# Tag Info

1

I don't understand why you must split up the integral, why not say $$\int_{-\infty}^x f(u) du = \int_{-1}^x f(u) du = \left.\frac{u + \alpha u^2/2}{2}\right|_{-1}^x ...$$

1

$$P(p) = \sum_{N_1 \geq N_2+2} \binom{4}{N_1}\binom{0}{N_2} p^{N_1+N_2}q^{4 - N_1 -N_2 }$$ $\binom{0}{x}$ is 1 if $x=0$ and 0 otherwise so the only $N_2$ That can affect the sum is $N_2 = 0$. similarly the other binomial coefficient is zero if $4=n_1<N_1$ so we don't have an infinite sum, $$P(p) = \sum_{ k= 2}^{4} \binom{4}{k} p^k q^{4-k}$$ Typing on ...

2

Take a look further to the front of the book, maybe you can find the definition of the generated sigma-algebra that the author uses. Typically, the generated sigma-algebra is defined just in the way you described. In this case you only need to prove the second assertion of this exercise. However, there are also other definitions. One possible definition is ...

1

Let $\omega\in\Omega$. If $\mathsf 1_{\limsup_{n\to\infty}E_n}(\omega)=1$, then $$\omega\in\limsup_{n\to\infty}\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty E_k.$$ Hence for each $n$, there exists $k\geqslant n$ such that $\omega\in E_k$. Therefore $$\limsup_{n\to\infty}\mathsf 1_{E_n}(\omega) = \lim_{n\to\infty}\sup_{k\geqslant n} \mathsf 1_{E_k}(\omega)=1.$$ ...

0

If you're using radians, then you have $U_1 \sim Unif(0, 2\pi).$ Then the density function must have height $1/2\pi$ over $(0, 2\pi).$ If you're using degrees, then you have $U_2 \sim Unif(0, 360),$ which has density $f_{U_2}(x) = 1/360,$ for $0 < x < 360\,$ (and $0$ elsewhere). If you define $U_3 = U_1/2\pi$ then you have $U_3 \sim Unif(0,1),$ which ...

0

You can depict this uniform distribution many different ways: (a) Angle measured in radian and divided by $2\pi$ results in a uniform distribution over the interval $[0,1)$. In this case the pdf is $$f_1(x)=\begin{cases} 1,& \text{ if } 0\le x <1\\ 0,& \text{ otherwise. } \end{cases}$$ (b) Angle measured in radian (not divided by $2\pi$) result ...

0

1) Remember, that for a continuous random variable, $Pr(X=x) = 0$. To derive the PDF write the CDF as : $$Pr[X \leq x] = \int_{0}^{x} \frac{1}{2\pi}dx' = \frac{x}{2\pi}$$ Hence PDF is: $$f(x) = \frac{1}{2\pi}$$ 2) It is inappropriate to think that there are exactly 360 points between 0 and 360 or 1 point between 0 and 1. 3) Even if you choose to ...

-3

If $p_1$false, then $p_3$ does not imply $p_2$ because assuming $p_1$ and $p_3$, we can't decide whether $X$ and $Y$ are dependent or independent. I <3 all Tim's.

2

Yes; though not in general, it is the case for identical and independent distributions (iid). (Sorry, missed that in the first reading.) $$\mathsf E[V_3\mid V_1<\max\{V_2,V_3\}, V_3=\max\{V_2, V_3\}] \;=\; \mathsf E[V_3\mid V_3=\max\{V_1,V_2, V_3\}]$$ This is not necessarily the same thing as $\mathsf E[\max\{V_1,V_2, V_3\}]$ Indeed: \begin{align} ... 2 For continuous random variables, the probability of a tie is immeasurably small. So we can say:V_1<\max\{V_2, V_3\} \iff V_2=\max\{V_1,V_2, V_3\} \cup V_3=\max\{V_1, V_2, V_3\}\quad\text{a.s.}$$Since the events in this union are disjoint (almost surely), then they partition the conditioned space:$$\begin{align}\mathsf E[\max\{V_2,V_3\}\mid ...

1

Hint: the probability that $X+Y\equiv n\pmod{10}$ is $$\sum_{k=0}^9\Pr[X=k]\Pr[Y\equiv n-k\pmod{10}]$$ Now consider the sum $$\sum_{k=0}^9\Pr[Y\equiv n-k\pmod{10}]$$

0

Let me just comment on your approach, and note the difference between what you do and the $\mod 10$ problem. Remember what the ranges of your random variables $X$ and $Y$ are: The $X$ can attain the values in the set $\{0,1,\ldots,9\}$, and $Y$ can attain the values $0,1,\ldots$ In your sum, $\text{Pr}(X=k)$ only has a non-zero value for $k\geq 0$, and ...

