# P z n µ 1 since they are two disjoint subsets and

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P ( Z n µ " ) = 1 since they are two disjoint subsets and probabilities for disjoint subsets sum to 1 by contruction. Since P ( Z n ² 0) = 1 for each n > 0 , this further indicates lim n !1 P (0 µ Z n µ " ) = 1 for every " > 0 , which is also true for very small " > 0 (i.e, all probability densities for a n th random variable Z n are located for Z n being in a small interval between 0 and " ). Therefore, we can say that Z n shrinks toward 0 as n grows. Figure 1 plots pdfs of the sequence of random variables Z n for n = 1 ; 2 and 5 and con°rms above result graphically; as n increases, we see the higher probability density for the random variable Z n being close to z = 0 ; i.e. Z n shrinks toward 0 as n grows. 3

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0 1 2 3 4 5 0 1 2 3 4 5 z f Zn (z) case 1: n=1 case 2 : n=2 case 3 : n=5 Figure 1: The probability distribution function of the random variable Z n for n = 1 ; 2 and 5 : 3. For any random variable X with pdf f X and any positive real number " > 0 ; we can prove Markov±s inequality as follows, P ( j x j > " ) = R j x j >" f X ( x ) dx = 1 " R j x j >" " ± f X ( x ) dx µ 1 " R j x j >" j x j ± f X ( x ) dx µ 1 " R 1 °1 j x j ± f X ( x ) dx = 1 " E ( j X j ) where the inequality comes from 1) j x j > " in the °rst inequality and 2) the domain of x 2 ( ´1 ; 1 ) includes the subset of x 2 fj x j > " g in the second inequality. The last directly comes from the de°nition, i.e. E ( j X j ) R 1 °1 j x j ± f X ( x ) dx: 4. Suppose that X 1 ; X 2 ; ::: are independent random variables, each of which is equal to zero 4
with probability 2 3 , and equal to one with probability 1 3 : (a) ° = 1 3 , ± 2 = 2 9 (See the answer for Q1 of the problem set 1) (b) Let Z n = 1 p n n P i =1 ( X i ´ ° ) : When n = 2 ; the probability that Z n µ 0 : 924 can be calculated as follows, P ( Z 2 µ 0 : 924) = P ° 1 p 2 2 P i =1 ( X i ´ ° ) µ 0 : 924 ± = P ° 1 p 2 ° X 1 ´ 1 3 + X 2 ´ 1 3 ± µ 0 : 924 ± = P ( X 1 + X 2 µ 1 : 9734) = 1 ´ P ( X 1 + X 2 > 1 : 9734) = 1 ´ P ( X 1 = 1 ; X 2 = 1) = 8 9 Note that we use P ( A ) = 1 ´ P ² A C ³

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