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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 ...

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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 ...

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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 ... 0 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 $$\frac ... 1 This is a kind of "uniform ergodic theorem" and extends naturally the case X_k=X for each k. Notice that X_k=X_k-X+X and by Birkhoff's ergodic theorem,$$\frac 1n\sum_{k=0}^{n-1}X\circ T^k\to \mathbb E[X\mid\mathcal I]\quad\mbox{ a.s.},$$where \mathcal I denotes the \sigma-algebra of invariant sets, that is, \mathcal I=\{A, T^{-1}(A)=A\}. If ... 0 The singletons look to me like an unnecessary detour. If F were generated by countably many of its elements, then it would also be generated by a family G of countably many countable sets, because you could just replace any co-countable sets among the original generators by their complements. The union U of all the generators in G is a countable ... 0 Counterexample Let$$S_n := \sum_{j=1}^n Y_j, \qquad n \in \mathbb{N} \tag{1}$$a simple random walk on \mathbb{Z}, i.e. Y_j \sim \frac{1}{2} (\delta_1+\delta_{-1}) are independent random variables. By Stirling's formula, we have$$\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$s$... 0 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 ... 0 Convergence in distribution is preserved by continuous maps, or more generally by maps for which the measure of discontinuity points are of null measure for the limiting law. Without this assumption, the result may fail. If we consider the case$k=1$, then we can take a simple example:$X_n:=1/n$,$X=0$and$H(x):=\chi_{(0,1)}(x)$, where$\chi_S$is the ... 0 (Not the best nor the shortest proof). By the definitions of$\lim \sup$and$\lim \inf$you have that $$\lim \sup (A_n\cap B_n):=\{x \in X: x \in A_n\cap B_n \text{ for infinitely many n}\}$$ From the given conditions you have that there exist two set of measure zero$N_A$and$N_B$such that $$x\in\lim\sup A_n=\{x \in X: x \in A_n\text{ for infinitely ... 1 Since it's a homework question, I will only give two intermediate steps. Notice that \mathbb P(\limsup_nA_n\cap \liminf_n B_n)=1. Show that \limsup_nA_n\cap \liminf_n B_n\subset\limsup_n(A_n\cap B_n). 0 Hint: You need to make the proof in two steps. First step, the open halflines (sets of the form (-\infty, x) generate the Borel \sigma-algebra in \mathbb{R}. Second step, use the portmanteau theorem - which provides conditions that are equivalent to the convergence in distribution - as stated in Thomas's answer to prove the convergence in distribution. ... 2 A sequence X_1,X_2,\ldots of random variables is said to converge in distribution, or converge weakly, or converge in law to a random variable X if$$ \lim_{n\to\infty}F_n(x)=F(x) $$for every number x\in\mathbb R at which F is continuous, where F_n(x)=\mathbb P(X_n\le x) and F(x)=\mathbb P(X\le x). Thus, we need to show that F(x) is ... 0 Start with the definition of what is the convergence. There is also a theorem (http://fr.wikipedia.org/wiki/Convergence_en_loi, can't find the english version) that says that you need to prove that P(X_n \in A) -> P(X \in A) for any A an open 2 If (X_i) is i.i.d., (Y_i) is i.i.d. and (X_i) and (Y_i) are independent, this follows from the central limit theorem applied to the i.i.d. sequence (Z_i) defined by$$ Z_i=X_i-Y_i-E(X_1)+E(Y_1). $$To wit, considering the events$$ A_N=[S_2-S_1\geqslant E(S_2)-E(S_1)], $$one gets$$ ... 2 Let$R$denote a property each$\omega$in$\Omega$may have or not, and$A=\{\omega\in\Omega\mid R(\omega)\}$, in your case$R(\omega)$is:$\exists n\geqslant1,X_n(\omega)=0$. Then,$P(R)$is simply$P(A)$, that is, $$P(\exists n\geqslant1,X_n=0)=P(\{\omega\in\Omega\mid \exists n\geqslant1,X_n(\omega)=0\}).$$ By the way, this is exactly for the same ... 4 Your computation is valid. Probably the easiest way to show it is to note that, for every$t$, $$E(\mathrm e^{\mathrm iuW_\tau}\mid\tau=t)=E(\mathrm e^{\mathrm iuW_t}\mid\tau=t)=E(\mathrm e^{\mathrm iuW_t}),$$ where the second identity holds thanks to the independence of$W_t$and$\tau$. One knows that, for every$t$,$E(\mathrm e^{\mathrm iuW_t})=\mathrm ...

