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Here we study

$$M_{t}=B_{A_{t}}\stackrel{d}{=}\int_{0}^{t}\sqrt{1+e^{W_{s}}}dW_{s}.$$

First, as mentioned Martingale Convergence here

Theorem 4 Let X be a continuous martingale. Then, almost surely, one of the following is satisfied

• $X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite.
• $\limsup_{t\rightarrow\infty}X_t=\infty$ and $\liminf_{t\rightarrow\infty}X_t=-\infty$. In this case, the process hits every value in $\mathbb R$ at arbitrarily large times.

By Itô isometry we have

$$E[M^{2}_{t}]=\int^t 1+Ee^{W_{s}}ds=\int^t 1+e^{\frac{s}{2}}ds \to +\infty,$$

which also shows that $<M>_{\infty}=\infty$. In general for continuous martingales we have that

"$X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite" iff $<X>_{\infty}<\infty.$

The left-right direction that is relevant here is as follows: we use the stopping time $\tau_{n}:=\inf\{t\geq 0: |X_{t}|\geq n\}$ and the bound $$E[<X>_{t\wedge \tau_{n}}]=E[X^{2}_{t\wedge \tau_{n}}]\leq n^{2}$$

to get that the event $\{\sup|X_t|<\infty\}=\bigcup_{n}\{\sup|X_t|<n\}$ implies that the finiteness $<X>_{\infty}<\infty$. (For the other direction we can use instead $\tau_{n}:=\inf\{t\geq 0: <X>_{t}\geq n\}$.)

So we can't have the first case in the above theorem, because it would give a contradiction and so we have the second case.

For the "in probability" part, we use conclusion of the second case due to the continuity that the process hits every value in $\mathbb R$ at arbitrarily large times. Therefore, given every $t>0$, we have that $X_{t}$ will hit zero in finite time , we write this as $\tau_{0}\circ t<\infty$. So even though we have that $t+\tau_{0}\circ t\to +\infty$, we have $X_{t+\tau_{0}\circ t}=0$.

Here we study

$$M_{t}=B_{A_{t}}\stackrel{d}{=}\int_{0}^{t}\sqrt{1+e^{W_{s}}}dW_{s}.$$

First, as mentioned Martingale Convergence here

Theorem 4 Let X be a continuous martingale. Then, almost surely, one of the following is satisfied

• $X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite.
• $\limsup_{t\rightarrow\infty}X_t=\infty$ and $\liminf_{t\rightarrow\infty}X_t=-\infty$. In this case, the process hits every value in $\mathbb R$ at arbitrarily large times.

By Itô isometry we have

$$E[M^{2}_{t}]=\int^t 1+Ee^{W_{s}}ds=\int^t 1+e^{\frac{s}{2}}ds \to +\infty,$$

which also shows that $<M>_{\infty}=\infty$. In general for continuous martingales we have that

"$X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite" iff $<X>_{\infty}<\infty.$

The left-right direction that is relevant here is as follows: we use the stopping time $\tau_{n}:=\inf\{t\geq 0: |X_{t}|\geq n\}$ and the bound $$E[<X>_{t\wedge \tau_{n}}]=E[X^{2}_{t\wedge \tau_{n}}]\leq n^{2}$$

to get that the event $\{\sup|X_t|<\infty\}=\bigcup_{n}\{\sup|X_t|<n\}$ implies that the finiteness $<X>_{\infty}<\infty$. (For the other direction we can use instead $\tau_{n}:=\inf\{t\geq 0: <X>_{t}\geq n\}$.)

So we can't have the first case in the above theorem, because it would give a contradiction and so we have the second case.

For the "in probability" part, we will use a reverse tail inequality: when $\mathbb {E} [X]=0,\,\mathbb {E} [X^{2}]=1$ then

$$\Pr(X\geq 0)\geq {\frac {2{\sqrt {3}}-3}{\mathbb {E} [X^{4}]}},$$

from Berger's "The Fourth Moment Method. By symmetry of Brownian motion we write

$$P[M_{t}\leq 0]=P[0\leq \tilde{M}_{t}:=\int_{0}^{t}\sqrt{1+e^{-W_{s}}}dW_{s}].$$

We still have $E[\tilde{M}_{t}]=0$ and we will need the fourth moment bound

$$E\left(\int f dW\right)^{4}\leq c_{2}\left(\int Ef^{2} ds\right)^{2},$$

from Corollary 4 in "Notes on the Itô Calculus" by Steven P. Lalley. Therefore, for $X_{t}:=\tilde{M}_{t}\frac{1}{\sqrt{E[\tilde{M}_{t}^{2}]}}$ and using the above bound we get

$$P[\tilde{M}_{t}\geq 0]\geq {\frac {2{\sqrt {3}}-3}{\mathbb {E} [X^{4}]}}\geq \frac {2{\sqrt {3}}-3}{c_{2}}>0$$

for all $t>0$.

Here we study

$$M_{t}=B_{A_{t}}\stackrel{d}{=}\int_{0}^{t}\sqrt{1+e^{W_{s}}}dW_{s}.$$

First, as mentioned Martingale Convergence here

Theorem 4 Let X be a continuous martingale. Then, almost surely, one of the following is satisfied

• $X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite.
• $\limsup_{t\rightarrow\infty}X_t=\infty$ and $\liminf_{t\rightarrow\infty}X_t=-\infty$. In this case, the process hits every value in $\mathbb R$ at arbitrarily large times.

