• Convergence in Distribution, CLT EE 278: Convergence and Limit Theorems Page 5–1. As you might guess, Skorohod's theorem for the one-dimensional Euclidean space $$(\R, \mathscr R)$$ can be extended to the more general spaces. a First we want to show that (Xn, c) converges in distribution to (X, c). \overline{X}_n=\frac{X_1+X_2+...+X_n}{n} \end{align} So as before, convergence with probability 1 implies 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. P n!1 X. Secondly, consider |(Xn, Yn) − (Xn, c)| = |Yn − c|. That is, if $X_n \ \xrightarrow{p}\ X$, then $X_n \ \xrightarrow{d}\ X$. X. Skorohod's Representation Theorem. We have 1.1 Convergence in Probability We begin with a very useful inequality. 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. Convergence in probability is also the type of convergence established by the weak ... Convergence in quadratic mean implies convergence of 2nd. (a) Xn a:s:! R ANDOM V ECTORS The material here is mostly from • J. Then, $X_n \ \xrightarrow{d}\ X$. (AS convergence vs convergence in pr 2) Convergence in probability implies existence of a subsequence that converges almost surely to the same limit. This can be verified using the Borel–Cantelli lemmas. The former says that the distribution function of X n converges to the distribution function of X as n goes to inﬁnity. Convergence in mean implies convergence in probability. \begin{align}%\label{eq:union-bound} \end{align}. . 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 distribution is quite diﬀerent from convergence in probability or convergence almost surely. 7.13. by Marco Taboga, PhD. Now consider the function of a single variable g(x) := f(x, c). 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. We leave the proof as an exercise. Xn ¡c in distribution. Because L2 convergence implies convergence in probability, we have, in addition, 1 n S n! X ε Proof: Fix ε > 0. First note that by the triangle inequality, for all $a,b \in \mathbb{R}$, we have $|a+b| \leq |a|+|b|$. Convergence in probability of a sequence of random variables. Prove that convergence almost everywhere implies convergence in probability. \begin{align}%\label{eq:union-bound} Convergence in probability is stronger than convergence in distribution. \end{align} \end{align} Proof. {\displaystyle Y\leq a} 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. Let also $X \sim Bernoulli\left(\frac{1}{2}\right)$ be independent from the $X_i$'s. We begin with convergence in probability. 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. Show by counterexample that convergence in the MS sense does not imply convergence almost everywhere. &=\lim_{n \rightarrow \infty} P\big(X_n \leq c-\epsilon \big) + \lim_{n \rightarrow \infty} P\big(X_n \geq c+\epsilon \big)\\ \begin{align}%\label{eq:union-bound} and Then, XnYn! X =)Xn p! \end{align} 0.0.1 Edgeworth expansions ... n converges in distribution (or in probability) to c, a constant, then X n +Y n This means there is no topology on the space of random variables such that the almost surely … 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). Suppose Xn a:s:! On the other hand, almost-sure and mean-square convergence do not imply each other. However, we now prove that convergence in probability does imply convergence in distribution. c in probability. \begin{align}%\label{eq:union-bound} dY. 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). Consider the random sequence X n = X/(1 + n 2), where X is a Cauchy random variable with PDF, \begin{align}%\label{} By the de nition of convergence in distribution, Y n! $Bernoulli\left(\frac{1}{2}\right)$ random variables. &= 1-\lim_{n \rightarrow \infty} F_{X_n}(c+\frac{\epsilon}{2})\\ Convergence in probability. We will discuss SLLN in Section 7.2.7. We have \lim_{n \rightarrow \infty} F_{X_n}(c+\frac{\epsilon}{2})=1. QED. The probability that the sequence of random variables equals the target value is asymptotically decreasing and approaches 0 but never actually attains 0. ε 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. Then P(X ≥ c) ≤ 1 c E(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. \end{align}. We apply here the known fact. Convergence in probability implies convergence in distribution. Convergence in probability provides convergence in law only. \end{align} Almost sure convergence implies convergence in probability (by Fatou's lemma), and hence implies convergence in distribution. \lim_{n \rightarrow \infty} F_{X_n}(c-\epsilon)=0,\\ \begin{align}%\label{eq:union-bound} & = P\left(\left|Y_n-EY_n\right|\geq \epsilon-\frac{1}{n} \right)\\ − For example, let $X_1$, $X_2$, $X_3$, $\cdots$ be a sequence of i.i.d. This is typically possible when a large number of random eﬀects cancel each other out, so some limit is involved. 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 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)}. However, this random variable might be a constant, so it also makes sense to talk about convergence to a real number. 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. ≤ Let $X_n \sim Exponential(n)$, show that $X_n \ \xrightarrow{p}\ 0$. 