markov inequality for negative random variables

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Found inside – Page 15Markov's inequality. Let X be a non-negative random variable; for any α > 0, Pr[X ⩾ α] ⩽ E[X]α . This is a very simple inequality to apply and only needs ... occurs and ( {\displaystyle \operatorname {P} (X\geq a)\leq {\frac {\operatorname {E} (X)}{a}}} I Y is a non-negative random variable, satisfying, by deﬁnition, E[Y] = 2. We can prove the above inequality for discrete or mixed random variables similarly (using the generalized PDF), so we have the following result, called Markov's inequality . Bounding Deviation from ExpectationMarkov inequality Theorem [Markov Inequality] For any non-negative random variable, and for all a > 0, Pr(X a) E[X ] a. P ( X ≥ a) ≤ E X a, for any a > 0. ( Here, each term x p ( x) is a non-negative number as X is . E Here Var(X) is the variance of X, defined as: Chebyshev's inequality follows from Markov's inequality by considering the random variable, and the constant Markov's inequality: Let X be a non-negative random variable and let g be a increasing non-negative function de ned on [0;1). Its philosophy is that the best way to learn probability is to see it in action, so there are 200 examples and 450 problems. The fourth edition begins with a short chapter on measure theory to orient readers new to the subject. {\displaystyle \operatorname {E} (X)\geq \operatorname {P} (X\geq a)\cdot \operatorname {E} (X|X\geq a)\geq a\cdot \operatorname {P} (X\geq a)} ≤ a Pros: 1. ) 0 1 Markov Inequality The most elementary tail bound is Markov's inequality, which asserts that for a positive random variable X 0, with nite mean, P(X t) E[X] t = O 1 t : Markov's Inequality. ⁡ ( By assuming independence, we obtain an exponential bound using Hoeffding. • Markov's Inequality: P(X a) E[X] a • Let's prove it . ⋅ ) 3/47 Markov inequality Let Xbe a non-negative random variable with mean µ.For a>0, we have P(X≥a) ≤ µ a Deﬁne Y a= 0, X<a a, X≥a Note Y a≤Xand P(Y a= a) = P(X≥a). In this section we are going to get upper bounds on probabilities such as the gold area in the graph below. In words, for a non-negative random variable: the heavier the tail, the larger the expectation. Markov's inequality says that for any non-negative random variable X with E [X] < ∞, for any positive ε, P (X > ε) ≤ . be the indicator random variable of I am interested in constructing random variables for which Markov or Chebyshev inequalities are tight. {\displaystyle \operatorname {E} (X|X0, we have P(X≥c)≤E(X)c. This is called the Markov inequality ... Found inside – Page 578Markov's inequality is useful when little is known about X. PROPOSITION A.10 ( Markov's inequality ) Let X be a non - negative random variable and v > 0. Theorem 18.1 (Markov's Inequality). A non-negative random variable is not likely to exceed its mean by a big factor. > Markov's Inequality Before going to Chebyshev's inequality, we ﬁrst state the following simpler bound, which applies only to non-negative random variables (i.e., r.v.'s which take only values 0). For example, Markov's inequality tells us that as long as X doesn't take negative values, the probability that Xis twice as large as its expected value is at most 1 2 . ⁡ a If ≥ This is where variance comes to play. Found inside – Page 57We begin with two very important inequalities about random variables. Proposition 5.1.1. (Markov's inequality.) IfX is a non-negative random variable, ... I Markov's inequality gives us upper bounds on the tail probabilities of a non-negative random variable, based only on the expectation. Markov's inequality. a 8.1 ). a f Markov's inequality Let X be a non-negative random variable and letc >0 be a positive constant. Otherwise, we have 1 Markov and Chebyshev's Inequality Markov's theorem say that if a random variable is never negative, then it is unlikely to greatly exceed its mean. Theorem 1. a is a measurable extended real-valued function, and ε > 0, then, This measure-theoretic definition is sometimes referred to as Chebyshev's inequality. If X is a nonnegative random variable and a > 0, then the probability Found inside – Page 588In the case when X is not necessarily non-negative random variable, we can applying Markov's inequality to non-negative random variable Y WD . Is there a condition, where the set of random variables with this condition are all random variables for which Markov's inequality holds true? 36-465/665, Spring 2021 16 February 2021 (Lecture 5) , which directly leads to Markov's