Integral of a Gaussian process
$begingroup$
Let $(Omega,Sigma,P)$ be a probability space and $X: [0,infty) times Omega to mathbb{R}$ be a Gaussian process (i.e. all finite linear combinations $sum_i a_i X_{t_i}$ are Gaussian random variables). If the
process is continuous, it seems to be clear that the process $Y_t (omega) = int_0^t X_s(omega) ds$ is a Gaussian process.
Is it true that $Y_t$ is a Gaussian process even if $X$ is only assumed to be measurable? I am working through the derivation of Kalman Filter in the text by Bernt Oksendal Eq 6.2.10 (fifth edition) and this seems to be a way to show that $M_t$ (in the book) is a Gaussian process.
probability-theory stochastic-processes normal-distribution stochastic-calculus
edited Oct 9 '15 at 9:02
saz
80k860125
asked Oct 9 '15 at 5:34
jpvjpv
889718
$endgroup$
add a comment |
$begingroup$
Let $(Omega,Sigma,P)$ be a probability space and $X: [0,infty) times Omega to mathbb{R}$ be a Gaussian process (i.e. all finite linear combinations $sum_i a_i X_{t_i}$ are Gaussian random variables). If the
process is continuous, it seems to be clear that the process $Y_t (omega) = int_0^t X_s(omega) ds$ is a Gaussian process.
Is it true that $Y_t$ is a Gaussian process even if $X$ is only assumed to be measurable? I am working through the derivation of Kalman Filter in the text by Bernt Oksendal Eq 6.2.10 (fifth edition) and this seems to be a way to show that $M_t$ (in the book) is a Gaussian process.
probability-theory stochastic-processes normal-distribution stochastic-calculus
edited Oct 9 '15 at 9:02
saz
80k860125
asked Oct 9 '15 at 5:34
jpvjpv
889718
$endgroup$
$begingroup$
Yes, the measurability in the sense of $([0,infty)times Omega, mathcal{B}([0,infty))otimes mathcal F) to (mathbb{R},mathcal{B}(mathbb{R}))$ is enough.
$endgroup$
– zhoraster
Oct 9 '15 at 7:13
$begingroup$
@zhoraster Why do you think so? For example, if $Z sim N(0,1)$, then $X_t := frac{1}{t} Z$, $t>0$, $X_0 := 0$, is Gaussian, but the integral $int_0^t X_s , ds$ is not well-defined. As far as I can see, we need some additional assumption on the integrability, e.g. $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for $T>0$.
$endgroup$
– saz
Oct 9 '15 at 7:38
$begingroup$
@saz, I assume that everything is well-defined. Of course, to this end we need some integrability.
$endgroup$
– zhoraster
Oct 9 '15 at 8:44
$begingroup$
@zhoraster I see.
$endgroup$
– saz
Oct 9 '15 at 8:48
add a comment |
$begingroup$
Let $(Omega,Sigma,P)$ be a probability space and $X: [0,infty) times Omega to mathbb{R}$ be a Gaussian process (i.e. all finite linear combinations $sum_i a_i X_{t_i}$ are Gaussian random variables). If the
process is continuous, it seems to be clear that the process $Y_t (omega) = int_0^t X_s(omega) ds$ is a Gaussian process.
Is it true that $Y_t$ is a Gaussian process even if $X$ is only assumed to be measurable? I am working through the derivation of Kalman Filter in the text by Bernt Oksendal Eq 6.2.10 (fifth edition) and this seems to be a way to show that $M_t$ (in the book) is a Gaussian process.
probability-theory stochastic-processes normal-distribution stochastic-calculus
edited Oct 9 '15 at 9:02
saz
80k860125
asked Oct 9 '15 at 5:34
jpvjpv
889718
$endgroup$
Let $(Omega,Sigma,P)$ be a probability space and $X: [0,infty) times Omega to mathbb{R}$ be a Gaussian process (i.e. all finite linear combinations $sum_i a_i X_{t_i}$ are Gaussian random variables). If the
process is continuous, it seems to be clear that the process $Y_t (omega) = int_0^t X_s(omega) ds$ is a Gaussian process.
