# Tag Info

5

Very roughly speaking, you can think of a stochastic process as a process that evolves in a random way. The randomness can be involved in when the process evolves, and also how it evolves. A very simple example of a stochastic process is the decay of a radioactive sample (with only one parent and one daughter product). Initially, it has some large number ...

4

E.g. let $\vec{Y}$ be a random vector with normal distribution on $\mathbb R^n$ with $\mathbb E\vec{Y}=\vec{0}$ and the identity matrix as covariance matrix. Then $\vec{X}:=\vec{Y}/||\vec{Y}||$ is a uniformly distributed unit-vector.

3

Collected from the comments. Fact 1. $f(y) = E[|X-y|^p]$ is continuous. Indeed, for any sequence $\{y_n,n\ge 1\}$ such that $y_n\to y_0$, $n\to\infty$, we have $f(y_n)\to f(y_0)$ in view of the dominated convergence theorem (thanks to convergence, $\{|y_n|\}$ is bounded by some $k$). Fact 2. $f\to+\infty$, $f\to\pm\infty$. Hint: use the Fatou lemma. ...

3

You're right that you use the scaling property, ie $\tilde B_t = c B_{t/c^2}$ ($c>0$) is a Brownian motion. Try this with some generic $c$, and then you should be able to guess what $c$ should be (in terms of $\lambda$). Recall also that "$\sup_{t\ge0} \iff \sup_{t/c^2\ge0}$", as $c^2 > 0$. If you're still stuck, I'll show you in the morning. (I'm ...

3

Question 1: Is $Y_t(\omega)$ well-defined? Answer: No, in general, $Y_t(\omega)$ is not well-defined; we need some additional assumption on the integrability of $X$ to ensure that $$\int_0^t |X_s(\omega)| \, ds <\infty$$ for $t>0$. This is e.g. satisfied if $X$ has continuous sample paths or $\sup_{t \leq T} \mathbb{E}(|X_t|)<\infty$ for any ...

3

For non-negative summands $\phi(y)\geq 0$, the uncountable sum is defined as $$\sum_{y\in Y} \phi(y)=\sup\left\{ \sum_{y\in F} \phi(y) : F \mbox{ is a finite subset of }Y\right\}.$$ In the countable case, Durrett gives a definition on page 20 just after Theorem 1.4.7.

3

As taking the inverse image is a cool operation. It commutes with practically everthing, here: with $\sigma(-)$. Lemma. Let $\Omega$, $R$ be any two sets, $X \colon \Omega \to R$ a map and $\mathcal A \subseteq \mathfrak P(R)$. Then $$\sigma\bigl(X^{-1}(\mathcal A)\bigr) = X^{-1}\bigl(\sigma(\mathcal A)\bigr)$$ Proof. To show $\subseteq$ note ...

2

As you observe, $P(X+Y=0)=\int_{\Bbb R}\mu(\{-y\})\,\nu(dy)$. The function $y\mapsto \mu(\{-y\})$ is non-zero for at most countably many values of $y$; call the set of those values $D$. We then have $$\int_{\Bbb R}\mu(\{-y\})\,\nu(dy)=\int_D \mu(\{-y\})\,\nu(dy)=\sum_{z\in D}\int_{\{z\}}\mu(\{-y\})\,\nu(dy)=\sum_{z\in D}\mu(\{-z\})\,\nu(\{z\}).$$ The sum ...

2

What is the probability of him not getting caught during the entire week? The probability of him getting caught on a given day is $0.2\cdot0.95=0.19$ The probability of him not getting caught on a given day is $1-0.19=0.81$ The probability of him not getting caught during the entire week is $0.81^5\approx35\%$ What is the probability of him ...

2

There's a major misconception here. First, what do we really mean by $P(X=Y)=1$? Well, we mean exactly what it says: We mean that $P(E)=1$, where $E$ is the event $$E=\{s\in S\,:\,X(s)=Y(s)\}.$$ I imagine you can concoct an example where that set $E$ is not measurable. (EDIT: For example, let $S=T=\{0,1\}$, and give both $S$ and $T$ the trivial ...

