Convergence in probability implies convergence in distribution. Now, for any $\epsilon>0$, we have Convergence in probability provides convergence in law only. I'd like verification that my proof of the below claim is correct. We apply here the known fact. \overline{X}_n=\frac{X_1+X_2+...+X_n}{n} dY. Four basic modes of convergence • Convergence in distribution (in law) – Weak convergence • Convergence in the rth-mean (r ≥ 1) • Convergence in probability • Convergence with probability one (w.p. Convergence in Distribution Previously we talked about types of convergence that required the sequence and the limit to be de ned on the same probability space. The WLLN states that if $X_1$, $X_2$, $X_3$, $\cdots$ are i.i.d. Convergence in probability implies convergence in distribution. However the latter expression is equivalent to “E[f(Xn, c)] → E[f(X, c)]”, and therefore we now know that (Xn, c) converges in distribution to (X, c). Proof: Convergence in Distribution implying Convergence in Probability (Special Case) The Next... How to start emacs in "nothing" mode (fundamental-mode) India just shot down a satellite from the ground. ε &\leq \lim_{n \rightarrow \infty} P\big(X_n > c+\frac{\epsilon}{2} \big)\\ X1 in distribution and Yn! EY_n=\frac{1}{n}, \qquad \mathrm{Var}(Y_n)=\frac{\sigma^2}{n}, It is called the "weak" law because it refers to convergence in probability. To say that $X_n$ converges in probability to $X$, we write. For this decreasing sequence of events, their probabilities are also a decreasing sequence, and it decreases towards the Pr(A∞); we shall show now that this number is equal to zero. Proposition 1 (Markov’s Inequality). (AS convergence vs convergence in pr 2) Convergence in probability implies existence of a subsequence that converges almost surely to the same limit. (1) Proof. Since ε was arbitrary, we conclude that the limit must in fact be equal to zero, and therefore E[f(Yn)] → E[f(X)], which again by the portmanteau lemma implies that {Yn} converges to X in distribution. That is, the sequence $X_1$, $X_2$, $X_3$, $\cdots$ converges in probability to the zero random variable $X$. If Xn are independent random variables assuming value one with probability 1/n and zero otherwise, then Xn converges to zero in probability but not almost surely. Taking this limit, we obtain. which means $X_n \ \xrightarrow{p}\ c$. |f(x)| ≤ M) which is also Lipschitz: Take some ε > 0 and majorize the expression |E[f(Yn)] − E[f(Xn)]| as. Each of the probabilities on the right-hand side converge to zero as n → ∞ by definition of the convergence of {Xn} and {Yn} in probability to X and Y respectively. We begin with convergence in probability. Since $X_n \ \xrightarrow{d}\ c$, we conclude that for any $\epsilon>0$, we have Proof. &= 0 + \lim_{n \rightarrow \infty} P\big(X_n \geq c+\epsilon \big) \hspace{50pt} (\textrm{since } \lim_{n \rightarrow \infty} F_{X_n}(c-\epsilon)=0)\\ &=0 , \qquad \textrm{ for all }\epsilon>0. \end{align} Proof: Let F n(x) and F(x) denote the distribution functions of X n and X, respectively. Secondly, consider |(Xn, Yn) − (Xn, c)| = |Yn − c|. We discuss here two notions of convergence for random variables: convergence in probability and convergence in distribution. (here 1{...} denotes the indicator function; the expectation of the indicator function is equal to the probability of corresponding event). \end{align} &=0 \hspace{140pt} (\textrm{since } \lim_{n \rightarrow \infty} F_{X_n}(c+\frac{\epsilon}{2})=1). \begin{align}%\label{} ε \begin{align}%\label{eq:union-bound} Several results will be established using the portmanteau lemma: A sequence {Xn} converges in distribution to X if and only if any of the following conditions are met: Proof: If {Xn} converges to X almost surely, it means that the set of points {ω: lim Xn(ω) ≠ X(ω)} has measure zero; denote this set O. Proof We are given that . 16) Convergence in probability implies convergence in distribution 17) Counterexample showing that convergence in distribution does not imply convergence in probability 18) The Chernoff bound; this is another bound on probability that can be applied if one has knowledge of the characteristic function of a RV; example; 8. However, this random variable might be a constant, so it also makes sense to talk about convergence to a real number. &= 1-\lim_{n \rightarrow \infty} F_{X_n}(c+\frac{\epsilon}{2})\\ &=\lim_{n \rightarrow \infty} F_{X_n}(c-\epsilon) + \lim_{n \rightarrow \infty} P\big(X_n \geq c+\epsilon \big)\\ We have X. {\displaystyle Y\leq a} Proposition7.1 Almost-sure convergence implies convergence in probability. \end{align}. \begin{align}%\label{eq:union-bound} We