Is it true that $Y_t$ is a Gaussian process even if $X$ is only assumed to be measurable? I am working through the derivation of Kalman Filter in the text by Bernt Oksendal Eq 6.2.10 (fifth edition) and this seems to be a way to show that $M_t$ (in the book) is a Gaussian process.
probability-theory stochastic-processes normal-distribution stochastic-calculus
probability-theory stochastic-processes normal-distribution stochastic-calculus
edited Oct 9 '15 at 9:02
saz
80k860125
asked Oct 9 '15 at 5:34
jpvjpv
889718
edited Oct 9 '15 at 9:02
saz
80k860125
asked Oct 9 '15 at 5:34
jpvjpv
889718
edited Oct 9 '15 at 9:02
saz
80k860125
edited Oct 9 '15 at 9:02
saz
80k860125
edited Oct 9 '15 at 9:02
saz
80k860125
80k860125
asked Oct 9 '15 at 5:34
jpvjpv
889718
asked Oct 9 '15 at 5:34
jpvjpv
889718
asked Oct 9 '15 at 5:34
jpvjpv
889718
889718
$begingroup$
Yes, the measurability in the sense of $([0,infty)times Omega, mathcal{B}([0,infty))otimes mathcal F) to (mathbb{R},mathcal{B}(mathbb{R}))$ is enough.
$endgroup$
– zhoraster
Oct 9 '15 at 7:13
$begingroup$
@zhoraster Why do you think so? For example, if $Z sim N(0,1)$, then $X_t := frac{1}{t} Z$, $t>0$, $X_0 := 0$, is Gaussian, but the integral $int_0^t X_s , ds$ is not well-defined. As far as I can see, we need some additional assumption on the integrability, e.g. $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for $T>0$.
$endgroup$
– saz
Oct 9 '15 at 7:38
$begingroup$
@saz, I assume that everything is well-defined. Of course, to this end we need some integrability.
$endgroup$
– zhoraster
Oct 9 '15 at 8:44
$begingroup$
@zhoraster I see.
$endgroup$
– saz
Oct 9 '15 at 8:48
add a comment |
$begingroup$
Yes, the measurability in the sense of $([0,infty)times Omega, mathcal{B}([0,infty))otimes mathcal F) to (mathbb{R},mathcal{B}(mathbb{R}))$ is enough.
$endgroup$
– zhoraster
Oct 9 '15 at 7:13
$begingroup$
@zhoraster Why do you think so? For example, if $Z sim N(0,1)$, then $X_t := frac{1}{t} Z$, $t>0$, $X_0 := 0$, is Gaussian, but the integral $int_0^t X_s , ds$ is not well-defined. As far as I can see, we need some additional assumption on the integrability, e.g. $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for $T>0$.
$endgroup$
– saz
Oct 9 '15 at 7:38
$begingroup$
@saz, I assume that everything is well-defined. Of course, to this end we need some integrability.
$endgroup$
– zhoraster
Oct 9 '15 at 8:44
$begingroup$
@zhoraster I see.
$endgroup$
– saz
Oct 9 '15 at 8:48
$begingroup$
Yes, the measurability in the sense of $([0,infty)times Omega, mathcal{B}([0,infty))otimes mathcal F) to (mathbb{R},mathcal{B}(mathbb{R}))$ is enough.
$endgroup$
– zhoraster
Oct 9 '15 at 7:13
$begingroup$
Yes, the measurability in the sense of $([0,infty)times Omega, mathcal{B}([0,infty))otimes mathcal F) to (mathbb{R},mathcal{B}(mathbb{R}))$ is enough.
$endgroup$
– zhoraster
Oct 9 '15 at 7:13
$begingroup$
@zhoraster Why do you think so? For example, if $Z sim N(0,1)$, then $X_t := frac{1}{t} Z$, $t>0$, $X_0 := 0$, is Gaussian, but the integral $int_0^t X_s , ds$ is not well-defined. As far as I can see, we need some additional assumption on the integrability, e.g. $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for $T>0$.
