Mixing
How to prove that the process $Y_t$ is martingale?
$begingroup$
We have SDE $$dX_t = X_t(1-X_t)dW_t$$ where $W$ is standard Brownian motion and $X_0 = x_0 in (0,1)$. Assume that holds $P(X_t in (0,1))=1.$
For any $u$, we can define $f(x) = (frac{x}{1-x})^u sqrt{x(1-x)}$.
So first i have to prove that if we take $lambda = u^2 - frac{1}{4}$, for any $u$ the process $$Y_t = e^{frac{-lambda t}{2}}f(X_t)$$ is local martingale.
That part i did it. But now i have to take complex $u$. If i take $ui$ with real part $u$, then the process $Y_t$ is bounded and martingale. How i prove this?
stochastic-processes stochastic-calculus martingales local-martingales
asked Jan 30 at 17:19
tjakratjakra
465
$endgroup$
add a comment |
$begingroup$
We have SDE $$dX_t = X_t(1-X_t)dW_t$$ where $W$ is standard Brownian motion and $X_0 = x_0 in (0,1)$. Assume that holds $P(X_t in (0,1))=1.$
For any $u$, we can define $f(x) = (frac{x}{1-x})^u sqrt{x(1-x)}$.
So first i have to prove that if we take $lambda = u^2 - frac{1}{4}$, for any $u$ the process $$Y_t = e^{frac{-lambda t}{2}}f(X_t)$$ is local martingale.
That part i did it. But now i have to take complex $u$. If i take $ui$ with real part $u$, then the process $Y_t$ is bounded and martingale. How i prove this?
stochastic-processes stochastic-calculus martingales local-martingales
asked Jan 30 at 17:19
tjakratjakra
465
$endgroup$
$begingroup$
I don't understand what you mean by "If I take $ui$ with real part $u$ [...]"? Do you mean imaginary part? Do you want to consider $u=iv$ with $v$ real...?
$endgroup$
– saz
Jan 30 at 18:59
$begingroup$
Yes, the last part if i take $u = vi$ where $v$ is real number.
$endgroup$
– tjakra
Jan 31 at 15:10
add a comment |
$begingroup$
We have SDE $$dX_t = X_t(1-X_t)dW_t$$ where $W$ is standard Brownian motion and $X_0 = x_0 in (0,1)$. Assume that holds $P(X_t in (0,1))=1.$
For any $u$, we can define $f(x) = (frac{x}{1-x})^u sqrt{x(1-x)}$.
So first i have to prove that if we take $lambda = u^2 - frac{1}{4}$, for any $u$ the process $$Y_t = e^{frac{-lambda t}{2}}f(X_t)$$ is local martingale.
That part i did it. But now i have to take complex $u$. If i take $ui$ with real part $u$, then the process $Y_t$ is bounded and martingale. How i prove this?
stochastic-processes stochastic-calculus martingales local-martingales
asked Jan 30 at 17:19
tjakratjakra
465
$endgroup$
We have SDE $$dX_t = X_t(1-X_t)dW_t$$ where $W$ is standard Brownian motion and $X_0 = x_0 in (0,1)$. Assume that holds $P(X_t in (0,1))=1.$
For any $u$, we can define $f(x) = (frac{x}{1-x})^u sqrt{x(1-x)}$.
So first i have to prove that if we take $lambda = u^2 - frac{1}{4}$, for any $u$ the process $$Y_t = e^{frac{-lambda t}{2}}f(X_t)$$ is local martingale.
That part i did it. But now i have to take complex $u$. If i take $ui$ with real part $u$, then the process $Y_t$ is bounded and martingale. How i prove this?
stochastic-processes stochastic-calculus martingales local-martingales
stochastic-processes stochastic-calculus martingales local-martingales
asked Jan 30 at 17:19
tjakratjakra
465
asked Jan 30 at 17:19
tjakratjakra
465
asked Jan 30 at 17:19
tjakratjakra
465
asked Jan 30 at 17:19
tjakratjakra
465
asked Jan 30 at 17:19
tjakratjakra
465
465
$begingroup$
I don't understand what you mean by "If I take $ui$ with real part $u$ [...]"? Do you mean imaginary part? Do you want to consider $u=iv$ with $v$ real...?