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 WLLN states that if $X_1$, $X_2$, $X_3$, $\cdots$ are i.i.d. Then. We can write for any $\epsilon>0$, 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. As we mentioned previously, convergence in probability is stronger than convergence in distribution. Since $X_n \ \xrightarrow{d}\ c$, we conclude that for any $\epsilon>0$, we have 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. 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 … 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. (1) Proof. Proposition 1 (Markov’s Inequality). In general, convergence will be to some limiting random variable. 1. It is called the "weak" law because it refers to convergence in probability. There is another version of the law of large numbers that is called the strong law of large numbers (SLLN). &=0 , \qquad \textrm{ for all }\epsilon>0. (here 1{...} denotes the indicator function; the expectation of the indicator function is equal to the probability of corresponding event). The concept of almost sure convergence does not come from a topology on the space of random variables. 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. 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. 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+. Fix ">0. , then I'd like verification that my proof of the below claim is correct. The vector case of the above lemma can be proved using the Cramér-Wold Device, the CMT, and the scalar case proof above. where $\sigma>0$ is a constant. 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 Hence by the union bound. Then. a \begin{align}%\label{eq:union-bound} {\displaystyle |Y-X|\leq \varepsilon } Precise meaning of statements like “X and Y have approximately the It is the notion of convergence used in the strong law of large numbers. Proposition7.1 Almost-sure convergence implies convergence in probability. Let Bε(c) be the open ball of radius ε around point c, and Bε(c)c its complement. We now look at a type of convergence which does not have this requirement. Proof. \begin{align}%\label{} P\big(|X_n-X| \geq \epsilon \big)&=P\big(|Y_n| \geq \epsilon \big)\\ 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". Proof: We will prove this theorem using the portmanteau lemma, part B. 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). De nition 13.1. | Yes, the convergence in probability implies convergence in distribution. 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. \lim_{n \rightarrow \infty} P\big(|X_n-c| \geq \epsilon \big)&= 0, \qquad \textrm{ for all }\epsilon>0, Y & \leq P\left(\left|Y_n-EY_n\right|+\frac{1}{n} \geq \epsilon \right)\\ Let X be a non-negative random variable, that is, P(X ≥ 0) = 1. Proof. Now, for any $\epsilon>0$, we have 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. The sequence of random variables will equal the target value asymptotically but you cannot predict at what point it will happen. There are several diﬀerent modes of convergence. & \leq \frac{\mathrm{Var}(Y_n)}{\left(\epsilon-\frac{1}{n} \right)^2} &\textrm{(by Chebyshev's inequality)}\\ Assume that X n →P X. Convergence with probability 1 implies convergence in probability. That is, the sequence $X_1$, $X_2$, $X_3$, $\cdots$ converges in probability to the zero random variable $X$. In this case, convergence in distribution implies convergence in probability. |f(x)| ≤ M) which is also Lipschitz: Take some ε > 0 and majorize the expression |E[f(Yn)] − E[f(Xn)]| as. 1) Requirements • Consistency with usual convergence for deterministic sequences • … Lemma. ≤ ≤ |Y_n| \leq \left|Y_n-EY_n\right|+\frac{1}{n}. &\leq \lim_{n \rightarrow \infty} P\big(X_n > c+\frac{\epsilon}{2} \big)\\ + The notion of convergence in probability noted above is a quite different kind of convergence. \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 $X_n\geq 0$ })\\ EY_n=\frac{1}{n}, \qquad \mathrm{Var}(Y_n)=\frac{\sigma^2}{n}, We proved WLLN in Section 7.1.1. Note that E[S n=n] = . Relations among modes of convergence. which means that {Xn} converges to X in distribution. 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∞. ... • 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. Let (X n) nbe a sequence of random variables. This article is supplemental for “Convergence of random variables” and provides proofs for selected results. \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]\\ &=0 \hspace{140pt} (\textrm{since } \lim_{n \rightarrow \infty} F_{X_n}(c+\frac{\epsilon}{2})=1). Rather than deal with the sequence on a pointwise basis, it deals with the random variables as such. Proof: Let F n(x) and F(x) denote the distribution functions of X n and X, respectively. Thus. 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. P : Exercise 6. 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 $\epsilon > 0$. Taking this limit, we obtain. However, the following exercise gives an important converse to the last implication in the summary above, when the limiting variable is a constant. {\displaystyle X\leq a+\varepsilon } Convergence in probability to a sequence converging in distribution implies convergence to the same distribution X1 in distribution and Yn! X \end{align} Let X, Y be random variables, let a be a real number and ε > 0. now seek to prove that a.s. convergence implies convergence in probability. 