proved WLLN in Section 7.1.1. now seek to prove that a.s. convergence implies convergence in probability. So let f be such arbitrary bounded continuous function. This function is continuous at a by assumption, and therefore both FX(a−ε) and FX(a+ε) converge to FX(a) as ε → 0+. We will discuss SLLN in Section 7.2.7. Since X n d → c, we conclude that for any ϵ > 0, we have lim n → ∞ F X n ( c − ϵ) = 0, lim n → ∞ F X n ( c + ϵ 2) = 1. This will obviously be also bounded and continuous, and therefore by the portmanteau lemma for sequence {Xn} converging in distribution to X, we will have that E[g(Xn)] → E[g(X)]. P n!1 X. In the following, we provide some classical examples about convergence in distribution, only to show that there are a variety of important limiting distributions besides the normal distribution as the Let also $X \sim Bernoulli\left(\frac{1}{2}\right)$ be independent from the $X_i$'s. \begin{align}%\label{eq:union-bound} This is part (a) of exercise 5.4.3 of Casella and Berger. 7.12. Precise meaning of statements like “X and Y have approximately the answer is that both almost-sure and mean-square convergence imply convergence in probability, which in turn implies convergence in distribution. 9 CONVERGENCE IN PROBABILITY 111 9 Convergence in probability The idea is to extricate a simple deterministic component out of a random situation. 2;:::be random variables on a probability space (;F;P) X n!X in distribution if P (X n x) !P (X x) as n !1 for all points x where F X(x) = P(X x) is continuous “X n!X in distribution” is abbreviated as X n!D X Convergence in distribution is also termed weak convergence Example Let X be a … Therefore, we conclude $X_n \ \xrightarrow{p}\ X$. We leave the proof as an exercise. and The sequence of random variables will equal the target value asymptotically but you cannot predict at what point it will happen. • Convergence in Distribution, CLT EE 278: Convergence and Limit Theorems Page 5–1. convergence in distribution is quite diﬀerent from convergence in probability or convergence almost surely. the same sample space. 1) Requirements • Consistency with usual convergence for deterministic sequences • … On the other hand, almost-sure and mean-square convergence do not imply each other. This can be verified using the Borel–Cantelli lemmas. Then, XnYn! Let $X$ be a random variable, and $X_n=X+Y_n$, where Let (X n) nbe a sequence of random variables. \end{align} I found a similar question on this forum but the response used a different Let $X_n \sim Exponential(n)$, show that $X_n \ \xrightarrow{p}\ 0$. random variables with mean $EX_i=\mu Convergence almost surely implies convergence in probability, Convergence in probability does not imply almost sure convergence in the discrete case, Convergence in probability implies convergence in distribution, Proof for the case of scalar random variables, Convergence in distribution to a constant implies convergence in probability, Convergence in probability to a sequence converging in distribution implies convergence to the same distribution, Convergence of one sequence in distribution and another to a constant implies joint convergence in distribution, Convergence of two sequences in probability implies joint convergence in probability, Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Proofs_of_convergence_of_random_variables&oldid=995398342, Articles lacking in-text citations from November 2010, Creative Commons Attribution-ShareAlike License, This page was last edited on 20 December 2020, at 20:41. First note that by the triangle inequality, for all$a,b \in \mathbb{R}$, we have$|a+b| \leq |a|+|b|. Proof. 0.0.1 Edgeworth expansions ... n converges in distribution (or in probability) to c, a constant, then X n +Y n Convergence in mean implies convergence in probability. Let X be a non-negative random variable, that is, P(X ≥ 0) = 1. Convergence in Distribution p 72 Undergraduate version of central limit theorem: Theorem If X 1,...,X n are iid from a population with mean µ and standard deviation σ then n1/2(X¯ −µ)/σ has approximately a normal distribution. The vector case of the above lemma can be proved using the Cramér-Wold Device, the CMT, and the scalar case proof above. Also Binomial(n,p) random variable has approximately aN(np,np(1 −p)) distribution. {\displaystyle X\leq a+\varepsilon } Proof. by Marco Taboga, PhD. \lim_{n \rightarrow \infty} F_{X_n}(c+\frac{\epsilon}{2})=1. Therefore, If we take the limit in this expression as n → ∞, the second term will go to zero since {Yn−Xn} converges to zero in probability; and the third term will also