$endgroup$
– saz
Oct 9 '15 at 7:38
$begingroup$
@zhoraster Why do you think so? For example, if $Z sim N(0,1)$, then $X_t := frac{1}{t} Z$, $t>0$, $X_0 := 0$, is Gaussian, but the integral $int_0^t X_s , ds$ is not well-defined. As far as I can see, we need some additional assumption on the integrability, e.g. $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for $T>0$.
$endgroup$
– saz
Oct 9 '15 at 7:38
$begingroup$
@saz, I assume that everything is well-defined. Of course, to this end we need some integrability.
$endgroup$
– zhoraster
Oct 9 '15 at 8:44
$begingroup$
@saz, I assume that everything is well-defined. Of course, to this end we need some integrability.
$endgroup$
– zhoraster
Oct 9 '15 at 8:44
$begingroup$
@zhoraster I see.
$endgroup$
– saz
Oct 9 '15 at 8:48
$begingroup$
@zhoraster I see.
$endgroup$
– saz
Oct 9 '15 at 8:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Question 1: Is $Y_t(omega)$ well-defined?
Answer: No, in general, $Y_t(omega)$ is not well-defined; we need some additional assumption on the integrability of $X$ to ensure that $$int_0^t |X_s(omega)| , ds <infty$$ for $t>0$. This is e.g. satisfied if $X$ has continuous sample paths or $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for any $T>0$. (To see that it $Y_t$ is in general not well-defined, just consider $X_t := t^{-1} Z$, $t>0$, for $Z sim N(0,1)$; then $X$ is Gaussian, but the integral $int_0^t X_t , ds$ does not exist.)
Question 2: Is $omega mapsto Y_t(omega)$ a random variable for fixed $t geq 0$?
Answer: If the process $X: (0,infty) times Omega to mathbb{R}$ is jointly measurable, then $Y_t$ is a random variable for each $t geq 0$. Otherwise, measurability of $Y_t$ might fail.
Question 3: Is $(Y_t)_{t geq 0}$ Gaussian?
Answer: If $t mapsto X_t(omega)$ is Riemann integrable, this follows by approximation the integral by Riemann sums; see e.g. this question. (Note that a bounded function $f:[0,T] to mathbb{R}$ is Riemann integrable if, and only if, the points in $[0,T]$ where $f$ is discontinuous is a Lebesgue null set.)
Edit: Okay, so somewhat more detailed: For any (Riemann) integrable function $f:[0,t] to mathbb{R}$ it is known that the (Riemann) integral
$$int_0^t f(s) , ds$$
can be approximated by Riemann sums
$$sum_{j=0}^{n-1} f(s_j) (t_{j+1}-t_j)$$
where $0=t_0 < ldots <t_n = t$ is a partition of the interval $[0,t]$ and $s_j in [t_j,t_{j+1}]$. In particular, if we choose $s_j = t_j := t frac{j}{n}$, we find
$$int_0^t f(s) , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} f left( t frac{j}{n} right).$$
Applying this in order (stochastic) setting, we get
$$int_0^t X_s , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} X_{t j/n};$$
and $frac{t}{n} sum_{j=0}^{n-1} X_{t j/n}$ is Gaussian because $X$ is Gaussian.
answered Oct 9 '15 at 9:01
sazsaz
80k860125
$endgroup$
$begingroup$
saz: Your answer to the linked question uses continuity of Brownian paths to prove this. I am not sure if I am missing something.
$endgroup$
– jpv
Oct 9 '15 at 17:35
$begingroup$
@jpv Yes, but the continuity is not used to prove that it is Gaussian. The point is simply that the Riemann sums are Gaussian (since they are finite linear combinations) and so is their limit.
