# Tag Info

## New answers tagged probability-theory

0

The following is an addendum to GEdgar's answer, aimed to clarify some points for my future reference, as well as for the sake of other readers who, like me, do not find these points self evident. The principal results are propositions 6 and 7 below. Notation We denote the borel field on the real line by $\mathcal{B}$, the Borel field on $\mathbb{R}^2$ by ...

1

Answer: Expected E (X) = 1$pq+2pq^2+3pq^3+4pq^4+\cdots$ E = $pq(1+2q+3q^2+4q^3+....)$ S = $(1+2q+3q^2+4q^3+....)$ qS = $q+2q^2+3q^3+....)$ S-qS = $1+q+q^2+q^3+\cdots$ S(1-q) = $\frac{1}{(1-q)}$ S = $\frac{1}{(1-q)^2}$ E =$\frac{pq}{(1-q)^2}$ E = $\frac{1-p}{p}$ That will be your answer This is a simple way to understand expectation. Thanks ...

0

As you already figured out, we have by Tonelli's theorem $$\mathbb{E}(X^p) = \int_0^{\infty} p \cdot r^{p-1} \mathbb{P}(X > r) \, dr.$$ Using the assumption $r \mathbb{P}(X >r) \leq \int_{X>r} Y \, d\mathbb{P}$ this yields \begin{align*} \mathbb{E}(X^p) &\leq \int_0^{\infty} p r^{p-2} \int_{X>r} Y \, d\mathbb{P} \, dr \\ &=p \int ... 2 The question has been answered, so we give a different derivation. If p\ne 0, our series clearly converges, so the expectation exists. Call it a. On the first trial, either we have a success, in which case the expectation is 0, or we have a failure. In that case, we have wasted a trial, and the expectation is 1+a. It follows thata=(1-p)(1+a).$$... 1 Hint: differentiate the equation: Sigma(k = 0 to infinity)(x^k) = 1/(1 - x) both sides, then multiply both sides with x. Then let x = 1 - p. 1 The conditional distribution of N given that K = k is a Poisson distribution with parameter q\lambda but displaced k to the right; that is, (conditionally) N is of the form M+k where M is Poisson(q\lambda) and so$$E[N\mid K = k] = E[M+k] = E[M]+k = q\lambda + k.$$Or, without spending time thinking about the matter, write m = n-k and the ... 1 Hint: We get the independence result if we can show that$$\mathbb{E}\left[\mathrm e^{\mathrm i\left(aN_1(s)+bN_2(t)\right)} \right] = \mathbb{E}\left[\mathrm e^{\mathrm i aN_1(s)} \right]\mathbb{E}\left[\mathrm e^{\mathrm i bN_2(t)} \right], $$for all s\leq t and a,b. Once the case s=t is done, the case s<t is formal, based on conditioning and ... 2 Hint: We can say :$$\omega \in C \Leftrightarrow (\forall m \in \Bbb N )(\exists p \in \Bbb N)( \forall k \geq p) \quad \left|\sum_{i=k+1}^{+\infty} X_k(\omega)\right| < \frac 1{m+1} $$That gives:$$C=\bigcap_{m \in \Bbb N}\bigcup_{p \in \Bbb N}\bigcap_{k \geq p} Y_{m,p,k}$$Where :$$Y_{m,p,k}=\left\{\omega \in \Omega / \left| \sum_{i=k+1}^{+\infty} ...

0

First, the power equation can be written as $$P = I^2 R$$ where you have $R = 1000 \Omega$ and $I$ is given by $\mu = 10mA$, $\sigma = 6mA$. The power exceeds $100 mW$ when $P = I^2 R \geq 0.1 W$ or in other words, when $$I \geq \sqrt{\frac{1000}{0.1}} = \sqrt{10000} = 100 A$$ So the question is, what is the probability that $I \geq 100 A$. This ...