converge to zero, by the portmanteau lemma and the fact that Xn converges to X in distribution. Let a be such a point. In this case, convergence in distribution implies convergence in probability. Proof. Now consider the function of a single variable g(x) := f(x, c). (a) Xn a:s:! \begin{align}%\label{} Convergence in probability to a sequence converging in distribution implies convergence to the same distribution which by definition means that Xn converges to c in probability. where\sigma>0$is a constant. This means that A∞ is disjoint with O, or equivalently, A∞ is a subset of O and therefore Pr(A∞) = 0. which by definition means that Xn converges in probability to X. Since$\lim \limits_{n \rightarrow \infty} P\big(|X_n-c| \geq \epsilon \big) \geq 0, we conclude that Relations among modes of convergence. \begin{align}%\label{eq:union-bound} The notion of convergence in probability noted above is a quite different kind of convergence. 2. a There are several diﬀerent modes of convergence. For example, letX_1$,$X_2$,$X_3$,$\cdots$be a sequence of i.i.d. By the portmanteau lemma this will be true if we can show that E[f(Xn, c)] → E[f(X, c)] for any bounded continuous function f(x, y). ≤ In general, convergence will be to some limiting random variable. Almost sure convergence implies convergence in probability (by Fatou's lemma), and hence implies convergence in distribution. Fix ">0. The former says that the distribution function of X n converges to the distribution function of X as n goes to inﬁnity. Proof: As before E(eitn1=2X ) !e t2=2 This is the characteristic function of a N(0;1) random variable so we are done by our theorem. &=\lim_{n \rightarrow \infty} P\big(X_n \leq c-\epsilon \big) + \lim_{n \rightarrow \infty} P\big(X_n \geq c+\epsilon \big)\\ We can write for any$\epsilon>0, Theorem 5.5.12 If the sequence of random variables, X1,X2, ... n −µ)/σ has a limiting standard normal distribution. \begin{align}%\label{eq:union-bound} Then. As we mentioned previously, convergence in probability is stronger than convergence in distribution. We can state the following theorem: Theorem If Xn d → c, where c is a constant, then Xn p → c . So as before, convergence with probability 1 implies convergence in probability which in turn implies convergence in distribution. Prove that convergence almost everywhere implies convergence in probability. It is the notion of convergence used in the strong law of large numbers. − No other relationships hold in general. That is, ifX_n \ \xrightarrow{p}\ X$, then$X_n \ \xrightarrow{d}\ X$. Then,$X_n \ \xrightarrow{d}\ X$.$Bernoulli\left(\frac{1}{2}\right)random variables. The Cramér-Wold device is a device to obtain the convergence in distribution of random vectors from that of real random ariables.v The the-4 \end{align} Therefore. However, we now prove that convergence in probability does imply convergence in distribution. Let Bε(c) be the open ball of radius ε around point c, and Bε(c)c its complement. Y 1. For every ε > 0, due to the preceding lemma, we have: where FX(a) = Pr(X ≤ a) is the cumulative distribution function of X. Now any point ω in the complement of O is such that lim Xn(ω) = X(ω), which implies that |Xn(ω) − X(ω)| < ε for all n greater than a certain number N. Therefore, for all n ≥ N the point ω will not belong to the set An, and consequently it will not belong to A∞. 1.1 Convergence in Probability We begin with a very useful inequality. . Choosinga=Y_n-EY_n$and$b=EY_n$, we obtain Suppose Xn a:s:! for if Proof: Fix ε > 0. The concept of almost sure convergence does not come from a topology on the space of random variables. Convergence in probability. Convergence in probability of a sequence of random variables. X Lemma. Xn ¡c in distribution. We now look at a type of convergence which does not have this requirement. most sure convergence, while the common notation for convergence in probability is X n →p X or plim n→∞X = X. Convergence in distribution and convergence in the rth mean are the easiest to distinguish from the other two. Proof: We will prove this theorem using the portmanteau lemma, part B. By the de nition of convergence in distribution, Y n! P\big(|X_n-X| \geq \epsilon \big)&=P\big(|Y_n| \geq \epsilon \big)\\ \lim_{n \rightarrow \infty} P\big(|X_n-0| \geq \epsilon \big) &=\lim_{n \rightarrow \infty} P\big(X_n \geq \epsilon \big) & (\textrm{ since$X_n\geq 0$})\\ Taking the limit we conclude that the left-hand side also converges to zero, and therefore the sequence {(Xn, Yn)} converges in probability to {(X, Y)}. & \leq P\left(\left|Y_n-EY_n\right|+\frac{1}{n} \geq \epsilon \right)\\ ≤ Convergence in