0

Your attempt almost solves the question, but indeed there is a reason for the mod $10$. Your solution does not take into account that $z$ can only be a number between $0$ and $9$. Here is my solution: Let $z \in \lbrace 0,..., 9 \rbrace$. Then we have due to independence: $$Pr(Z= z ) = Pr(X+Y = z) = \sum_{i=0}^{9}{Pr(X = i) Pr(Y = z- i \textrm{ mod } 10)} ... 4 There is a mistake in the definition, it should surely be \rho(C) = \lim \limits_{n \to \infty} \frac{1}{n} \# \{k \mid 1 \le k \le n, k \in C\}. You use the index n twice, when you should actually use two different indices.$$\rho(C_n) = \lim \limits_{m \to \infty} \frac{1}{m}\#\{k \mid 1 \le k \le m, k \in C_n\}$$Now note that$$\frac{1}{m}\#\{k ...

0

Borel $\sigma$ algebra $B[0,1])$ of the set $[0,1]$ is infinite but it is not generated by any partition of the interval $[0.1]$. Assume the contrary and let $(X_i)_{i \in I}$ be such a partition.Since singletons belong to $B[0,1])$ we deduce that each $X_i$ must be singleton. Then $[0,1/2]$ does not belong to $B[0,1])$ because it is not presented as ...

1

Let $\epsilon > 0$ and consider the stopping time $$\tau_\epsilon = \inf\{t \in [\tau_0, T] : X_t \ge \epsilon\} \wedge T.$$ Since $\tau_0 \le \tau_\epsilon$ almost surely, optional stopping gives $E[X_{\tau_\epsilon}] \le E[X_{\tau_0}]$. Now let consider the events $\{\tau_0 < T\}$ and $\{\tau_0 = T\}$ and write $$E[X_{\tau_\epsilon} ; \tau_0 < T] ... 2 Not sure if this is what you are asking, but: Given any two members of the equivalence class E(X|\mathcal N), each is measurable with respect to \mathcal N, so the set where the members differ (which has probability zero) belongs to the sigma algebra \mathcal N. 0 A conditional expectation is an equivalence class of random variables, which functions are defined on the sample space, not a \sigma-algebra. 0 Here are two approaches, one more elementary than the other. Modify your argument to show that X(r) = 0 a.s on the event \{\tau_0< r\} for each r\in (0,T]. Deduce from this that X(r)=0 for every rational in (\tau_0,T), a.s. on the event \{\tau_0<T\}. Now invoke the continuity of X. Let B=\{\omega: \tau_0(\omega)<T, X(t,\omega)>0 ... 2 You just need to look at more stopping times. For any t\geq 0, you have$${E}[\mathbf{1}_{\{\tau_0<T\}}X(T \wedge (\tau_0+t))] ={E}[\mathbf{1}_{\{\tau_0<T\}}E[X(T\wedge (\tau_0+t))|F_{\tau_0}]] \le {E}[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0.$$This gives$$P\biggl(X(T\wedge(\tau_0+t))=0\mbox{ for all }t\in\mathbb{Q}\cap[0,T]; \tau_0<T\biggr) ...

2

You can generalize your derivation from: $$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$ to: $$E[\mathbf{1}_{\{\tau_0=s\}}X(t)]=E[\mathbf{1}_{\{\tau_0=s\}}E[X(t)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0=s\}}X(\tau_0)]=0$$ where $t > s$. As you mentioned, using the fact ...

0

The vertical axis $\pi_3=p(Q)$ is being bounded. Two of the four planes are vertical and thus don't yield a bound on $\pi_3$. In any case the two vertical planes correspond to the trivial constraints $\pi_1\ge0$ and $\pi_2\le1$, and the other two are the interesting ones.

3

The first thing you must do is find $a,b$ from the given mean ($7$) and variance ($4$).   Only then can you try to find $\mathsf P(a\leq X\leq 6\mid 4\leq X\leq b)$ The mean and variance of a uniform discrete distribution, $X\sim\mathcal U\{a..b\}$ are: $$\mathsf E(X) = \frac{a+b}{2} = 7\\\mathsf{Var}(X)= \frac{(a-b+1)^2-1}{12} = 4$$

2

Note that $\mathbb{E}[f]$ is a constant (assuming it exists). All constant functions are measurable with respect to ANY $\sigma$-algebra. It has nothing to do with $f$ being independent of the $\sigma$-algebra.

0

According to this study by Jean-Philippe Villeneuve (in French): http://culturemath.ens.fr/histoire%20des%20maths/htm/villeneuve2009/theories-de-la-mesure.html the essential property of sigma-additivity that is used to define sigma-algebra was first used both by Émile Borel in 1898 for his work on the theory of functions in which there is a chapter on the ...