1

Consider the following two scenarios: 1) You have two independent Poisson processes $N_1(t)$ and $N_2(t)$ with rates $\lambda p$ and $\lambda (1-p)$. $N(t) = N_1(t) + N_2(t)$ is their sum. This is a Poisson process with rate $\lambda$. 2) You have a Poisson process $N(t)$ with rate $\lambda$. Each time an event occurs, a coin is tossed: with ...

1

Correct answer with simple proof: $\frac{1}{2}$. Notice that the entire situation is symmetric. Everything happens with the same probability to white as it does to red. Thus $P(Red)=P(white)=\frac{1}{2}$ What's wrong with your answer: "We prove that if all that jugs have same amount of red balls and the same amount of white balls (each one), the probability ...

1

It's obviously $\frac{1}{2}$, due to the symmetry in the problem : if you paint all white balls red and all red balls white, this is the same problem, so the probability of getting a red ball is equal to the probability to get a white ball.

1

The identity to prove is that, for every function $f$ such that $f(0)=0$, $$E(f(X)-qf(X+1))=0.$$ To show this, one computes $E(f(X))$ and $E(f(X+1))$, using the definition of the distribution of $X$, and one watches the simplification occur... A more general formula, valid when $f(0)\ne0$, is $$E(f(X)-qf(X+1))=pf(0).$$

2

Exercise: Let $H:\mathbb R\to\mathbb R$ denote a non-decreasing function. Show that $H$ is continuous from the right at $a$ if and only if $H(x_n)\to H(a)$ for at least one sequence $(x_n)$ such that $x_n\geqslant a$ for every $n$ and $x_n\to a$ when $n\to\infty$, if and only if $H(a+1/n)\to H(a)$ when $n\to\infty$. Recall that $H$ being continuous from ...

1

I assume that you mean that $x,y$ are two (fixed) elements of $[0,1)$ and that $M$ is the collection of all sets that either contain both $x$ and $y$ or contain neither $x$ nor $y$. We will show that $M$ is a $\sigma$-algebra. It is immediate that $\Omega \in M$ as $\Omega$ contains both $x$ and $y$. Furthermore, if $A \in M$ then $A^c \in M$ since if $A$ ...

0

No, take the set of odd natural numbers and even ones $$A = \bigcup_{n=1}^{\infty} \{ 2n+1 \}$$ $$B = \bigcup_{n=1}^{\infty} \{ 2n \}$$ they are both infinite sets hence $\mathbb{P}(A) = \mathbb{P}(B) = 1$.But $\mathbb{N} = A\cup B$ and $A\cap B = \emptyset$. while $$1 = \mathbb{P}\big( \mathbb{N} \big) \neq 2 =\mathbb{P}(A) + \mathbb{P}(B)$$

6

There is a Borel set $E$ in $\mathbb R^2$ such that $F := \{x-y\colon (x,y) \in E\}$ is not a Borel set. Let $A := \{f \in \mathbf{C}\colon (f(0), f(1)) \in E\}$. Then $A \in \mathcal{B}_{\left[0,\infty\right)}$. How about $T(A)$? In fact $$T(A) = \{g \in \mathbf{C}\colon g(0)=0, g(1) \in F\}$$ and is not Borel.

-2

Hi I think it is true that the operator $T$ that you defined is such that the image of any measurable set of $\mathcal{B}([0,+\infty))$ is a measurable set or as claimed that $T$ preserves measurability. The idea is to view things in a "topological" way first. If you "examine" carefully $T$ I think that you would shortly realize that it is a continuous ...

1

Rewrite this as $z_n=z_{n-1}y_n$ where $z_n=f(x_n)$ and $y_n=1+\alpha/m+\beta\epsilon_n/\sqrt{m}$, thus, $(y_n)$ is i.i.d. such that $E(y_1)=\gamma$ with $\gamma=1+\alpha/m$ and $\mathrm{var}(y_1)=\beta^2/m$. If $x_0$ is independent of $(\epsilon_n)$, one gets $E(z_n)=E(z_{n-1})E(y_n)$, that is, $$E(f(x_n))=\gamma^nE(f(x_0)).$$ Likewise, ...