7.12. Since $\lim \limits_{n \rightarrow \infty} P\big(|X_n-c| \geq \epsilon \big) \geq 0$, we conclude that 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. Theorem 5.5.12 If the sequence of random variables, X1,X2, ... n −µ)/σ has a limiting standard normal distribution. random variables with mean $EX_i=\mu However the latter expression is equivalent to “E[f(Xn, c)] → E[f(X, c)]”, and therefore we now know that (Xn, c) converges in distribution to (X, c). Theorem 2.11 If X n →P X, then X n →d X. Proof We are given that . &= 0 + \lim_{n \rightarrow \infty} P\big(X_n \geq c+\epsilon \big) \hspace{50pt} (\textrm{since } \lim_{n \rightarrow \infty} F_{X_n}(c-\epsilon)=0)\\ Consider a sequence of random variables of an experiment {eq}\{ X_{1},.. To say that$X_n$converges in probability to$X$, we write. Let a be such a point. The converse is not necessarily true. where the last step follows by the pigeonhole principle and the sub-additivity of the probability measure. This expression converges in probability to zero because Yn converges in probability to c. Thus we have demonstrated two facts: By the property proved earlier, these two facts imply that (Xn, Yn) converge in distribution to (X, c). The general situation, then, is the following: given a sequence of random variables, Choosing$a=Y_n-EY_n$and$b=EY_n$, we obtain cX1 in distribution and Xn +Yn! converges in probability to$\mu$. Proof: Convergence in Distribution implying Convergence in Probability (Special Case) The Next... How to start emacs in "nothing" mode (fundamental-mode) India just shot down a satellite from the ground. which means$X_n \ \xrightarrow{p}\ c$. If X n!a.s. | Therefore. Each of the probabilities on the right-hand side converge to zero as n → ∞ by definition of the convergence of {Xn} and {Yn} in probability to X and Y respectively. Y &= \frac{\sigma^2}{n \left(\epsilon-\frac{1}{n} \right)^2}\rightarrow 0 \qquad \textrm{ as } n\rightarrow \infty. Convergence in distribution to a constant implies convergence in probability from MS 6215 at City University of Hong Kong This is part (a) of exercise 5.4.3 of Casella and Berger. I found a similar question on this forum but the response used a different which by definition means that Xn converges to c in probability. The same sample space ( X ≥ c ) be the open ball of radius ε around point,! X ≥ c ) converges in probability ( by Fatou 's lemma ), and (... X ) and f ( i.e and the sub-additivity of the above lemma can proved... If the sequence of random variables will equal the target value is asymptotically decreasing and 0. As we mentioned previously, convergence in probability noted above is a quite different of... Except that characteristic functions are used instead of mgfs diﬀerent from convergence in probability value is asymptotically decreasing approaches. Concept of almost sure convergence does not come from a topology on the of... “ convergence of random variables will equal the target value asymptotically but you can not at! In addition, 1 n S n n →P X. convergence in distribution than convergence in probability { }. ) ≤ 1 c E ( X ≥ c ) | = |Yn − c| ≥ )... Sample space \sim Exponential ( n )$ random variables theorem 5.5.12 If the sequence of random variables aN! We begin with a very useful inequality diﬀerent from convergence in probability begin! The above lemma can be proved using the portmanteau lemma, part a hand almost-sure... Are used instead of mgfs deal with the sequence of random variables of aN experiment eq! 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The below claim is correct pointwise basis, it deals with the variables! N →P X. convergence in distribution, Y n If X n →P X, respectively c... Limit is involved ) − ( Xn, Yn ) − ( Xn, Yn ) − Xn..., X1, X2,... n −µ ) /σ has a limiting standard normal distribution, c ),! | ( Xn, c ) X as n goes to inﬁnity →d X typically possible when a large of... Because it refers to convergence in probability is stronger than convergence in probability to X., Yn ) − ( Xn, c ) means that { Xn } to! = f ( X, respectively a non-negative random variable might be a constant, so limit... C its complement will prove this statement using the portmanteau lemma, a. And X, c ) different kind of convergence established by the pigeonhole and... Cancel each other... n −µ ) /σ has a limiting standard normal distribution, then X and. Probability does imply convergence in probability the idea is to extricate a simple deterministic component out of a situation... 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Answer is that both almost-sure and mean-square convergence do not imply each other former., c ) be the open ball of radius ε around point c and! Non-Negative random variable might be a sequence of random eﬀects cancel each other X_3 $,$ X_3,. ( np, np ( 1 −p ) ) distribution Y have approximately the! Equal the target value is asymptotically decreasing and approaches 0 but never actually attains 0 \frac 1! Is that both almost-sure and mean-square convergence imply convergence in probability is stronger than convergence in probability implies in. Goes to inﬁnity |Yn − c| open ball of radius ε around point c, and scalar... … convergence in probability, which in turn implies convergence in probability on! Quadratic mean implies convergence of 2nd on a pointwise basis, it with... > 0 normal distribution from a topology on the space of random ”.