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3

We have that $$E(B(t)W(1))=E \left( (W(t)-tW(1))W(1) \right) =E(W(t)W(1)-tW(1)W(1)) $$ Using the linearity of the expectation we also have that this expression is equal to $$ E(W(t)W(1))-E(tW(1)W(1))=E(W(t)W(1))-tE(W(1)W(1))$$ As @PetiteEtincelle said, $E(W(t)W(s))=\min(t,s)$, so finally we conclude that $$ E(B(t)W(1))= \min (t,1)-t\min(1,1)=t-t=0$$


3

If $D$ is a domain in $\mathbb{R}^n$, and $x_0 \in D$ is a point, then the distribution of Brownian motion started at $x_0$ stopped at the time when it leaves $D$ is the harmonic measure of $\partial D$ with base point $x_0$. In the plane you can often use conformal invariance to calculate harmonic measure. Since you know that the harmonic measure of the ...


2

A simple approach (which avoids the task of checking the nonnegativity of the function in your question) is to compare the hitting times $\tau_1$ and $\tau_2$ with the hitting time $\theta$ of $\pm2\pi$ by the process $$dY=\mathrm dW-\mathrm{sgn}(Y)\,\mathrm dt,$$ starting from some $|y|\leqslant2\pi$. Since $|\sin|\leqslant1$, $\tau_1$ and $\tau_2$ are ...


2

Basically, the definition of expected value gives $$\sum_{n=0}^\infty n p_n(t)=\vec{\mu}(t),$$ and the definition of a probability distribution gives $$\sum_{n=0}^\infty p_n(t)=1.$$ In particular, $$\sum_{n=0}^\infty (n+1)p_n(t) = \vec{\mu}(t)+1.$$ Being explicit: \begin{align*} &\phantom{{}={}} \lambda \sum_{n = 0}^\infty(n+1)p_n(t) - \lambda ...


2

You can study this using a Markov chain with two states (1=post-win, 2 = post-loss), and transition matrix $$ \pmatrix{ p & 1-p\cr q & 1-q\cr}$$ The equilibrium probabilities are $q/(q+1-p)$ and $(1-p)/(q+1-p)$. The long term proportion of wins is then $q/(q+1-p)$.


1

Hint: The random variables $S_A$ and $S_{A+B}-S_A$ are independent and $S_{A+B}-S_A$ is distributed like $S_B$. Now, you ought to know the distribution of each $S_A$ hence you ought to be able to finish this.


1

$\dfrac{\log X_n}{n} = \dfrac{\sum_{m \le n} \log Y_m}{n} \to E\log Y_1$ almost surely by the strong law of large number. And by Jensen's inequality, $E\log Y_1 < \log EY_1 =0$ since $P(Y = 1) < 1$ So we get that $\dfrac{\log X_n}{n}$ converges to a strictly negative number alomost surely, thus $\log X_n \to -\infty$, i.e. $X_n \to 0$


1

It’s a little confusing to have both a constant $\mu$ and a function $\mathbf{\mu}$, so I’m going to replace the former by $\nu$ for this answer. In addition, there appears to be a typo in the first line: the argument makes sense only if the expression on the righthand side of the first equals sign is $$\lambda\sum_{n\ge 1}np_{n-1}(t)-\lambda\sum_{n\ge ...


1

The joint distribution of the running maximum $M_1 = \max_{0 \leq t \leq 1}B_t$ and $B_1$ is $$f(m,b) = \frac{2(2m-b)}{\sqrt{2\pi}}\exp(-\frac{(2m-b)^2}{2}), \qquad m\geq 0, b \leq m $$ Hence, $P(M_1 = B_1) = 0$ Note : In your approach, you replaced $M_1$ with $\vert B_1 \vert $ which is not true! They have same distribution but the random variables are ...


1

$P(B_1 = \max_{t\in[0,1]}B_t) = P(\max_{t\in[0,1]}(B_t-B_1) = 0)= P(\max_{t\in[0,1]}(B_{1-t}-B_1) = 0)$ You can easily check that $(B_{1-t} - B_1)_{t\in [0,1]}$ is itself a Brownian motion, so $P(\max_{t\in[0,1]}(B_{1-t}-B_1) = 0) = P(\max_{t\in[0,1]}B_t = 0) = P(|B_1| = 0) = 0$


1

Hint: Use $$\left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k \right\} = \bigcup_{\ell \in \mathbb{N}} \left\{ \limsup_{t \to 0+} \frac{B_t}{\sqrt{t}} > k+ \frac{1}{\ell} \right\}.$$


1

Not sure I understand the bafflement here... Let $X=\limsup\limits_{t \to 0^{+} } B_t/\sqrt{t}$. For every positive $s$, $X$ is equal to the same limsup restricted to the values of $t$ in the interval $(0,s)$ hence $X$ is $\mathcal F_s$-measurable, that is, for every Borel subset $A$, $[X\in A]$ belongs to $\mathcal F_s$. Thus, $[X\in A]$ belongs to ...


1

Let $N(t)=M(t+\tau)$ and $G(t)=F(t+\tau)$ then, for every nonnegative $t$, $$N(t)=G(t)+\int_0^tN(t-x)\,\mathrm dF(x)$$ hence $$\tilde{N}(s)=\frac{\tilde{G}(s)}{1-\tilde{f}(s)}.$$


1

First of all, your second and third paragraphs. Recall that if $\xi$ is $F$-measurable and $F\subseteq G$ then $\xi$ is $G$-measurable: that follows from the definition of measurability. Now, if $\Bbb F = (F_n)_{n\geq 0}$ is a filtration and $\xi = (\xi_n)_{n\geq 0}$ is $\Bbb F$-adapted, then for each $n$ we have $\xi_n$ is $F_n$-measurable. Since ...


1

Note that your series for $G_n$ is a geometric series. You can find a simple closed form for the sum. Then take the derivative.


1

If $G_n'(s) =\sum_{r=1}^\infty \frac{n^{r-1}}{(n+1)^{r+1}}\frac{d}{dx}\left[s^r\right] = \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}}s^{r-1} $, then $G_n'(1) = \sum_{r=1}^\infty\frac{rn^{r-1}}{(n+1)^{r+1}} = \frac1{n(n+1)}\sum_{r=1}^\infty\frac{rn^{r}}{(n+1)^{r}} = \frac1{n(n+1)}\sum_{r=1}^\infty rz^r $, where $z = \frac{n}{n+1} $. Since $\sum_{r=1}^\infty ...


1

In addition to your comment: Let $P$ be a stochastic, irreducible matrix. Since $P$ is irreducible it will be either aperiodic or periodic. $P$ aperiodic$\iff\displaystyle\lim_{k\to\infty}P^k$ exists. $P$ periodic matrix with period $d>1$ $\iff \displaystyle\lim_{k\to\infty}P^k$ does not exist. We can prove the latter using cyclic subclasses.



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