By Itô isometry we have

$$E[M^{2}_{t}]=\int^t 1+Ee^{W_{s}}ds=\int^t 1+e^{\frac{s}{2}}ds \to +\infty,$$

which also shows that $<M>_{\infty}=\infty$. In general for continuous martingales we have that

"$X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite" iff $<X>_{\infty}<\infty.$

The left-right direction that is relevant here is as follows: we use the stopping time $\tau_{n}:=\inf\{t\geq 0: |X_{t}|\geq n\}$ and the bound $$E[<X>_{t\wedge \tau_{n}}]=E[X^{2}_{t\wedge \tau_{n}}]\leq n^{2}$$

to get that the event $\{\sup|X_t|<\infty\}=\bigcup_{n}\{\sup|X_t|<n\}$ implies that the finiteness $<X>_{\infty}<\infty$. (For the other direction we can use instead $\tau_{n}:=\inf\{t\geq 0: <X>_{t}\geq n\}$.)

So we can't have the first case in the above theorem, because it would give a contradiction and so we have the second case.

Here we study

$$M_{t}=B_{A_{t}}\stackrel{d}{=}\int_{0}^{t}\sqrt{1+e^{W_{s}}}dW_{s}.$$

First, as mentioned Martingale Convergence here

Theorem 4 Let X be a continuous martingale. Then, almost surely, one of the following is satisfied

• $X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite.
• $\limsup_{t\rightarrow\infty}X_t=\infty$ and $\liminf_{t\rightarrow\infty}X_t=-\infty$. In this case, the process hits every value in $\mathbb R$ at arbitrarily large times.

By Itô isometry we have

$$E[M^{2}_{t}]=\int^t 1+Ee^{W_{s}}ds=\int^t 1+e^{\frac{s}{2}}ds \to +\infty,$$

which also shows that $<M>_{\infty}=\infty$. In general for continuous martingales we have that

"$X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite" iff $<X>_{\infty}<\infty.$

The left-right direction that is relevant here is as follows: we use the stopping time $\tau_{n}:=\inf\{t\geq 0: |X_{t}|\geq n\}$ and the bound $$E[<X>_{t\wedge \tau_{n}}]=E[X^{2}_{t\wedge \tau_{n}}]\leq n^{2}$$

to get that the event $\{\sup|X_t|<\infty\}=\bigcup_{n}\{\sup|X_t|<n\}$ implies that the finiteness $<X>_{\infty}<\infty$. (For the other direction we can use instead $\tau_{n}:=\inf\{t\geq 0: <X>_{t}\geq n\}$.)

So we can't have the first case in the above theorem, because it would give a contradiction and so we have the second case.

For the "in probability" part, we use conclusion of the second case due to the continuity that the process hits every value in $\mathbb R$ at arbitrarily large times. Therefore, given every $t>0$, we have that $X_{t}$ will hit zero in finite time , we write this as $\tau_{0}\circ t<\infty$. So even though we have that $t+\tau_{0}\circ t\to +\infty$, we have $X_{t+\tau_{0}\circ t}=0$.

Here we study

$$M_{t}=B_{A_{t}}\stackrel{d}{=}\int_{0}^{t}\sqrt{1+e^{W_{s}}}dW_{s}.$$

First, as mentioned Martingale Convergence here

Theorem 4 Let X be a continuous martingale. Then, almost surely, one of the following is satisfied

• $X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite.
• $\limsup_{t\rightarrow\infty}X_t=\infty$ and $\liminf_{t\rightarrow\infty}X_t=-\infty$. In this case, the process hits every value in $\mathbb R$ at arbitrarily large times.

By Itô isometry we have

$$E[M^{2}_{t}]=\int^t 1+Ee^{W_{s}}ds=\int^t 1+e^{\frac{s}{2}}ds \to +\infty,$$

which also shows that $<M>_{\infty}=\infty$, and so we have the second case.

Here we study

$$M_{t}=B_{A_{t}}\stackrel{d}{=}\int_{0}^{t}\sqrt{1+e^{W_{s}}}dW_{s}.$$

First, as mentioned Martingale Convergence here

Theorem 4 Let X be a continuous martingale. Then, almost surely, one of the following is satisfied

• $X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite.
• $\limsup_{t\rightarrow\infty}X_t=\infty$ and $\liminf_{t\rightarrow\infty}X_t=-\infty$. In this case, the process hits every value in $\mathbb R$ at arbitrarily large times.

By Itô isometry we have

$$E[M^{2}_{t}]=\int^t 1+Ee^{W_{s}}ds=\int^t 1+e^{\frac{s}{2}}ds \to +\infty,$$

which also shows that $<M>_{\infty}=\infty$. In general for continuous martingales we have that

"$X_\infty=\lim_{t\rightarrow\infty}X_t$ exists and is finite" iff $<X>_{\infty}<\infty.$

The left-right direction that is relevant here is as follows: we use the stopping time $\tau_{n}:=\inf\{t\geq 0: |X_{t}|\geq n\}$ and the bound $$E[<X>_{t\wedge \tau_{n}}]=E[X^{2}_{t\wedge \tau_{n}}]\leq n^{2}$$

to get that the event $\{\sup|X_t|<\infty\}=\bigcup_{n}\{\sup|X_t|<n\}$ implies that the finiteness $<X>_{\infty}<\infty$. (For the other direction we can use instead $\tau_{n}:=\inf\{t\geq 0: <X>_{t}\geq n\}$.)

So we can't have the first case in the above theorem, because it would give a contradiction and so we have the second case.