$endgroup$
– saz
Oct 9 '15 at 17:38
$begingroup$
saz: When the process is not continuous, the Riemann sums might not converge to the integral. Isn't that so?
$endgroup$
– jpv
Oct 9 '15 at 18:06
$begingroup$
@jpv No, the sums convergence whenever the integrand is integrable; this follows directly from the definition of the Riemann integral (or Lebesgue integral, if you prefer it; it doesn't matter).
$endgroup$
– saz
Oct 9 '15 at 18:40
$begingroup$
saz: I am assuming that $Y_t(omega) = int_0^t X_s(omega) ds$ exists for all $omega,t$ in the Lebesgue sense. I think that the Lebesgue integral cannot be approximated by Riemann sums. The way to approximate the Lebesgue integral is by using $Y_t^+,Y_t^-$ and using the fact that these non-negative random variables can be approximated by step functions. However, I cannot see how this entire process of approximation will preserve Gaussianity.
$endgroup$
– jpv
Oct 10 '15 at 3:42
|
show 8 more comments
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$begingroup$
Question 1: Is $Y_t(omega)$ well-defined?
Answer: No, in general, $Y_t(omega)$ is not well-defined; we need some additional assumption on the integrability of $X$ to ensure that $$int_0^t |X_s(omega)| , ds <infty$$ for $t>0$. This is e.g. satisfied if $X$ has continuous sample paths or $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for any $T>0$. (To see that it $Y_t$ is in general not well-defined, just consider $X_t := t^{-1} Z$, $t>0$, for $Z sim N(0,1)$; then $X$ is Gaussian, but the integral $int_0^t X_t , ds$ does not exist.)
Question 2: Is $omega mapsto Y_t(omega)$ a random variable for fixed $t geq 0$?
Answer: If the process $X: (0,infty) times Omega to mathbb{R}$ is jointly measurable, then $Y_t$ is a random variable for each $t geq 0$. Otherwise, measurability of $Y_t$ might fail.
Question 3: Is $(Y_t)_{t geq 0}$ Gaussian?
Answer: If $t mapsto X_t(omega)$ is Riemann integrable, this follows by approximation the integral by Riemann sums; see e.g. this question. (Note that a bounded function $f:[0,T] to mathbb{R}$ is Riemann integrable if, and only if, the points in $[0,T]$ where $f$ is discontinuous is a Lebesgue null set.)
Edit: Okay, so somewhat more detailed: For any (Riemann) integrable function $f:[0,t] to mathbb{R}$ it is known that the (Riemann) integral
$$int_0^t f(s) , ds$$
can be approximated by Riemann sums
$$sum_{j=0}^{n-1} f(s_j) (t_{j+1}-t_j)$$
where $0=t_0 < ldots <t_n = t$ is a partition of the interval $[0,t]$ and $s_j in [t_j,t_{j+1}]$. In particular, if we choose $s_j = t_j := t frac{j}{n}$, we find
$$int_0^t f(s) , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} f left( t frac{j}{n} right).$$
Applying this in order (stochastic) setting, we get
$$int_0^t X_s , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} X_{t j/n};$$
and $frac{t}{n} sum_{j=0}^{n-1} X_{t j/n}$ is Gaussian because $X$ is Gaussian.
answered Oct 9 '15 at 9:01
sazsaz
80k860125
$endgroup$
$begingroup$
saz: Your answer to the linked question uses continuity of Brownian paths to prove this. I am not sure if I am missing something.
$endgroup$
– jpv
Oct 9 '15 at 17:35
$begingroup$
@jpv Yes, but the continuity is not used to prove that it is Gaussian. The point is simply that the Riemann sums are Gaussian (since they are finite linear combinations) and so is their limit.
$endgroup$
– saz
Oct 9 '15 at 17:38
$begingroup$
saz: When the process is not continuous, the Riemann sums might not converge to the integral. Isn't that so?
$endgroup$
– jpv
Oct 9 '15 at 18:06
$begingroup$
@jpv No, the sums convergence whenever the integrand is integrable; this follows directly from the definition of the Riemann integral (or Lebesgue integral, if you prefer it; it doesn't matter).
