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4

Hint: The (homogeneous) Poisson process has stationary independent increments, so write: $$N_t=N_s+(N_t-N_s)=N_s+N_{t-s}$$ and use that $E[N_s \mid F_s]=N_s$, since $N_s$ is known at time $s$ $E[N_{t-s}\mid F_s]=E[N_{t-s}]$, since $N_{t-s}$ is independent from what happened up to time $s$ $N_{t-s} \sim$ Poisson$(λ(t-s))$, in order to calculate ...

4

This martingale is also known as De Moivre's martingale. As you correctly observed, it is non-negative and hence converges a.s. to some $T$. Depending on the values of $p,q$ there are $3$ cases: $p=q$. You have correctly treated this case, by observing that $T_n$ is constant and equal to $1$ and hence uniformly integrable. $p>q$. Now $S_n\to +\infty$ ...

3

Observe that $$\max_{1\leq m \leq n} X_m \geq \lambda \Leftrightarrow \max_{1\leq m \leq n} X_m + c \geq \lambda + c \implies \max_{1\leq m \leq n} \lvert X_m + c \rvert \geq \lambda + c$$ $$\max_{1\leq m \leq n} \lvert X_m + c \rvert \geq \lambda + c \Leftrightarrow \max_{1\leq m \leq n} \lvert X_m + c \rvert^2 \geq (\lambda + c)^2$$ The second relation ...

2

Yes, if $c$ is such that $\mathbb{E}[c^{W_{t+1}} \mid F_t] = c^{W_t}$ then it will also be true that $E[c^{W_{t+1}-W_t} \mid F_t] = 1$. This is simply the well-known property of conditional expectation that if $Y$ is measurable with respect to $\mathcal{G}$, then $Y E[X \mid \mathcal{G}] = E[XY \mid \mathcal{G}]$. If I am not mistaken, you will find that ...

2

It's better to split it up the other way round, i.e. to note that it suffices to show that $$\mathbb{E}(Y_{n+1} \mid \mathcal{F}_n) \leq x \tag{1}$$ and $$\mathbb{E}(Y_{n+1} \mid \mathcal{F}_n) \leq X_n \tag{2}$$ To prove $(1)$ use that $Y_{n+1} \leq x$; for $(2)$ use $Y_{n+1} \leq X_{n+1}$ and the super-martingale property of $(X_n)_{n \in \mathbb{N}}$. ...

2

The process $\{Y_m\}_{m\ge 1}$ is a Poisson process that counts the events up to time $m \in \Bbb R_+$. The pexpression $$\tilde Y_1:=Y_{m+1}-Y_m$$ counts the number of events in the time period $(m, m+1]$ and due to the memoryless property of the exponential distribution it can be proven that it has again the Poisson distribution with Property (P1): ...

2

Hint: Use the optional sampling theorem for the submartingale $|M_n|$. Direct proof: First, note that it suffices to show $\mathbb{E}(|M_S|) \leq \mathbb{E}(|M_n|)$ for any stopping time $S$ such that $S \leq n$. Since $(M_n)_{n \in \mathbb{N}}$ is a martingale, we know from Jensen's inequaliy that $(|M_n|)_{n \in \mathbb{N}}$ is a submartingale, i.e. ...

1

To answer your first question, it is common probabilist short-hand to write things like $\{T = n\}$ to mean $\{\omega \in \Omega : T(\omega) = n\}$. Thus the event $\{T=n\}$ is the outcomes for which the stopping time evaluates to $n$. For your second question, recall that a martingale satisfies $E[Z_n] = E[Z_0]$ for all $n$. It turns out that under certain ...

1

This is a settled question but I like to add another solution :-) We have $W_t \mid \mathcal{F}_s \sim \mathcal{N}(W_s,t-s)$ hence using this we have $$\mathbb{E}(W_t^3 \mid \mathcal{F}_s)=W_s^3+3W_s(t-s).$$ Therefore \begin{align} \mathbb{E}(W_t^3 - 3t W_t \mid \mathcal{F}_s)&=\mathbb{E}(W_t^3 \mid \mathcal{F}_s)-3t\mathbb{E}( W_t \mid ...

1

Since $$X_s^2 = (X_{s-}+\Delta X_s)^2 = X_{s-}^2+ 2 X_{s-} \Delta X_s + (\Delta X_s)^2$$ we have $$\Delta (X_s^2) \stackrel{\text{def}}{=} X_s^2-X_{s-}^2 = 2 X_{s-} \Delta X_s + (\Delta X_s)^2.$$ Hence, $$\Delta (X_s^2) - 2 X_{s-} \Delta X_s - (\Delta X_s)^2 = 0$$ for all $s$. This means that the sum $$\sum_{s \leq t} \left( \Delta (X_s^2) - 2 X_{s-} ... 1$$ X_{n+1} = X_n + T_{n+1} $$where T_{n+1} is independent of X_1,\ldots,X_n and$$ \Pr(T_{n+1}\in A) = \int_A \frac{dt}{(2+t^2)^{3/2}}  for every measurable set $A$. (Thus $T_{n+1}$ has a t-distribution with $2$ degrees of freedom.)

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