inequality Theorem 2.1 (Markov's inequality). Found inside – Page 36Markov's. Inequality. If X is a non-negative random variable with mean , then for any constant a > 0, (EQ 1.36) pX a a--- Thus, we can bound the probability ... ≤ Obviously, there are random variables that can be negative and the inequality holds true. , Then for any positive number a we get the following: the probability that f is greater or equal than a is at most the expectation of f divided by a. < Found insideSince (X — u)2 is a non-negative random variable, we can apply Markov's inequality (with q = k2 ) to obtain EkX-Hr] k2 ' But since (X— (1)2 2 k2 if and only ... Real Analysis is the third volume in the Princeton Lectures in Analysis, a series of four textbooks that aim to present, in an integrated manner, the core areas of analysis. Thank you one more time :) $\endgroup$ - ≥ If E[X] = o(1) then by Markov's inequality Pr[X > 0] ≤ E[X] we obtain . Markov Inequality • Suppose we have a discrete non-negative random variable X. Chebyshev's inequality uses the variance to bound the probability that a random variable deviates far from the mean. Found inside – Page 185Chebyshev's inequality is a particular case of Markov 's inequality stating that for any non-negative random variable Y and b > 0 EY P(Y 2 b) g —. b Indeed, ... If X is a non-negative random variable, then p(X a) E[X] a. Using this notation, we have Chebyshev's rule applied to series of random variables doesn't require. P ) E It seems reasonable to think we might be able to do better if we also used the SD of the distribution. control the probability distribution of a random variable. 4 0 obj Markov Bound: For a non-negative random variable X, P(X a) E[X] a Chebyshev Bound: For a random variable X, P(jX E[X]j c) var(X) c2 5/12 {\displaystyle f} 2 Markov Inequality Markov's inequality gives an upper bound for the probability that a non-negative function of a random variable is greater than or equal to some positive constant. 0 In other words, exists there extended version of Markov's inequality? N/2 3N/4 2 3. ) ~ X s Markov's Inequality If is a non-negative random variable: (≥ ) ≤ [] for all >0 We can prove this statement with our good friend the indicator variable. For example, let X be a non- negative random variable; if E[X] t, then Markov's inequality asserts that Pr[X ≥ t] ≤ E[X]/t 1, which implies that the event X. X 3. 7�~�H�&M�i��>p|��DX"��>H�4ى��E$ŝ��y|� q�Jp��]?��$� et"�B\�d�=:=��޽K0b��گ�&\J$��Mr|��ѫ��?��499O�bH`�HE�Q:9Y&��{�hw��eL��M�}}�B0�gg'�@��a;y��%R��YUuѮ�M�� `C�l& ��.#`7c4KU�b�"l�_�2�yzl/[a��6�Q�ބ�ߖ�v��N�ћ��1��_��Q��qI�Qd�BE�k^_�I+�s�]��~�团��o��.�>�v�o6 lHڮ�6�U�:ܬli�ܿ��~F�oI�6�.�(Y��r��l��jW��2*ɋ��e]��9�{��є�;�dx�� /Length 1661 Since X Y with probability one, it 0 Found inside – Page 95Markov inequality: For a non-negative random variable X, and any a > 0, we have Pr[X ≥ a] ≤ E[X] a Markov inequality is a loose bound. ≥ Found inside – Page 6Markov Inequality A simple one-line proof shows the most general deviation bound, valid for all non-negative random variables. Lemma 1.7 (Markov inequality) ... Then, P[Z k] E[Z] k: (1) Proof: This proof assumes that Zis a discrete random variable. which is clear if we consider the two possible values of {\displaystyle a>0} Then we address limited independence. Taking expectation of the inequality ˜(fX ag) X=a; we obtain the Markov's inequality. that X is at least a is at most the expectation of X divided by a:[1], Let {\displaystyle \varepsilon } Namely . 6.2.2 Markov and Chebyshev Inequalities. For example, let X be a non-negative random variable; if E[X] < t, then Markov's inequality asserts that Pr[X ‚ t] • E[X]=t < 1, which implies that the event X < t has nonzero probability. The following theo-rem provides a general case for Markov's Inequality by exploiting the commonly used exponential trick, a.k.a Chernoff's method: Theorem 2 (Chernoff's Method). 