$endgroup$
– saz
Oct 9 '15 at 18:40
$begingroup$
saz: I am assuming that $Y_t(omega) = int_0^t X_s(omega) ds$ exists for all $omega,t$ in the Lebesgue sense. I think that the Lebesgue integral cannot be approximated by Riemann sums. The way to approximate the Lebesgue integral is by using $Y_t^+,Y_t^-$ and using the fact that these non-negative random variables can be approximated by step functions. However, I cannot see how this entire process of approximation will preserve Gaussianity.
$endgroup$
– jpv
Oct 10 '15 at 3:42
|
show 8 more comments
$begingroup$
Question 1: Is $Y_t(omega)$ well-defined?
Answer: No, in general, $Y_t(omega)$ is not well-defined; we need some additional assumption on the integrability of $X$ to ensure that $$int_0^t |X_s(omega)| , ds <infty$$ for $t>0$. This is e.g. satisfied if $X$ has continuous sample paths or $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for any $T>0$. (To see that it $Y_t$ is in general not well-defined, just consider $X_t := t^{-1} Z$, $t>0$, for $Z sim N(0,1)$; then $X$ is Gaussian, but the integral $int_0^t X_t , ds$ does not exist.)
Question 2: Is $omega mapsto Y_t(omega)$ a random variable for fixed $t geq 0$?
Answer: If the process $X: (0,infty) times Omega to mathbb{R}$ is jointly measurable, then $Y_t$ is a random variable for each $t geq 0$. Otherwise, measurability of $Y_t$ might fail.
Question 3: Is $(Y_t)_{t geq 0}$ Gaussian?
Answer: If $t mapsto X_t(omega)$ is Riemann integrable, this follows by approximation the integral by Riemann sums; see e.g. this question. (Note that a bounded function $f:[0,T] to mathbb{R}$ is Riemann integrable if, and only if, the points in $[0,T]$ where $f$ is discontinuous is a Lebesgue null set.)
Edit: Okay, so somewhat more detailed: For any (Riemann) integrable function $f:[0,t] to mathbb{R}$ it is known that the (Riemann) integral
$$int_0^t f(s) , ds$$
can be approximated by Riemann sums
$$sum_{j=0}^{n-1} f(s_j) (t_{j+1}-t_j)$$
where $0=t_0 < ldots <t_n = t$ is a partition of the interval $[0,t]$ and $s_j in [t_j,t_{j+1}]$. In particular, if we choose $s_j = t_j := t frac{j}{n}$, we find
$$int_0^t f(s) , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} f left( t frac{j}{n} right).$$
Applying this in order (stochastic) setting, we get
$$int_0^t X_s , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} X_{t j/n};$$
and $frac{t}{n} sum_{j=0}^{n-1} X_{t j/n}$ is Gaussian because $X$ is Gaussian.
answered Oct 9 '15 at 9:01
sazsaz
80k860125
$endgroup$
$begingroup$
saz: Your answer to the linked question uses continuity of Brownian paths to prove this. I am not sure if I am missing something.
$endgroup$
– jpv
Oct 9 '15 at 17:35
$begingroup$
@jpv Yes, but the continuity is not used to prove that it is Gaussian. The point is simply that the Riemann sums are Gaussian (since they are finite linear combinations) and so is their limit.
$endgroup$
– saz
Oct 9 '15 at 17:38
$begingroup$
saz: When the process is not continuous, the Riemann sums might not converge to the integral. Isn't that so?
$endgroup$
– jpv
Oct 9 '15 at 18:06
$begingroup$
@jpv No, the sums convergence whenever the integrand is integrable; this follows directly from the definition of the Riemann integral (or Lebesgue integral, if you prefer it; it doesn't matter).