$endgroup$
– saz
Jan 30 at 18:59
$begingroup$
Yes, the last part if i take $u = vi$ where $v$ is real number.
$endgroup$
– tjakra
Jan 31 at 15:10
add a comment |
$begingroup$
I don't understand what you mean by "If I take $ui$ with real part $u$ [...]"? Do you mean imaginary part? Do you want to consider $u=iv$ with $v$ real...?
$endgroup$
– saz
Jan 30 at 18:59
$begingroup$
Yes, the last part if i take $u = vi$ where $v$ is real number.
$endgroup$
– tjakra
Jan 31 at 15:10
$begingroup$
I don't understand what you mean by "If I take $ui$ with real part $u$ [...]"? Do you mean imaginary part? Do you want to consider $u=iv$ with $v$ real...?
$endgroup$
– saz
Jan 30 at 18:59
$begingroup$
I don't understand what you mean by "If I take $ui$ with real part $u$ [...]"? Do you mean imaginary part? Do you want to consider $u=iv$ with $v$ real...?
$endgroup$
– saz
Jan 30 at 18:59
$begingroup$
Yes, the last part if i take $u = vi$ where $v$ is real number.
$endgroup$
– tjakra
Jan 31 at 15:10
$begingroup$
Yes, the last part if i take $u = vi$ where $v$ is real number.
$endgroup$
– tjakra
Jan 31 at 15:10
add a comment |
1 Answer
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$begingroup$
If $y>0$ is a positive number and $u in mathbb{C}$ complex, then
$$y^u = exp(u log(y)).$$
Now if $u = i v$ for $v in mathbb{R}$, then this implies that
$$|y^u| = |exp(i v log(y))| leq 1 quad text{for all $y>0$} tag{1}$$
as $|e^{ix}|=1$ for any real $x$. Since the stochastic process satisfies, by assumption, $X_t in (0,1)$ almost surely, we can use $(1)$ for $y=X_t/(1-X_t)$ to obtain that $|f(X_t)| leq 1$ almost surely. In particular,
$$|Y_t| leq |f(X_t)| leq 1,$$
and this shows that $(Y_t)_{t geq 0}$ is bounded. Since any bounded local martingale is a "true" martingale you are done if you can show that $(Y_t)_{t geq 0}$ is a local martingale. To prove this, you can use exactly the same reasoning as you did in the real-valued case (i.e. apply Itô's formula and check that the process can be written as a stochastic integral with respect to Brownian motion).
answered Jan 31 at 18:43
sazsaz
82.1k862131
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add a comment |
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1 Answer
1
active
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1 Answer
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active
oldest
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oldest
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oldest
votes
$begingroup$
If $y>0$ is a positive number and $u in mathbb{C}$ complex, then
$$y^u = exp(u log(y)).$$
Now if $u = i v$ for $v in mathbb{R}$, then this implies that
$$|y^u| = |exp(i v log(y))| leq 1 quad text{for all $y>0$} tag{1}$$
as $|e^{ix}|=1$ for any real $x$. Since the stochastic process satisfies, by assumption, $X_t in (0,1)$ almost surely, we can use $(1)$ for $y=X_t/(1-X_t)$ to obtain that $|f(X_t)| leq 1$ almost surely. In particular,
$$|Y_t| leq |f(X_t)| leq 1,$$
and this shows that $(Y_t)_{t geq 0}$ is bounded. Since any bounded local martingale is a "true" martingale you are done if you can show that $(Y_t)_{t geq 0}$ is a local martingale. To prove this, you can use exactly the same reasoning as you did in the real-valued case (i.e. apply Itô's formula and check that the process can be written as a stochastic integral with respect to Brownian motion).