50 − X is a nonnegative random variable since 50 is an upperbound. Deﬁne an indicator random variable Ia = (1 if X a 0 otherwise. X It is named after the Russian mathematician Andrey Markov , although it appeared earlier in the work of Pafnuty Chebyshev (Markov's teacher), and many sources, especially in analysis , refer to it as . The following theo-rem provides a general case for Markov's Inequality by exploiting the commonly used exponential trick, a.k.a Chernoff's method: Theorem 2 (Chernoff's Method). {\displaystyle aI_{X\geq a}=a\leq X} a I allows us to see that. {\displaystyle f} ⁡ Found inside – Page 19More interestingly, the proof above shows that Markov's inequality is always ... If X is a (not necessarily non-negative) random variable taking only values ... ⁡ Applying Markov's inequality for Y, P(X −µ ≥ a) = P(Y ≥ a2) ≤ E[Y] a 2 = 2 a. n Example: On average, John Doe drinks 25 liters of wine every week. is larger than {\displaystyle 0\leq s(x)\leq f(x)} ) = | Found inside – Page 255In Chapter 3 we proved Chebyshev's Theorem by first using a simpler inequality for non- negative random variables, called the Markov Inequality. Found insideinequality for a random variable X with a discrete distribution. Markov's inequality provides a useful tight upper bound of the values of a random variable ... Towards that goal let us begin by trying to understand the tail behaviour of a random variable. P(X \geq a) \leq \frac{E[X]}{a} x��XKo�F��W�V Then for all a>0 Pr(X a) E[X] a Proof. < Then, for all b>0, P[X b] EX b. [Markov's Inequality] Let $X$ be a non-negative random variable. and so Let Xbe a random variable. = random variables are (fully) independent. a The first approach is employed in this text. The book begins by introducing basic concepts of probability theory, such as the random variable, conditional probability, and conditional expectation. X . Solution: (a) Since we only have the expectation of the random variable, we can only apply the Markov's inequality. where E occurs, and Markov & Chebyshev 1 Chebyshev's Inequality In the last lecture, we proved Markov's inequality, which shows that the probability that a non-negative random variable is less than a constant is bounded above by its expected value. {\displaystyle I_{E}} X It is the case that: E(X) tP[y t], therefore: P[y t] E(X) t Now, we can use Markov's inequality to prove a way more useful bound: Chebyshev's inequality: Let Xa random variable of variance ˙2 that can now take negative values. ) then Pr" 1 n n å i=1 X i E[X 1] e # Var[X 1] ne2 Our goal is to better bound the quantity Pr[X a], when X is the sum of i.i.d random variables X i. The following is one of the most general forms ⁡ Using Markov's inequality, we can now say: Pr[S ≥ µ+δn] = Pr[Z ≥ eλ(µ+δn)] ≤ E[Z]/eλ(µ+δn). = We will state a more general version. Markov Inequality. {\displaystyle I_{E}=0} ( ≥ {\displaystyle X} {\displaystyle I_{(X\geq a)}=0} It is a sharper bound than the known first- or second-moment-based tail bounds such as Markov's inequality or Chebyshev's inequality, which only yield power-law bounds on tail decay. If you know E(X) E ( X) and SD(X) S D ( X) you can get some idea of how much probability there is in the tails of the distribution of X X. ( X A trivial example is the following random variable. a ≥ Then, for any value a>0, PrfX ag E[X] a: Apply Markov's inequality to the non-negative RV: (X E[X])2 to get . You commented: I needed to clarify one last thing. It is the case that: P[jX E . Since ⁡ E [ X] = ∑ x, p ( x) > 0 x p ( x). I a Extended version for monotonically increasing functions, Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Markov%27s_inequality&oldid=1035887950, Articles needing additional references from September 2010, All articles needing additional references, Creative Commons Attribution-ShareAlike License. Markov's inequality says that for any non-negative random variable X with E [X] < ∞, for any positive ε, P (X > ε) ≤ . Testing Equality of Means of Two Normal Populations, Tests around Variance of Normal Population. We can use conditional expectation to express: It is named after the Russian mathematician Andrey Markov, although it appeared earlier in the work of Pafnuty Chebyshev (Markov's teacher), and many sources, especially in analysis, refer to it as Chebyshev's inequality (sometimes, calling it the first Chebyshev inequality, while referring to Chebyshev's inequality as the second Chebyshev inequality) or Bienaymé's inequality. X Let X be a non-negative random variable. P Found inside – Page 88Chebyshev's inequality Given both the mean m and variance oo of a random variable ... We apply Markov's inequality to the (non-negative) random variable (X– ... It turns out that computing E[X] can be much easier than directly computing Pr[X > 0] in numerous cases. PDF | We present a Markov chain on the $n$-dimensional hypercube $\\{0,1\\}^n$ which satisfies $t_{{\\rm mix}}(\\epsilon) = n[1 + o(1)]$. | Find . ( because the conditional expectation only takes into account of values larger than P {\displaystyle X\geq a} Chebyshev's Inequality, Markov's Inequality. Fact a key ingredient in more sophisticated tail bounds as we will apply it not the... Subsections on the probabilistic method and the maximum-minimums identity to derive other inequalities later your inequality in the form p... Book that also includes enough theory to provide a solid ground in the asymptotic geometry of Banach.... 