$endgroup$
– saz
Oct 9 '15 at 18:40
$begingroup$
saz: I am assuming that $Y_t(omega) = int_0^t X_s(omega) ds$ exists for all $omega,t$ in the Lebesgue sense. I think that the Lebesgue integral cannot be approximated by Riemann sums. The way to approximate the Lebesgue integral is by using $Y_t^+,Y_t^-$ and using the fact that these non-negative random variables can be approximated by step functions. However, I cannot see how this entire process of approximation will preserve Gaussianity.
$endgroup$
– jpv
Oct 10 '15 at 3:42
|
show 8 more comments
$begingroup$
Question 1: Is $Y_t(omega)$ well-defined?
Answer: No, in general, $Y_t(omega)$ is not well-defined; we need some additional assumption on the integrability of $X$ to ensure that $$int_0^t |X_s(omega)| , ds <infty$$ for $t>0$. This is e.g. satisfied if $X$ has continuous sample paths or $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for any $T>0$. (To see that it $Y_t$ is in general not well-defined, just consider $X_t := t^{-1} Z$, $t>0$, for $Z sim N(0,1)$; then $X$ is Gaussian, but the integral $int_0^t X_t , ds$ does not exist.)
Question 2: Is $omega mapsto Y_t(omega)$ a random variable for fixed $t geq 0$?
Answer: If the process $X: (0,infty) times Omega to mathbb{R}$ is jointly measurable, then $Y_t$ is a random variable for each $t geq 0$. Otherwise, measurability of $Y_t$ might fail.
Question 3: Is $(Y_t)_{t geq 0}$ Gaussian?
Answer: If $t mapsto X_t(omega)$ is Riemann integrable, this follows by approximation the integral by Riemann sums; see e.g. this question. (Note that a bounded function $f:[0,T] to mathbb{R}$ is Riemann integrable if, and only if, the points in $[0,T]$ where $f$ is discontinuous is a Lebesgue null set.)
Edit: Okay, so somewhat more detailed: For any (Riemann) integrable function $f:[0,t] to mathbb{R}$ it is known that the (Riemann) integral
$$int_0^t f(s) , ds$$
can be approximated by Riemann sums
$$sum_{j=0}^{n-1} f(s_j) (t_{j+1}-t_j)$$
where $0=t_0 < ldots <t_n = t$ is a partition of the interval $[0,t]$ and $s_j in [t_j,t_{j+1}]$. In particular, if we choose $s_j = t_j := t frac{j}{n}$, we find
$$int_0^t f(s) , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} f left( t frac{j}{n} right).$$
Applying this in order (stochastic) setting, we get
$$int_0^t X_s , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} X_{t j/n};$$
and $frac{t}{n} sum_{j=0}^{n-1} X_{t j/n}$ is Gaussian because $X$ is Gaussian.
answered Oct 9 '15 at 9:01
sazsaz
80k860125
$endgroup$
Question 1: Is $Y_t(omega)$ well-defined?
Answer: No, in general, $Y_t(omega)$ is not well-defined; we need some additional assumption on the integrability of $X$ to ensure that $$int_0^t |X_s(omega)| , ds <infty$$ for $t>0$. This is e.g. satisfied if $X$ has continuous sample paths or $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for any $T>0$. (To see that it $Y_t$ is in general not well-defined, just consider $X_t := t^{-1} Z$, $t>0$, for $Z sim N(0,1)$; then $X$ is Gaussian, but the integral $int_0^t X_t , ds$ does not exist.)
Question 2: Is $omega mapsto Y_t(omega)$ a random variable for fixed $t geq 0$?
Answer: If the process $X: (0,infty) times Omega to mathbb{R}$ is jointly measurable, then $Y_t$ is a random variable for each $t geq 0$. Otherwise, measurability of $Y_t$ might fail.
Question 3: Is $(Y_t)_{t geq 0}$ Gaussian?
Answer: If $t mapsto X_t(omega)$ is Riemann integrable, this follows by approximation the integral by Riemann sums; see e.g. this question. (Note that a bounded function $f:[0,T] to mathbb{R}$ is Riemann integrable if, and only if, the points in $[0,T]$ where $f$ is discontinuous is a Lebesgue null set.)