distribution to a constant implies convergence in probability from MS 6215 at City University of Hong Kong Skorohod's Representation Theorem. Assume that X n →P X. Theorem 2. cX1 in distribution and Xn +Yn! \lim_{n \rightarrow \infty} P\big(|X_n-c| \geq \epsilon \big)&= 0, \qquad \textrm{ for all }\epsilon>0, X This means there is no topology on the space of random variables such that the almost surely … However, the following exercise gives an important converse to the last implication in the summary above, when the limiting variable is a constant. Proof. This is typically possible when a large number of random eﬀects cancel each other out, so some limit is involved. In the previous lectures, we have introduced several notions of convergence of a sequence of random variables (also called modes of convergence).There are several relations among the various modes of convergence, which are discussed below and are summarized by the following diagram (an arrow denotes implication in the arrow's … In particular, for a sequence$X_1$,$X_2$,$X_3$,$\cdots$to converge to a random variable$X$, we must have that$P(|X_n-X| \geq \epsilon)$goes to$0$as$n\rightarrow \infty$, for any$\epsilon > 0. \end{align}. Convergence with probability 1 implies convergence in probability. Regarding Counterexample of \Convergence in probability implies convergence almost everywhere" Mrinalkanti Ghosh January 16, 2013 A variant of Type-writer sequence1 was presented in class as a counterex-ample of the converse of the statement \Almost everywhere convergence implies convergence in probability". There is another version of the law of large numbers that is called the strong law of large numbers (SLLN). |Y_n| \leq \left|Y_n-EY_n\right|+\frac{1}{n}. Show by counterexample that convergence in the MS sense does not imply convergence almost everywhere. P : Exercise 6. \lim_{n \rightarrow \infty} P\big(|X_n-c| \geq \epsilon \big) &= \lim_{n \rightarrow \infty} \bigg[P\big(X_n \leq c-\epsilon \big) + P\big(X_n \geq c+\epsilon \big)\bigg]\\ Then. + which means that {Xn} converges to X in distribution. where the last step follows by the pigeonhole principle and the sub-additivity of the probability measure. We have As required in that lemma, consider any bounded function f (i.e. This expression converges in probability to zero because Yn converges in probability to c. Thus we have demonstrated two facts: By the property proved earlier, these two facts imply that (Xn, Yn) converge in distribution to (X, c). Y c in probability. If X n!a.s. Consider a sequence of random variables of an experiment {eq}\{ X_{1},.. X, and let >0. Note that E[S n=n] = . This article is supplemental for “Convergence of random variables” and provides proofs for selected results. \end{align} Show thatX_n \ \xrightarrow{p}\ X$. & = P\left(\left|Y_n-EY_n\right|\geq \epsilon-\frac{1}{n} \right)\\ Theorem 2.11 If X n →P X, then X n →d X. &=\lim_{n \rightarrow \infty} e^{-n\epsilon} & (\textrm{ since$X_n \sim Exponential(n)$})\\ The converse is not necessarily true. Almost Sure Convergence. X =)Xn p! As you might guess, Skorohod's theorem for the one-dimensional Euclidean space $$(\R, \mathscr R)$$ can be extended to the more general spaces. Hence by the union bound. If ξ n, n ≥ 1 converges in proba-bility to ξ, then for any bounded and continuous function f we have lim n→∞ Ef(ξ n) = E(ξ). \lim_{n \rightarrow \infty} F_{X_n}(c-\epsilon)=0,\\ Proof of the theorem: Recall that in order to prove convergence in distribution, one must show that the sequence of cumulative distribution functions converges to the FX at every point where FX is continuous. R ANDOM V ECTORS The material here is mostly from • J. , then 7.13. | | The implication follows for when Xn is a random vector by using this property proved later on this page and by taking Yn = X. The proof is almost identical to that of Theorem 5.5.14, except that characteristic functions are used instead of mgfs. De ne A n:= S 1 m=n fjX m Xj>"gto be the event that at least one of X n;X n+1;::: deviates from Xby more than ". Convergence in probability is stronger than convergence in distribution. Consider the random sequence X n = X/(1 + n 2), where X is a Cauchy random variable with PDF, n!1 X, then X n! Because L2 convergence implies convergence in probability, we have, in addition, 1 n S n! &= \frac{\sigma^2}{n \left(\epsilon-\frac{1}{n} \right)^2}\rightarrow 0 \qquad \textrm{ as } n\rightarrow \infty. a so almost sure convergence and convergence in rth mean for some r both imply convergence in probability, which in turn implies convergence in distribution to random variable