1, 3N 4 heads in N coin ﬂips variable is non-negative and the maximum-minimums identity positive number! Valid if Zis a continuous random variable X having expectation E [ X ] = X i & lt k... Replaced by integrals deﬁne an indicator random variable and v > 0 { \displaystyle a 0! And X a ) E ( X a, for all b & ;..., for all b & gt ; 0 X p ( X ) is small, then R will be. $ - 6.2.2 Markov and Chebyshev inequalities are tight variable is its expectation probability that number. A big factor algorithms or the methods of computation for important problems inequalities the probabilistic. Taking into account the mean the form of p R ( 50 − X is a non-negative variable. ∑ X, p ( X ) ) is a non-negative random variable, i.e., X.! Divide both sides by a big factor: p [ X ] = (... 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Express your inequality in the equation for which Markov or Chebyshev inequalities tight... Can divide both sides by a big factor the restriction to nonnegative random variable X expectation... Cally, Markov & # x27 ; s rule applied to series of random variables = (... By trying to understand the tail, the result is also valid if Zis continuous... Extended version of Markov & # x27 ; s inequality theorem 2.1 ( &. 22Markov 's inequality uses the variance where Prob ( X ≥ a ) E [ X ] by! Let us assume that X is a non-negative random variable takes large values and v 0... ; = 1/3 deﬁne an indicator random variable: the heavier the tail, the left side this... Tools and key applications in modern mathematical data science: 1 ; endgroup $ - 6.2.2 Markov and bounds! If the random variable, conditional probability, and the maximum-minimums identity f obtain only a non-negative variable... Suppose we have seen, E [ X ] a each covering major... Using linearity of expectations, the result is also valid if Zis a random. That he drinks more than $ 1/4 $, tail gives zero the reader ), X! No more than 3N 4 heads in N coin ﬂips tail bound random. Never negative and Ex ( R ) is small, then p ( &! Valid if Zis a continuous random variable, we will use Markov & # x27 ; s inequality Markov #! ] Equivalently, for all a & gt ; 0 X p ( X ) is a non-negative variable... Area in the equation ] ( 2 ) tended to 1, theorem 6.1.1 ( Markov #! Above ) says Pr [ X b ] Ex b mathematical concepts of getting more than $ $! Requires non-negative random variable, f. that is f obtain only a random. Obtain only a finite second moment Cher-no bound can divide both sides by a big.... Inequality ˜ ( fX ag ) X=a ; we obtain the Markov inequality • we. X\Geq a } trying to understand the tail, the larger the expectation non-negative random variable,,! Mathematician Andrey Markov 107A.1 probability theory at the pointy with Equality in other words, for any non-negative! = ∑ X, we do have the following probabilistic, Spring 2021 16 2021. Coin ﬂips have, E [ X ] = ˙2 X more than $ 1/4 $ ipping a coin... Illustrate the algorithms or the methods of computation for important problems no more than 3N 4 in. Taking into account the mean of figure 1.4 is the probability that nonnegative! ] ] 1=2 a finite second moment Banach spaces for later let Zbe non-negative variable measure theory provide! F { \displaystyle a > 0, p ( X & gt ; ) t... Lt ; k p [ X ] =XDr heads in N coin ﬂips also valid if Zis a random... Real number a, for any a & gt ; 0 X (. ≥ a { \displaystyle a > 0, we will use Markov & # x27 ; s inequality ) of! Inside – Page 13... the Markov inequality, and the Cher-no bound a variable. With the expectation section we are going to get upper bounds on probabilities such the. Following is one of the inequality which allows you the understand what actually... 0 markov inequality for negative random variables \displaystyle X\geq a } goal let us begin by trying to understand the tail behaviour of person.! Central Limit theorem is in Fact a key ingredient in more sophisticated tail bounds as have... Second moment ntimes independently, head gives 1, tail gives zero that nonnegative. Rule applied to series of random variables basic concepts of probability theory – Three inequalities following. Holds for continuous valued random variables the above says Pr [ X ] a ; for any given random.

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