Edit: Okay, so somewhat more detailed: For any (Riemann) integrable function $f:[0,t] to mathbb{R}$ it is known that the (Riemann) integral
$$int_0^t f(s) , ds$$
can be approximated by Riemann sums
$$sum_{j=0}^{n-1} f(s_j) (t_{j+1}-t_j)$$
where $0=t_0 < ldots <t_n = t$ is a partition of the interval $[0,t]$ and $s_j in [t_j,t_{j+1}]$. In particular, if we choose $s_j = t_j := t frac{j}{n}$, we find
$$int_0^t f(s) , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} f left( t frac{j}{n} right).$$
Applying this in order (stochastic) setting, we get
$$int_0^t X_s , ds = lim_{n to infty} frac{t}{n} sum_{j=0}^{n-1} X_{t j/n};$$
and $frac{t}{n} sum_{j=0}^{n-1} X_{t j/n}$ is Gaussian because $X$ is Gaussian.
answered Oct 9 '15 at 9:01
sazsaz
80k860125
edited Dec 8 '18 at 19:45
answered Oct 9 '15 at 9:01
sazsaz
80k860125
answered Oct 9 '15 at 9:01
sazsaz
80k860125
answered Oct 9 '15 at 9:01
sazsaz
80k860125
80k860125
$begingroup$
saz: Your answer to the linked question uses continuity of Brownian paths to prove this. I am not sure if I am missing something.
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– jpv
Oct 9 '15 at 17:35
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@jpv Yes, but the continuity is not used to prove that it is Gaussian. The point is simply that the Riemann sums are Gaussian (since they are finite linear combinations) and so is their limit.
$endgroup$
– saz
Oct 9 '15 at 17:38
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saz: When the process is not continuous, the Riemann sums might not converge to the integral. Isn't that so?
$endgroup$
– jpv
Oct 9 '15 at 18:06
$begingroup$
@jpv No, the sums convergence whenever the integrand is integrable; this follows directly from the definition of the Riemann integral (or Lebesgue integral, if you prefer it; it doesn't matter).
$endgroup$
– saz
Oct 9 '15 at 18:40
$begingroup$
saz: I am assuming that $Y_t(omega) = int_0^t X_s(omega) ds$ exists for all $omega,t$ in the Lebesgue sense. I think that the Lebesgue integral cannot be approximated by Riemann sums. The way to approximate the Lebesgue integral is by using $Y_t^+,Y_t^-$ and using the fact that these non-negative random variables can be approximated by step functions. However, I cannot see how this entire process of approximation will preserve Gaussianity.
$endgroup$
– jpv
Oct 10 '15 at 3:42
|
show 8 more comments
$begingroup$
saz: Your answer to the linked question uses continuity of Brownian paths to prove this. I am not sure if I am missing something.
$endgroup$
– jpv
Oct 9 '15 at 17:35
$begingroup$
@jpv Yes, but the continuity is not used to prove that it is Gaussian. The point is simply that the Riemann sums are Gaussian (since they are finite linear combinations) and so is their limit.
$endgroup$
– saz
Oct 9 '15 at 17:38
$begingroup$
saz: When the process is not continuous, the Riemann sums might not converge to the integral. Isn't that so?
$endgroup$
– jpv
Oct 9 '15 at 18:06
$begingroup$
@jpv No, the sums convergence whenever the integrand is integrable; this follows directly from the definition of the Riemann integral (or Lebesgue integral, if you prefer it; it doesn't matter).
$endgroup$
– saz
Oct 9 '15 at 18:40
$begingroup$
saz: I am assuming that $Y_t(omega) = int_0^t X_s(omega) ds$ exists for all $omega,t$ in the Lebesgue sense. I think that the Lebesgue integral cannot be approximated by Riemann sums. The way to approximate the Lebesgue integral is by using $Y_t^+,Y_t^-$ and using the fact that these non-negative random variables can be approximated by step functions. However, I cannot see how this entire process of approximation will preserve Gaussianity.