2

A stochastic process is a way of representing the evolution of some situation that can be characterized mathematically (by numbers, points in a graph, etc.) over time. They are of greatest help when you either don't know the exact rules of that evolution over time, or when the exact rule of that evolution is too complicated or costly to compute precisely. ...

2

I'd like to add that many types of noise are also modeled as stochastic processes. Whenever a person collects data, such a voltage signals, audio signals, or image signals, there will always be small disturbances in the data caused by imperfections in the equipment or the environment. The signal we measure might be given by the formula: $$measured\ signal = ... 2 The notation 1-F(x) \sim \frac{C}{x^\alpha \Gamma(1-\alpha)} here means$$\lim_{x\to \infty} \frac{1-F(x)}{\frac{C}{x^\alpha \Gamma(1-\alpha)}} = 1$$Since 0\leq 1-F(x) \leq 1, and x>0,\;\Gamma(1-\alpha)>0, we can rearrange to get:$$\lim_{x\to \infty} (1-F(x))x^\alpha \Gamma(1-\alpha) = C$$However, C can be any positive value, so it is ... 2 Since the distribution of X is symmetric, Y has exactly the same distribution as X. Indeed: If we denote by p the density of X, i.e.$$p(x) = \frac{1}{\sqrt{2\pi}} \exp \left(- \frac{x^2}{2} \right),$$then p(x)=p(-x). This implies$$\mathbb{P}(X \in C) = \mathbb{P}(-X \in C) \tag{1}$$for any Borel set C. Consequently,$$\begin{align*} ...

2

For clarity, I'll write $E_Y(E_X(X|Y))$ to indicate that the inner expectation is taken over $X$ and the outer over $Y$. The inner expectation, $E_X(X|Y)$ is equal either to $\frac{1}{Y}$ or $\frac{1}{Y}-1$, depending on the particular "flavour" of the geometric distribution you're referring to. The outer expectation, $E_Y(\frac{1}{Y})$ or ...

2

You obtain $f_X$ by "integrating away" $y$. From the diagram, this will be over $[-2;2]$ when $-1\leq x\leq 1$ and over $[-1;1]$ when $1< x\leq 2$. $$f_X(x) = \begin{cases} \int\limits_{-2}^2 f_{X,Y}(x,y)\operatorname d y & : x\in [-1;1] \\[1ex] \int\limits_{-1}^1 f_{X,Y}(x,y)\operatorname d y & : x\in (1;2] \end{cases}$$ Take it from there.

2

Try this physical model: Take 5 black ball and 36 white balls (yes, a total of 41 balls!). Glue each black ball to a white ball; you now have 5 identical glued ball pairs, and 31 identical white balls. Arrange the 36 items in a straight line, left to right; always place a black-white pair with the black on the left. The position of each black ball, ...

1

Because of the monotonicity and continuity of $\Phi$, we have $$\{x; \Phi^{-1}(x) \leq c\} = \{y; y \leq \Phi(c)\}$$ for any constant $c$. This implies $$\mathbb{P}(X_j \leq c) = \mathbb{P}(\Phi^{-1}(U_j) \leq c) = \mathbb{P}(U_j \leq \Phi(c)).$$ Now use that $U_j$ is uniformly distributed on $[0,1]$ to finish the proof.

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Yes, that's correct. Since, for any set $C \in \mathcal{B}$, we have $$\pi_t^{-1}(C) \in \mathcal{A} \subseteq \sigma(A)$$ (just choose $I = \{t\}$), we find that $\pi_t$ is measurable with respect to $\sigma(\mathcal{A})$. As $t$ is arbitrary and $\mathcal{B}^{[0,T]}$ is the smallest $\sigma$-algebra (on $S^{[0,T]}$) such that $\pi_t$ is measurable for ...

1

Start at the origin, i.e., $S_0=0$. Since the random walk is symmetric, $p(x)^2=p(x)p(-x)$ is the probability that you choose the increment $x$ followed by increment $-x$. This implies $S_2=0$, so $p(x)^2\leq \mathbb{P}\{S_2=0\}$.

