) way to compute probabilities of the form $mathbb{P}(X_t^{(1)}=X_t^{(2)})$ for two weak solutions $(X_t^{(1)})_{t geq 0}$ and $(X_t^{(2)})_{t geq 0}$, and therefore we cannot even attempt to talk about pathwise uniqueness. For the same reason, it doesn't make sense to talk about pointwise initial conditions $xi$ for weak solutions (... for this we would need to fix some probability space on which $xi$ lives...); instead we only prescribe the initial distribution of $X_0$.
The next example shows that we cannot expect to have pathwise uniqueness even if the weak solutions are defined on the same probability space.
Example 2: Let $(W_t)_{t geq 0}$ be a Brownian motion. It follows from Example 1 that $X_t^{(1)} := W_t$ and $X_t^{(2)} := -W_t$ are weak solutions to the SDE $$dX_t = dB_t, qquad X_0 =0.$$ Clearly, $mathbb{P}(X_t^{(1)} = X_t^{(2)}) = mathbb{P}(W_t=0)=0$.
The "good" notion of uniqueness for weak solutions is weak uniqueness, i.e. uniqueness in distribution (= the solutions have the same finite-dimensional distributions).
Typically it is much easier to prove the existence (and/or uniqueness of) a weak solution the the existence (and/or uniqueness) of a strong solution.
Example 3: The SDE $$dX_t = - text{sgn},(X_t) , dB_t, qquad X_0 = 0 tag{2}$$ has a weak solution but no strong solution.
Let's prove that the SDE has a weak solution. Let $(X_t,mathcal{F}_t)_{t geq 0}$ be some Brownian motion and define $$W_t := -int_0^t text{sgn} , (X_s) , dX_s.$$ It follows from Lévy's characterization that $(W_t,mathcal{F}_t)$ is also a Brownian motion. Since $$dW_t = - text{sgn} , (X_t) , dX_t$$ implies $$dX_t = - text{sgn} , (X_t) , dW_t$$ this means that $(X_t)_{t geq 0}$ is a weak solution to $(2)$. For a proof that a strong solution does not exist see e.g. Example 19.16 in the book by Schilling & Partzsch on Brownian motion.
Let me finally mention that weak solutions are closely related to martingale problems; in this answer I tried to give some insights on the connection between the two notions.
edited 3 hours ago
answered Nov 8 at 19:04
saz
76k755116
76k755116
Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way.
– Monty
3 hours ago
@Monty Thank you; I fixed it.
– saz
3 hours ago
add a comment |
Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way.
– Monty
3 hours ago
@Monty Thank you; I fixed it.
– saz
3 hours ago
Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way.
– Monty
3 hours ago
Why is the text after your first two definitions. not formatting properly... Thanks for your answer by the way.
– Monty
3 hours ago
@Monty Thank you; I fixed it.
– saz
3 hours ago
@Monty Thank you; I fixed it.
– saz
3 hours ago
add a comment |
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});
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StackExchange.ready(function () {
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});
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