1

When an event, call it $T$, is a tail event, it is in $\cap_{n=1}^{\infty}\sigma(X_{n},X_{n+1},...)$, so it is an element is every one of the $\sigma$-algebras $\sigma(X_{n},.X_{n+1},...)$. The reason we can interpret this as not depending on any finite set $\{X_{1},X_{2},...,X_{n}\}$ is because while the event is in $\sigma(X_{1},X_{2},...)$, it is not ...

1

The concrete example I like to think of is where $\Omega = [0,1]^2$ with Lebesgue measure, and $\mathcal F_0 = \{A \times [0,1]: \text{$A$is a Lebesgue measurable subset of$[0,1]$}\}$. Then $E(f\mid\mathcal F_0) = g$, where $$g(x,y) = \int_0^1 f(x,\eta) \, d\eta .$$ In fact, every conditional expectation can be thought of in this form. I have a paper ...

0

Yes the variables $t_i$ are independent and identically distributed. The independence follows directly from the Strong Markov property. That gives that $t_i$ is independent of $t_j$ for all $j < i$. The Strong Markov property and time-homogeneity together imply that $$P[X_{s + t_i} \in A | t_i] = E[X_s \in A| X_0 = y]$$ for all $s \geq 0$. That yields ...

1

The conditional expectation $Y = E[X\mid\mathcal{F}]$ can be interpreted as the best $\mathcal{F}$-measurable approximation for $X$. More specifically: If $X$ is square-integrable, then the conditional expectation is the orthogonal projection of $X$ to the subspace of $L^2$ that consists of $\mathcal{F}$-measurable functions. Note that the conditional ...

0

To elaborate on @Dominik's answer, $\mathbb P(X\leqslant x)$ is really shorthand for $$\mathbb P\left(X^{-1}((-\infty,x]) \right) = \mathbb P\left(\{\omega\in\Omega : X(\omega)\leqslant x\}\right),$$ where we are working in a probability space $(\Omega, \mathcal F,\mathbb P)$. That is, $\Omega$ is a nonempty set called the sample space, $\mathcal F$ is a ...

0

As you've stated the problem, the answer is "no". You wrote: "Given a random variable $X$ defined in a probability space, is it possible to have more than one distribution function $F$?" A probability space is a triple $(\Omega,\mathcal F, P)$, where $\Omega$ is the set of outcomes, $\mathcal F$ is the set of events (i.e. measurable subsets of $\Omega$, ...

4

The CDF $F$ of a random variable $X$ is unique, since it is defined as $$F(x) = P(X \le x).$$

0

For a discrete random variable, $X$, the probability mass function, denoted $f_X$ satisfies the following $$\mathbb{P}[X\in\{a_1,a_2,\dots\}]=\sum_{n=1}^\infty f_X(a_n).$$ For a continuous random variable, $Y$, the analogue to PMF is called the probability density function, denoted $f_Y$. And it satisfies for any measurable set $U$, $$\mathbb{P}[Y\in ... 1 You have been given that$$f_{X,Y}(x,y) = \begin{cases}4xy & : 0 \leq x \leq 1,\; 0 \leq y \leq 1\\ 0 & : \text{ elsewhere}\end{cases}$$You wish to know where f_{X,Y}(x, z-x) is supported (ie: not zero) with respect to x, for values of z where 0\leq z\leq 2. (Since z=x+y, then 0+0\leq z\leq 1+1.)$$f_{X,Y}(x,z-x) = ...

2

You could require $f$ to only be continuous, but the idea is to have as few test functions as possible to satisfy the definition. Another way to say this is, the above possible values for $\{...\}$ are equivalent to the requirement that the equality holds for continuous functions. Here is a proof of that for the continuous bounded characterization. On one ...

1

While sample set could also be worthwhile terminology, it is worth recalling that any space in mathematics is always a set with some sort of operations defined. For example, in the theory of stochastic processes, we could let $\Omega = \mathcal{D}(\mathbb{R}_+, \mathbb{R})$, where $\mathcal{D}(\mathbb{R}_+, \mathbb{R})$ is the Skorokhod space of real-valued ...

2

An atom in a measurable space $(M,\mathcal{M})$ is a nonempty set $A\in\mathcal{M}$ such that $B\in\mathcal{M}$ and $B\subseteq A$ implies that either $B=\emptyset$ or $B=A$. If $\mathcal{M}$ is generated by a countable family of sets, so it is countably generated, every point in $M$ is in a unique atom and the atoms form a partition of $M$. A countably ...