1

Use $E(Y)=E(E(Y|X))=E(E(Fs(X)))=E(1/X)$ $Var(Y)=E(Var(Y|X))+Var(E(Y|X))=E((1-X)/X^2)+Var(1/X)$. Where $Var(Z)=E(Z^2)-E(Z)^2$ and $E(g(X))=\int g(x)f_X(x)dx$

1

So we have some probability space $\{\Omega,\mathbb{P}\}$, and vector-valued random variables: $A,B:\Omega\to\mathbb{R}^n$. In general, for every $x\in\mathbb{R}^n$, $$\hbox{Tr}(xx^T)=\langle x,x\rangle = ||x||^2$$ where $\langle,\cdot,\rangle, ||\cdot||$ denote the Euclidean inner-product and the Euclidean norm, respectively. So we can write: ...

1

We are given that the number of correctly answered questions $X$ is a binomial random variable with parameters $n = 40$ and $p = 0.5$. We are also given that $$\Pr[X > N] > 0.1, \quad \Pr[X > N+1] < 0.1.$$ Now approximate $X$ as $$Y \sim \operatorname{Normal}(\mu = np = 20, \sigma^2 = np(1-p) = 10),$$ we have $$\Pr[X > N] \approx \Pr[Y > ... 1 If \mu(\Omega)>0 then every nonnegative measurable function f:\Omega\rightarrow\mathbb R that satisfies \int fd\mu=1 induces a probability measure on \mathbb F prescribed by:$$B\mapsto\int f1_Bd\mu=\int_Bfd\mu$$Your construction is a special case of this construction. It only works if also \mu(\Omega)<\infty and rolls out if the function ... 1 If you do not have any further assumption on the values X can take (e.g., is it lower bounded a.s.?), then you cannot get any meaningful lower bound. For any \varepsilon\in[0,1] (and wlog the case \mu=0), consider the random variable defined by$$ X = \begin{cases} -x\frac{1-\varepsilon}{\varepsilon} & \text{ w.p. } \varepsilon \\ x & \text{ ...

1

The Kolmogorov extension theorem states that there is a probability distribution on $(E^I, \mathcal E^{\otimes I})$ such that $(1)$ holds. I claim that this distribution is uniquely determined. There are two ways to see this, both appealing to the definition of $\mathcal E^{\otimes I}$. In fact, it is the $\sigma$-algebra generated by cylindrical sets ...

1

In your attempt, the first step is correct, that is really $\left\vert X \right\vert$ is distributed as $U \sim Uniform(0,a)$ with distribution function $$F_U(u) = \begin{cases} u/a & \mbox{ for }0 \leq u \leq a, \\ 0 & \mbox{ for }u < 0, \\ 1 & \mbox{ for }u > a. \end{cases}$$ However, then you must obtain the distribution of $Z = U/a$, ...

1

Hint: First of all, the only ways to roll a 24 product are: $1, 1, 4, 6$ and $1, 2, 2, 6$ and $2, 2, 2, 3$ and $1, 2, 3, 4$. Second thing - there are not 4! ways to arrange for example $1,1,4,6$ or $1,2,2,6$. There are only $12$ ways to do so (you have to pick a place for two numbers and then the remaining spots get filled in). Similarly there are only ...

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The "ways to arrange" must take into account the undistinguishable outcomes To count them, say you have $n$ objects, divided into $k$ disjoint classes of undistinguishable objects, and say that class $C_i$ ($0≤i<k$) has size $m_i$, where $\sum_{i<k} m_i = n$, the number of different arrangements is $$\frac{n!}{\prod m_i!}$$ When all objects are ...

1

For your version of the geometric, the values taken on by $X^2$ are $0,1,4,9,\dots$. Let $Y=X^2$. The probability that $Y=y$ is $p(1-p)^{\sqrt{y}}$ whenever $y$ is a perfect square, and $\Pr(Y=y)=0$ for all $y$ that are not perfect squares. Remark: There are two versions of the geometric distribution. The one you seem to be using is that $X$ is the number ...

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