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Greyearl
Greyearl
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Let $W$ be a standard one dimensional Brownian motion and let $A$ be the process defined by : $$\forall \ t\geq 0: \quad A_t := \int_0^t\left(1 + e^{W_s}\right)\mathrm{d}s$$

and for $t\geq 0$, we define the following stopping time: $$\sigma_t := \inf \{s\geq 0: \ A_s > t\}$$

Notice that a.s: $$\forall \ t\geq 0: \quad \sigma_t \leq t$$

Now, define the martingale $M$ as a time changed Brownian motion, i.e. $M := \left(W_{\sigma_t}\right)_{t\geq 0}$$M := \left(W_{A_t}\right)_{t\geq 0}$. Can we prove that $M$ goes to infinity in probability ? Thanks.

Let $W$ be a standard one dimensional Brownian motion and let $A$ be the process defined by : $$\forall \ t\geq 0: \quad A_t := \int_0^t\left(1 + e^{W_s}\right)\mathrm{d}s$$

and for $t\geq 0$, we define the following stopping time: $$\sigma_t := \inf \{s\geq 0: \ A_s > t\}$$

Notice that a.s: $$\forall \ t\geq 0: \quad \sigma_t \leq t$$

Now, define the martingale $M$ as a time changed Brownian motion, i.e. $M := \left(W_{\sigma_t}\right)_{t\geq 0}$. Can we prove that $M$ goes to infinity in probability ? Thanks.

Let $W$ be a standard one dimensional Brownian motion and let $A$ be the process defined by : $$\forall \ t\geq 0: \quad A_t := \int_0^t\left(1 + e^{W_s}\right)\mathrm{d}s$$

and for $t\geq 0$, we define the following stopping time: $$\sigma_t := \inf \{s\geq 0: \ A_s > t\}$$

Notice that a.s: $$\forall \ t\geq 0: \quad \sigma_t \leq t$$

Now, define the martingale $M$ as a time changed Brownian motion, i.e. $M := \left(W_{A_t}\right)_{t\geq 0}$. Can we prove that $M$ goes to infinity in probability ? Thanks.

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Greyearl
Greyearl
  • 63
  • 5

Let $W$ be a standard one dimensional Brownian motion and let $A$ be the process defined by : $$\forall \ t\geq 0: \quad A_t := \int_0^t\left(1 + e^{W_t}\right)\mathrm{d}t$$$$\forall \ t\geq 0: \quad A_t := \int_0^t\left(1 + e^{W_s}\right)\mathrm{d}s$$

and for $t\geq 0$, we define the following stopping time: $$\sigma_t := \inf \{s\geq 0: \ A_s > t\}$$

Notice that a.s: $$\forall \ t\geq 0: \quad \sigma_t \leq t$$

Now, define the martingale $M$ as a time changed Brownian motion, i.e. $M := \left(W_{\sigma_t}\right)_{t\geq 0}$. Can we prove that $M$ goes to infinity in probability ? Thanks.

Let $W$ be a standard one dimensional Brownian motion and let $A$ be the process defined by : $$\forall \ t\geq 0: \quad A_t := \int_0^t\left(1 + e^{W_t}\right)\mathrm{d}t$$

and for $t\geq 0$, we define the following stopping time: $$\sigma_t := \inf \{s\geq 0: \ A_s > t\}$$

Notice that a.s: $$\forall \ t\geq 0: \quad \sigma_t \leq t$$

Now, define the martingale $M$ as a time changed Brownian motion, i.e. $M := \left(W_{\sigma_t}\right)_{t\geq 0}$. Can we prove that $M$ goes to infinity in probability ? Thanks.

Let $W$ be a standard one dimensional Brownian motion and let $A$ be the process defined by : $$\forall \ t\geq 0: \quad A_t := \int_0^t\left(1 + e^{W_s}\right)\mathrm{d}s$$

and for $t\geq 0$, we define the following stopping time: $$\sigma_t := \inf \{s\geq 0: \ A_s > t\}$$

Notice that a.s: $$\forall \ t\geq 0: \quad \sigma_t \leq t$$

Now, define the martingale $M$ as a time changed Brownian motion, i.e. $M := \left(W_{\sigma_t}\right)_{t\geq 0}$. Can we prove that $M$ goes to infinity in probability ? Thanks.

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Greyearl
Greyearl
  • 63
  • 5

On the convergence of a martingale

Let $W$ be a standard one dimensional Brownian motion and let $A$ be the process defined by : $$\forall \ t\geq 0: \quad A_t := \int_0^t\left(1 + e^{W_t}\right)\mathrm{d}t$$

and for $t\geq 0$, we define the following stopping time: $$\sigma_t := \inf \{s\geq 0: \ A_s > t\}$$

Notice that a.s: $$\forall \ t\geq 0: \quad \sigma_t \leq t$$

Now, define the martingale $M$ as a time changed Brownian motion, i.e. $M := \left(W_{\sigma_t}\right)_{t\geq 0}$. Can we prove that $M$ goes to infinity in probability ? Thanks.