1

Your notation is a bit confusing (at least to me), since I would usually expect $D_1$ to $D_3$ to refer to specific doors (e.g. the first, second and third door from the left), whereas if I understand your calculations correctly, you're using them to refer, respectively, to "the door I picked", "the door the presenter opened" and "the door I could switch ...

1

Extended Comment: I think it would be helpful for you to use more specific language: (a) When you say "the probability that I achieve 5 successes within 10 trials," you need to say 'exactly 5' or 'at least 5'. Even if it is clear to you what you mean, it is not necessarily clear to others. (b) Also, you need to distinguish between what you 'control' for the ...

1

As @user2566092 suggested, we have $$\mathbb E\left[|X-X_n|^r\right]^{\frac pr} \leqslant \mathbb E\left[|X-X_n|^{\frac{rp}r}\right] = \mathbb E\left[|X-X_n|^p\right].$$ Hence $$\mathbb E\left[|X-X_n|^r\right] \leqslant \mathbb E\left[|X-X_n|^p\right]^{\frac rp}\stackrel{n\to\infty}\longrightarrow 0,$$ so that $X_n\stackrel{L^r}\longrightarrow X$. Note ...

1

Hint: Try to use Jensen's inequality with the convex function $f(Y_n)$ of the variable $Y_n = |X - X_n|^r$ defined by $f(Y_n) = {Y_n}^{p/r}$.

0

Don't be misled by formality, rigor, or philosophy. Probability theory is as perfectly sensible and interesting as every other topic in mathematics. Like many others (e.g., "algebra"), it is poorly named. It is about those random mechanisms (processes) that act unpredictably from action to action, called "trials," but deterministically when repeated over and ...

1

Hint: If $f_{n}:=\sum_{m=1}^{\infty}m2^{-n}1_{E_{m.n}}$ then $f_{n}\left(x\right)=\lfloor f\left(x\right)2^{n}\rfloor2^{-n}$ for a fixed $x\in\mathbb{R}$. In general $2\lfloor a\rfloor\leq\lfloor2a\rfloor$ for each $a\in\mathbb R$ so that: $$f_{n}\left(x\right)=\lfloor f\left(x\right)2^{n}\rfloor2^{-n}\leq\lfloor ... 1 Well, let's review a few helpful definitions that will clarify how to calculate the Kullback-Leibler divergence here. By definition the summation of the parameters of the mutlinomial distribution is 1; i.e.,$$\sum_{m=1}^k\theta_m=1$$, where \theta_m is the probability of the m^{th} outcome occuring. The probability mass function (PMF) of the ... 2 The L.H.S. is increasing with n Use the equality$$E_{m,n}=E_{2m,n+1} \biguplus E_{2m+1,n+1}$$to get$$\sum_{m=1}^{\infty }\frac{m}{2^n} \mu (E_{n,m}) \le \sum_{m=1}^{\infty }\frac{m}{2^{n+1}} \mu (E_{n+1,m}). The limit is indeed the R.H.S. Prove that if $f$ is any simple function, you will have $\sum_{m=1}^{\infty }\frac{m}{2^n} \mu (E_{n,m}) \ge ... 3 The integral is defined as the supremum of functions of bounded functions with finite support. I would work directly from this definition to show the result. If your function had been bounded and of finite support, then your above analysis is almost complete. Each$f_n$would be composed of a finite set of functions that as$n$goes to infinity, ... 0 If there are N raffle tickets sold to N different people, then each ticket holder has the same probability of winning the prize. The distribution is uniform. 0 Indeed, by the Kolmogorov extension theorem any systems of random variables (stochastic processes) can be represented by a suitable set of finite dimensional cumulative distribution functions. However, handling (just to mention one important example) the so called tail events, that is, events relating to the limiting behavior of infinite series of random ... 0 Here is the proof of Question 1: first arrow, left to right: \begin{eqnarray} P(X, A, B) & = & P(X, A\ |\ B)P(B)\\ & = & P(X\ |\ B)P(A\ |\ B)P(B)\ \ \ \ \ because\ X\ and\ A\ are\ conditionally\ independent\ on\ B\\ & = & P(X\ |\ B)P(AB)\\ & = & P(X)P(AB)\ \ \ \ \ because\ X\ and\ B\ are\ independent \end{eqnarray} first ... 1 Assume that there are$k$elements in$A_1$. There are$\binom{n}{k}$ways of choosing such an$A_1$. Now for a given$A_1$count the number of admissible$A_2$. Well,$k$elements must not belong to$A_2$, so the others can be chosen or not in$2^{n-k}$ways. Letting$k$run from$0$to$n$, we get the total number of possible combinations of$(A_1,A_2)\$ ...

Top 50 recent answers are included