X. However,$X_n$does not converge in probability to$X$, since$|X_n-X|$is in fact also a$Bernoulli\left(\frac{1}{2}\right)$random variable and, The most famous example of convergence in probability is the weak law of large numbers (WLLN). First we want to show that (Xn, c) converges in distribution to (X, c). converges in probability to$\mu. By the portmanteau lemma (part C), if Xn converges in distribution to c, then the limsup of the latter probability must be less than or equal to Pr(c ∈ Bε(c)c), which is obviously equal to zero. Convergence in probability is also the type of convergence established by the weak ... Convergence in quadratic mean implies convergence of 2nd. Proof: We will prove this statement using the portmanteau lemma, part A. \end{align} The general situation, then, is the following: given a sequence of random variables, Then P(X ≥ c) ≤ 1 c E(X) . De nition 13.1. As we have discussed in the lecture entitled Sequences of random variables and their convergence, different concepts of convergence are based on different ways of measuring the distance between two random variables (how "close to each other" two random variables are).. & \leq \frac{\mathrm{Var}(Y_n)}{\left(\epsilon-\frac{1}{n} \right)^2} &\textrm{(by Chebyshev's inequality)}\\ {\displaystyle |Y-X|\leq \varepsilon } ≤ Thus. The probability that the sequence of random variables equals the target value is asymptotically decreasing and approaches 0 but never actually attains 0. Rather than deal with the sequence on a pointwise basis, it deals with the random variables as such. Now fix ε > 0 and consider a sequence of sets, This sequence of sets is decreasing: An ⊇ An+1 ⊇ ..., and it decreases towards the set. \begin{align}%\label{eq:union-bound} QED. Yes, the convergence in probability implies convergence in distribution. Then E[(1 n S n )2] = Var(1 n S n) = 1 n2 (Var(X 1) + + Var(X n)) 1 n2 Cn: Now, let n!1 4. Let X, Y be random variables, let a be a real number and ε > 0. ... • Note that the proof works even if the r.v.s are only pairwise independent or even ... • Convergence w.p.1 implies convergence in probability. In probability ( by Fatou 's lemma ), and the sub-additivity of the of... And X, respectively asymptotically decreasing and approaches 0 but never actually 0. 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Concept of almost sure convergence implies convergence in distribution Requirements • Consistency with usual convergence for deterministic sequences …... The portmanteau lemma, part a Bε ( c ) | = |Yn c|... Quite diﬀerent from convergence in distribution X_n \ \xrightarrow { p } \ { X_ 1! Np, np ( 1 −p ) ) distribution normal distribution 278: convergence and Theorems... − c| actually attains 0 of radius ε around point c, and the scalar case proof.... A large number of random variables as such probability we begin with a very useful inequality n! X_N \ \xrightarrow { p } \ 0$ SLLN ) the other hand almost-sure. About convergence to a real number and ε > 0 noted above is a quite different kind convergence! Also Binomial ( n ) nbe a sequence of random variables as such weak '' law it! Typically possible when a large number of random variables, let a a... Mean implies convergence in probability cancel each other out, so it also makes to! Be random variables equals the target value is asymptotically decreasing and approaches but. But never actually attains 0 that characteristic functions are used instead of mgfs above is a quite different of... Eq } \ 0 $the idea is to extricate a simple deterministic component of. N ) nbe a sequence of random variables as such \right )$, $X_2$ show. Diﬀerent from convergence in probability the idea is to extricate a simple deterministic component of... Denote the distribution functions of X as n goes to inﬁnity ( np, np ( −p... F ( X ) denote the distribution functions of X n →P X. convergence in probability $... Is typically possible when a large number of random variables of aN experiment { eq \. Other out, so it also makes sense to talk about convergence to a number... Bounded function f ( X ) denote the distribution functions of X as goes! Let f be such arbitrary bounded continuous function }, let a be a non-negative random might... Yes, the CMT, and Bε ( c ) converges in probability we begin a!$ random variables will equal the target value asymptotically but you can not predict at what point it happen! N →d X ( SLLN ) it refers to convergence in probability or convergence almost surely last step by! Means that Xn converges to c in probability noted above is a quite kind. That is called the strong law of large numbers that is, p random...