1

Rewriting this in terms of $X\setminus A_j$ and using $\mu(X)=1$, the inequality is equivalent to $$\mu\left(\bigcup_{j=1}^n(X\setminus A_j)\right)\leqslant \sum_{j=1}^n\mu(X\setminus A_j).$$ This can be handled integrating the inequality $$\chi\left(\bigcup_{j=1}^nB_j\right)\leqslant \sum_{j=1}^n\chi(B_j)$$ valid for any collection $(B_j)_{j=1}^n$ of ...

1

Neither (2) nor (3) are necessary. Note that, by definition, $E(T_1;T_1\lt T_2)$ is $$\int_0^\infty uf_{T_1}(u)\int_u^\infty f_{T_2}(v)\mathrm dv\mathrm du=\int_0^\infty \lambda_1u\mathrm e^{-\lambda_1 u}P(T_2\gt u)\mathrm du.$$ Can you finish this?

0

Outline: Suppose that a program has $n$ lines. Then the number $X$ of errors in the program has Poisson distribution with parameter $\lambda=n(0.015)$. We want $\Pr(X\le 1)\le 0.10$. So we want $e^{-\lambda}(1+\lambda)\le 0.10$. This inequality does not have a nice algebraic solution for $\lambda$. However, a little calculator experimentation will yield a ...

0

The Central Limit Theorem examines sums of random variables from which we subtract the mean of the sum and then divide the whole by the standard deviation of the sum: let $Y_1,...,Y_n$ be random variables and define $S_n \equiv \sum_{i=1}^nY_i$. Then the CLT examines the random variable $$Z_n = \frac {S_n - E[S_n]}{\sqrt {\text {Var}(S_n)}}$$ and what is ...

