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5

Neither of them holds, in general, for stochastic integrals. The trouble already starts if you consider measures which need not to be non-negative, i.e. signed measures. For a signed measure $\mu: (\Omega,\mathcal{A}) \to \mathbb{R}$ we cannot expect that the triangle inequality $$\left| \int f(x) \, \mu(dx) \right| \leq \int |f(x)| \, \mu(dx) \tag{1}$$ ...


2

For an approximation, which is rather good because there are a lot of white balls, each black ball when drawn has $\frac 12$ chance to go to either player. The winning sequences are $AAB, ABB, ABA, BAB$, so the chance of your event is $\frac 12$. The correct value will be a little higher because $A$ is more likely than $B$ to get the first black ball.


2

Fix $f \in C(E)$ and $\eta \in E$ such that $$f(\eta) = \min_{\zeta \in E} f(\zeta). $$ Then $$g(\zeta) := f(\zeta)-f(\eta) \in C(E)$$ defines a non-negative function and therefore, since $T$ is a positive operator, $$Tg \geq 0. \tag{1}$$ On the other hand, it follows from the linearity of $T$ and $T1 = 1$ that $$Tg = Tf - f(\eta). \tag{2}$$ Combining ...


2

Let the sample space be $[0,1]$ and let each element of the filtration be the borel $\sigma-algebra$. Let $I$ be some non-measurable subset and $\tau_x=\mathbb{1}(x), x\in [0,1]$. Then $\{\tau_x<t\}$ is the complement of a singleton for $x\le 1$ and $[0,1]$ otherwise so it belongs to each $F_t$, but $\{sup_{x\in I}\tau_x<t\}=\cap_{x\in I}\{\tau_x ...


1

First, it is not the same as the probability of having $k/0.65$ customers walk into the shop. For example, taking $k = 5$, it is really unlikely that exactly $2/0.65 \approx 7.692$ customers will walk into the shop. (Well, maybe less unlikely if the shop is located in King's Landing...) What you have is a rate of customers walking in and buying coffee of ...


1

Firstly, exponential distributions are continuous, which means you're dealing with probability densities rather than masses. $$f_X(x)=\lambda~\mathsf e^{-\lambda x}~\mathbf 1_{x\in(0;\infty)}\textrm{ and }f_Y(y)\textrm{ likewise.}$$ Secondly $Z=X+Y$ so $f_{X, Z}(x, z) = f_X(x)~f_Y(z-x)$ and $f_Z(z) = \int_\Bbb R f_X(x)~f_Y(z-x)\operatorname d x$ Thusly: ...


1

If you mean that $L = M$, then $$\operatorname{E}[Z] = \operatorname{E}[\operatorname{E}[Z \mid M]] = \operatorname{E}\left[\left. \sum_{k=1}^M \operatorname{E}[X_k] \;\right| M \right] = \operatorname{E}[M \mu] = \mu \operatorname{E}[M] = \mu \lambda.$$ The second moment is more difficult, since $\operatorname{E}[Z^2]$ is messy to deal with using the above ...


1

Yeah, applying Etemadi's inequality is a good idea. The following identity, which holds for any non-negative random variable $X$, will also be useful: $$\mathbb{E}(X) = \int_0^{\infty} \mathbb{P}(X > r) \, dr.$$ Because of the monotonicity of $r \mapsto \mathbb{P}(X > r)$ this implies $$b \sum_{k=0}^{\infty} \mathbb{P}(X > (k+1)b) \leq ...


1

So the thing is that what you are looking for is exactly the content of lemma 3. I report the claim here : Lemma 3 : Let $X$ be a càdlàg square integrable martingale and ${\xi}$ be a bounded predictable process. Then, $\int\xi\,dX$ is a square integrable martingale. In your case $W$ is a Brownian motion, so you fit the conditions for the lemma as a ...


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.)


1

Fix $\epsilon>0$. Since $(X_t)_{t \geq 0}$ satisfies the SLLN, there exists for almost all $\omega \in \Omega$ a constant $S=S(\omega)>0$ such that $$\left| \frac{X_s(\omega)}{s} - \mathbb{E}(X_1) \right| \leq \epsilon \quad \text{for all $s \geq S$}.$$ As $U_t \uparrow \infty$, we have $U_t(\omega) \geq S$ for $t \geq T=T(\omega)$ sufficiently ...


1

Hints: 1) Law of total probability, in particular, the second formula. Note that since $Y$ is continuous, you will use integrals instead of sums. 2 and 3): Use what you get from #1. $N$ is poisson distributed, so use $$\mathbb{E}[g(N)] = \sum_{n=0}^{\infty}g(n)f_N(n)\text{.}$$


1

If there is someone at your university or in your area who is an applied statistician, you should begin by having some discussions on these topics with him or her. Someone who knows you personally can give you a level of advice you can't get here. I'm not sure what you might gain from participating in competitions because they are intentionally beyond the ...


1

$E[X|\{ \emptyset, \Omega \}]$ is defined as the a.s. unique random variable in $\sigma(\{ \emptyset, \Omega \})$ such that for any set $A\in\sigma(\{ \emptyset, \Omega \})$, $E[E[X|\{ \emptyset, \Omega \}];A]=E[X;A]$. $\sigma(\{ \emptyset, \Omega \})$ is just $\{ \emptyset, \Omega \}$ and any r.v. measurable wrt this collection is a.s. constant. Take ...


1

I use book Applied Stochastic Control and Jump Diffusion (2nd edittion) written by Bernt Oksendal and Agnes Sulem. You can find more details therein. Let $$dX_{t}=a(t,X_{t})dt+b(t,X_{t})dB_{t}+\int_{\mathbb{R}}\gamma(t,X_{t},z)N(dz,dt)$$ then infinitesimal generator is of the form $$\mathcal{A}f(t,x) = \frac{\partial f}{\partial t}(t,x) + ...



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