answered Jan 31 at 18:43
sazsaz
82.1k862131
$endgroup$
add a comment |
$begingroup$
If $y>0$ is a positive number and $u in mathbb{C}$ complex, then
$$y^u = exp(u log(y)).$$
Now if $u = i v$ for $v in mathbb{R}$, then this implies that
$$|y^u| = |exp(i v log(y))| leq 1 quad text{for all $y>0$} tag{1}$$
as $|e^{ix}|=1$ for any real $x$. Since the stochastic process satisfies, by assumption, $X_t in (0,1)$ almost surely, we can use $(1)$ for $y=X_t/(1-X_t)$ to obtain that $|f(X_t)| leq 1$ almost surely. In particular,
$$|Y_t| leq |f(X_t)| leq 1,$$
and this shows that $(Y_t)_{t geq 0}$ is bounded. Since any bounded local martingale is a "true" martingale you are done if you can show that $(Y_t)_{t geq 0}$ is a local martingale. To prove this, you can use exactly the same reasoning as you did in the real-valued case (i.e. apply Itô's formula and check that the process can be written as a stochastic integral with respect to Brownian motion).
answered Jan 31 at 18:43
sazsaz
82.1k862131
$endgroup$
add a comment |
$begingroup$
If $y>0$ is a positive number and $u in mathbb{C}$ complex, then
$$y^u = exp(u log(y)).$$
Now if $u = i v$ for $v in mathbb{R}$, then this implies that
$$|y^u| = |exp(i v log(y))| leq 1 quad text{for all $y>0$} tag{1}$$
as $|e^{ix}|=1$ for any real $x$. Since the stochastic process satisfies, by assumption, $X_t in (0,1)$ almost surely, we can use $(1)$ for $y=X_t/(1-X_t)$ to obtain that $|f(X_t)| leq 1$ almost surely. In particular,
$$|Y_t| leq |f(X_t)| leq 1,$$
and this shows that $(Y_t)_{t geq 0}$ is bounded. Since any bounded local martingale is a "true" martingale you are done if you can show that $(Y_t)_{t geq 0}$ is a local martingale. To prove this, you can use exactly the same reasoning as you did in the real-valued case (i.e. apply Itô's formula and check that the process can be written as a stochastic integral with respect to Brownian motion).
answered Jan 31 at 18:43
sazsaz
82.1k862131
$endgroup$
If $y>0$ is a positive number and $u in mathbb{C}$ complex, then
$$y^u = exp(u log(y)).$$
Now if $u = i v$ for $v in mathbb{R}$, then this implies that
$$|y^u| = |exp(i v log(y))| leq 1 quad text{for all $y>0$} tag{1}$$
as $|e^{ix}|=1$ for any real $x$. Since the stochastic process satisfies, by assumption, $X_t in (0,1)$ almost surely, we can use $(1)$ for $y=X_t/(1-X_t)$ to obtain that $|f(X_t)| leq 1$ almost surely. In particular,
$$|Y_t| leq |f(X_t)| leq 1,$$
and this shows that $(Y_t)_{t geq 0}$ is bounded. Since any bounded local martingale is a "true" martingale you are done if you can show that $(Y_t)_{t geq 0}$ is a local martingale. To prove this, you can use exactly the same reasoning as you did in the real-valued case (i.e. apply Itô's formula and check that the process can be written as a stochastic integral with respect to Brownian motion).
answered Jan 31 at 18:43
sazsaz
82.1k862131
answered Jan 31 at 18:43
sazsaz
82.1k862131
answered Jan 31 at 18:43
sazsaz
82.1k862131
answered Jan 31 at 18:43
sazsaz
82.1k862131
82.1k862131
add a comment |
add a comment |
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$begingroup$
I don't understand what you mean by "If I take $ui$ with real part $u$ [...]"? Do you mean imaginary part? Do you want to consider $u=iv$ with $v$ real...?
$endgroup$
– saz
Jan 30 at 18:59
$begingroup$
Yes, the last part if i take $u = vi$ where $v$ is real number.
$endgroup$
– tjakra
Jan 31 at 15:10