1

It's probably simpler to prove that the complement of $\limsup_{n\to+\infty}E_n$ is contained in $C$. Indeed, if $\omega$ is not in $\limsup_{n\to+\infty}E_n$, then there is $N=N(\omega)$ such that for $n\geqslant N$, $\omega\notin E_n$. We thus have $|X_n(\omega)|\leqslant q^n$ for these $n$, which proves the convergence of the series $\sum_{n\geqslant ... 2 If$X_n\to 0$in distribution, then any$\varepsilon$does the job. The converse is harder. Here it's the proof of Levy's continuity theorem which will be used. Denoting by$\varphi_n$the characteristic function of$X_n$and$\mu_n$its distribution, we indeed have the equality ... 1 The lack of memory of the exponential distribution can be used to produce conceptual proofs that, for every$n\geqslant2$,$G$is distributed as the maximum of$(n-1)$i.i.d. random variables each exponentially distributed with parameter$a$. Since, however, the OP failed to explain their background, here is a direct, hands-on, approach. Consider ... 1 It seems Doob-Dynkin theorem is the answer to your question. Let$X$be a real valued random variable,$\sigma(X):=\{X^{-1}(B), B\in\mathcal B(\mathbb R)\}$. The random variable$Y$is$\sigma(X)$-measurable if and only if there exists$f\colon\mathbb R\to\mathbb R$Borel measurable such that$Y=f(X)$. But there are restrictions on$f$in order to ... 2 In order to make the inner product well-defined, we talk about$L^2(\Omega,\mathcal F,\mu)$, where$(\Omega,\mathcal F,\mu)$is the underlying probability space. But we then extend condition expectation to integrable random variables. We use a projection over the closed subspace$L^2(\Omega,\mathcal N,\mu)$, that is, the vector subspace which consists of ... 2 Let$(X_t)_{t \geq 0}$be a non-negative solution of the SDE $$X_t - x = 3t + 2 \int_0^t \sqrt{X_s} \, dB_s \tag{1}$$ for$x \geq 0$. Applying Itô's formula to$f(y) = \frac{1}{\sqrt{y}}$, we find $$\frac{1}{\sqrt{X_t}} - \frac{1}{\sqrt{x}} = - \int_0^t \frac{1}{X_s} \, dB_s.$$ For$\tau_{a,b} := \inf\{t \geq 0; X_t \notin (a,b)\}$,$0<a<b$, this ... 1 I realize what I did in the other answer was maybe overcomplicated. Define$D_l:=\sup_{k\geqslant l}|X_k-X|$(which is integrable for each$l$), and notice that by Birkhoff's ergodic theorem, $$\frac 1n\sum_{j=0}^{n-1}D_l\circ T^j\to\mathbb E[D_l\mid\mathcal I]\quad \mbox{a.e.}$$ with the same notations as in the other answer. Since $$\frac ... 2 This is a kind of "uniform ergodic theorem" and extends naturally the case X_k=X for each k. Notice that X_k=X_k-X+X and by Birkhoff's ergodic theorem,$$\frac 1n\sum_{k=0}^{n-1}X\circ T^k\to \mathbb E[X\mid\mathcal I]\quad\mbox{ a.s.},$$where \mathcal I denotes the \sigma-algebra of invariant sets, that is, \mathcal I=\{A, T^{-1}(A)=A\}. If ... 1 The singletons look to me like an unnecessary detour. If F were generated by countably many of its elements, then it would also be generated by a family G of countably many countable sets, because you could just replace any co-countable sets among the original generators by their complements. The union U of all the generators in G is a countable ... 2 Counterexample Let$$S_n := \sum_{j=1}^n Y_j, \qquad n \in \mathbb{N} \tag{1}$$a simple random walk on \mathbb{Z}, i.e. Y_j \sim \frac{1}{2} (\delta_1+\delta_{-1}) independent identically distributed random variables. By Stirling's formula, we have$$\mathbb{P}(S_{2n}=0) = 2^{-2n} {2n \choose n} \sim \frac{1}{\sqrt{\pi n}} \qquad \text{and} \qquad ... -1 There is a Markova property as$\min A_i\sim exp$And it is correct with your step as$\max A \le min\implies\max=min$. I think the difficulty is how to get the distribution function from the wrok you have done 1 There are many ways. Differentiation is one of them, not the simplest. We have $$(1+x)^n=\sum_0^n \binom{n}{k} x^k.$$ Differentiate. We get $$n(1+x)^{n-1}=\sum_0^n k\binom{n}{k}x^{k-1}.$$ Set$x=\frac{p}{1-p}$. Then the left-hand side is$n \frac{1}{(1-p)^{n-1}}$. Multiply through by$(1-p)^{n-1}$. We get $$\sum_0^n k\binom{n}{k}p^{k-1} (1-p)^{n-k}=n.$$ ... 2 You don't need to use induction or take derivatives; just note that$k\binom{n\vphantom{1}}{k}=n\binom{n-1}{k-1}\begin{align} \sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k} &=\sum_{k=0}^nn\binom{n-1}{k-1}p^k(1-p)^{n-k}\\ &=np\sum_{k=0}^n\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}\\ &=np\,\Big(p+(1-p)\Big)^{n-1}\\[12pt] &=np \end{align} 4 Let $$f(x)=\sum_{k=0}^{n} \binom nk x^k y^{n-k}= (x+y)^n$$ then $$x\frac{\partial f}{\partial x}= \sum_{k=0}^{n} k \binom nk x^k y^{n-k}=nx(x+y)^{n-1}$$ so withy=1-p$and$x=p$we find the desired result. 