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Let us interpret the event for the case where $\mathcal{F}$ is the sigma-algebra generated by a partition $P = {A_1,\dots,A_n}$, $A_i \cap A_j = \emptyset$, and $\cup_{i=1}^n A_i = \Omega$, just to get some intuition. We set $\mathcal{F} := \sigma(P)$. The conditional expectation then takes the simple form: $$E[X | \mathcal{F}] = \sum_{i=1}^n \frac{E[X ... 2 I would do it using approximations. First suppose that g \in C_c(\mathbb R^n). Construct a sequence of measures \mu_k that are absolutely continuous w.r.t. Lebesgue measure (or alternatively are finite linear combinations of dirac delta functions) such that \mu_k converges weakly to \mu. Then for each x \in \mathbb R^n, we have \mu_k*g(x) \to ... 1 If people vote reflecting their preferences (i.e. voting for their first preference candidate) then somebody who gets over 50% of votes would be the Condorcet candidate. There are other issues: in particular simple plurality systems may discourage some voters from voting for their first preference candidate, and if this happens, then somebody who gets ... 2 Formulas involving the existential and universal quantifiers \exists and \forall can be rephrased in terms of unions and intersections. In this case, you can write:$$P\left[\exists n\geq 1: X_{n}=0\right] = P\left[\bigcup_{n\geq 1}\{\omega: X_{n}(\omega)=0\}\right]$$1 It is saying that there is some index n such that X_n will be at the origin with probability 1. This statement makes no claims as to what range of values that n might take; rather, that with probability 1 there is some finite value n such that X_n = 0. 1 Hint: \lceil X\rceil\geq k\iff X>k-1 so:$$P\left\{ \lceil X\rceil\geq k\right\} =P\left\{ X>k-1\right\} $$1 Let us assume that X_1 and X_2 are independent. Then E(X_1X_2)=E(X_1)E(X_2)=\lambda^2. Now we deal with X_1^2. It is a standard fact about the Poisson with parameter \lambda that it has variance \lambda. It follows that E(X_1^2)=\lambda+\lambda^2, so X_1^2 is not an unbiased estimator of \lambda^2. Remark: If the "standard fact" cannot ... 0 I think it' strictly linked to the self-avoiding path problem (which is slightly different as the circuit doesn't necessary ends where it started); there is also a detailed study , and a partial solution to what you called N_n(x_{\left\lfloor\frac{n-1}{2}\right\rfloor\left\lfloor\frac{n-1}{2}\right\rfloor}), all circuits which visit the origin. Anyway ... 2 Recall that$$\mathbb{E}e^{c B_t} = e^{\frac{1}{2} c^2 t} \tag{1}$$as B_t is Gaussian with mean 0 and variance t. In particular, we see that$$M_t := 4^{B_t} = \exp \bigg( B_t \cdot \log 4 \bigg)$$is not a martingale since$$\mathbb{E}M_t \stackrel{(1)}{=} \exp \left( \frac{1}{2} (\log 4)^2 \cdot t \right)$$is not constant. In fact, by the ... 0 I assume B is a random variable, let's call it X for the sake of clarity (B is usually a set name). Then if the X has some probability density f, then P(X \in dy) means f(y)dy. The intuition comes from the fact that if for a set A \subset \mathbb{R} then$$ P(X \in A) = \int_A f(y)dy$$thus f(y)dy measures the probability of X being ... 0$$ \operatorname E(XZ)=\operatorname E(X^2+XY)=\operatorname EX^2+\operatorname E(XY) $$and, since \operatorname{Var}X=\operatorname EX^2-(\operatorname EX)^2=\lambda, we have that$$ \operatorname E(XZ)=\lambda+3\lambda^2. $$Finally,$$ \operatorname{Cov}(X,Z)=E(XZ)-\operatorname EX\operatorname EZ=\lambda+3\lambda^2-3\lambda^2=\lambda. $$Edit. As ... 1 I don't understand the fetish for calculus. There should be a way without it (in fact, personally I prefer avoiding it if necessary): Order the data as follows: x_1 < x_2 \ldots <x_{n}. Let's suppose that x \in [x_i,x_{i+1}]. Then we have the following the$$\|(x_1, \ldots x_n) - (x,x\ldots x) \|_1 = \sum_{j \leq i}(x-x_j) + \sum_{j>i} ...

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