$endgroup$
– jpv
Oct 10 '15 at 3:42
$begingroup$
saz: Your answer to the linked question uses continuity of Brownian paths to prove this. I am not sure if I am missing something.
$endgroup$
– jpv
Oct 9 '15 at 17:35
$begingroup$
saz: Your answer to the linked question uses continuity of Brownian paths to prove this. I am not sure if I am missing something.
$endgroup$
– jpv
Oct 9 '15 at 17:35
$begingroup$
@jpv Yes, but the continuity is not used to prove that it is Gaussian. The point is simply that the Riemann sums are Gaussian (since they are finite linear combinations) and so is their limit.
$endgroup$
– saz
Oct 9 '15 at 17:38
$begingroup$
@jpv Yes, but the continuity is not used to prove that it is Gaussian. The point is simply that the Riemann sums are Gaussian (since they are finite linear combinations) and so is their limit.
$endgroup$
– saz
Oct 9 '15 at 17:38
$begingroup$
saz: When the process is not continuous, the Riemann sums might not converge to the integral. Isn't that so?
$endgroup$
– jpv
Oct 9 '15 at 18:06
$begingroup$
saz: When the process is not continuous, the Riemann sums might not converge to the integral. Isn't that so?
$endgroup$
– jpv
Oct 9 '15 at 18:06
$begingroup$
@jpv No, the sums convergence whenever the integrand is integrable; this follows directly from the definition of the Riemann integral (or Lebesgue integral, if you prefer it; it doesn't matter).
$endgroup$
– saz
Oct 9 '15 at 18:40
$begingroup$
@jpv No, the sums convergence whenever the integrand is integrable; this follows directly from the definition of the Riemann integral (or Lebesgue integral, if you prefer it; it doesn't matter).
$endgroup$
– saz
Oct 9 '15 at 18:40
$begingroup$
saz: I am assuming that $Y_t(omega) = int_0^t X_s(omega) ds$ exists for all $omega,t$ in the Lebesgue sense. I think that the Lebesgue integral cannot be approximated by Riemann sums. The way to approximate the Lebesgue integral is by using $Y_t^+,Y_t^-$ and using the fact that these non-negative random variables can be approximated by step functions. However, I cannot see how this entire process of approximation will preserve Gaussianity.
$endgroup$
– jpv
Oct 10 '15 at 3:42
$begingroup$
saz: I am assuming that $Y_t(omega) = int_0^t X_s(omega) ds$ exists for all $omega,t$ in the Lebesgue sense. I think that the Lebesgue integral cannot be approximated by Riemann sums. The way to approximate the Lebesgue integral is by using $Y_t^+,Y_t^-$ and using the fact that these non-negative random variables can be approximated by step functions. However, I cannot see how this entire process of approximation will preserve Gaussianity.
$endgroup$
– jpv
Oct 10 '15 at 3:42
|
show 8 more comments
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$begingroup$
Yes, the measurability in the sense of $([0,infty)times Omega, mathcal{B}([0,infty))otimes mathcal F) to (mathbb{R},mathcal{B}(mathbb{R}))$ is enough.
$endgroup$
– zhoraster
Oct 9 '15 at 7:13
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@zhoraster Why do you think so? For example, if $Z sim N(0,1)$, then $X_t := frac{1}{t} Z$, $t>0$, $X_0 := 0$, is Gaussian, but the integral $int_0^t X_s , ds$ is not well-defined. As far as I can see, we need some additional assumption on the integrability, e.g. $sup_{t leq T} mathbb{E}(|X_t|)<infty$ for $T>0$.
$endgroup$
– saz
Oct 9 '15 at 7:38
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@saz, I assume that everything is well-defined. Of course, to this end we need some integrability.
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– zhoraster
Oct 9 '15 at 8:44
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@zhoraster I see.
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– saz
Oct 9 '15 at 8:48