1 You can get rid of the$k$by cancelling it with$k!from the binomial coefficient and simplifying like this: \begin{align} \sum_{k=0}^{n} k \binom nk p^k (1-p)^{n-k} &= \sum_{k=1}^{n} k\frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} = n\sum_{k=1}^{n} \frac{(n-1)!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}\\ &= n\sum_{k=1}^{n} \binom{n-1}{k-1} p^k ... 2 This is saying that the average number of coin flips if you flip n fair coins each with heads probability p is np (you are calculating the mean of a Binomial(n,p) distribution). You can use linearity of expectation - if Y = \sum_i X_i where X_i is Bernoulli(p) and there are n such X_i, then Y is Binomial(n,p) and has that distribution. ... 1 Let A be the event it rains on Saturday, and B the event it rains Sunday. You want bounds on \Pr(A\cup B). It is clear that \Pr(A\cup B)\le 1. We cannot do better. For example, the rain god could choose an integer at random between 1 and 10. If the number is between 1 and 6, it rains on Saturday. If it is between 4 and 10, it rains on ... 0 For such a small set you can do it by brute force. Consider any sigma field or Dynkin system of {1,2,3,4} We know it's got to have {} and {1,2,3,4} in it. What other sets can it have? Some combination of: \{1\}, \{2\}, \{3\}, \{4\}, \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\}, \{1,2,3\}, \{1,3,4\}, \{1,2,4\}, \{2,3,4\} Start by choosing ... 0 So there are 2^5 possible sequences in terms of break or not break. 16 of those 32 have 3 or more breaks. More precisely, there are 10 ways to get 3 breaks, 5 ways to get 4 breaks, and 1 way to get 5 breaks. However, these sequences are weighted. The easiest is the 1 way to get 5 breaks, this probability is 0.05^5. The second easiest is 5 ways to get ... 1 There is a simple algorithm: let P the transition matrix. let A=I + P, j = 1 replace each non 0 entry by 1 replace A by A^2, j by 2j Go back to 2. until j\ge n Then the graph is connected iff every entry of A is 1. At each iteration, at the end of step 4. the non zero entries of A are the state you can go to with \le j ... 1 There are some common (and very useful) results concerning recurrence and transitivity of states which you might know and/or could use: All states in an irreducible set are either all recurrent or all transient. State i is transient if and only if $$\sum_{n=1}^{\infty}p_{ii}^{(n)} < \infty$$ State i is recurrent if and ... 1 Firstly separate the states in communication classes. There are two communication classes which are determined as follows. Start from state 0. Which states can you visit and then return to state 0? You can make following transitions0\xrightarrow{0.5} 2 \xrightarrow{0.9} 0$$So 0 and 2 communicate, which means that they belong to the same ... 4 Note that the series lasts 7 games if and only if each team wins 3 times in the first 6 games; so the probability is$$\binom{6}{3}\cdot{\left (\frac{1}{2}\right )}^{6}$$If the probabilities of winning for A were p then the result is$$\binom{6}{3}\cdot p^{3}\cdot (1-p)^{3}$$0 It follows from the very definition of (pointwise) convergence that$$\{\lim_{n \to \infty} X_n = X\} = \bigcap_{n \in \mathbb{N}} \bigcup_{m \in \mathbb{N}} \left\{ \sup_{k \geq m} |X_k-X| < \frac{1}{n} \right\}.$$Hence, by the continuity of the measure \mathbb{P},$$\begin{align*} \mathbb{P}(\lim_{n \to \infty} X_n = X) &= \lim_{n \to \infty} ... 0 Only if: SupposeX$is Feller. Take$x\in\mathbb{N}$. For$y\in\mathbb{N}$, let$f(y) = \delta_{x,y}. \begin{align}\lim_{t\to 0}p_t(x, \{x\}) &= \lim_{t\to 0}\mathbb{E}_x[f(X_t)]\\ &= f(x)\\ &= 1\end{align} Fixy\in\mathbb{N}$and$t>0$. Suppose $$\lim_{x\to\infty}p_t(x,\{y\}) > 0.$$ Then there is$\epsilon>0$and an infinite ... 1 Some words might be missing and sloppy notations do not help but the linearity is the following. Assume that the bivariate distribution$p$of$(X,Y)$is such that$p(x,y)=r(x)q(y\mid x)$for every$(x,y)$in the state space of$(X,Y)$, where$r$is the distribution of$X$. (Then$q$is the conditional distribution of$Y$conditionally on$X$.) Let$s\$ ...

1

The question, in my opinion, extends to the debate "Bayesian vs classical (or frequentist) statistics". According to Wikipedia there are two major differences in the frequentist and Bayesian approaches to inference: In a frequentist approach to inference, unknown parameters are often, but not always, treated as having fixed but unknown values that are not ...

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