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Unformatted text preview: a ) ≤ E [X ]
, for all a > 0.
a Chance of being abnormally large is small. Proof: Deﬁne a random variable Y as (draw a ﬁgure to compare the
distributions of Y and X )
Y = 0,
a, if X < a;
if X ≥ a. Clearly, E [Y ] ≤ E [X ] and P (Y = a) = P (X ≥ a). Thus
E [Y ] ≤ E [X ] ⇒ 0 · P (Y = 0) + a · P (Y = a) = a · P (X ≥ a) ≤ E [X ].
M. Chen (IE@CUHK) ENGG2430C lecture 10 5 / 29 Markov Inequality If a random variable X takes only nonnegative values, then
P (X ≥ a ) ≤ E [X ]
, for all a > 0.
a Example (Umbrella problem):
Each of n students chooses one out of n umbrellas at random. What is
the probability that at least 3 students get their own umbrellas?
Deﬁne a r.v. Y to be the number of students who choose their own
umbrellas. We already know E [Y ] = 1. Then
P (Y ≥ 3) ≤ M. Chen (IE@CUHK) E [Y ] 1
=.
3
3 ENGG2430C lecture 10 6 / 29 Markov Inequality Markov inequality may give loose bound. [Example 7.1 in textbook]
Let X be a r.v. uniformly distributed in [0, 4]
E [X ] = 2
Applying Markov inequality, we have...
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