Let X X 1 X 2 X n be the number of cards that have never been chosen after 1 n

Let x x 1 x 2 x n be the number of cards that have

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Let X = X 1 + X 2 + ... + X n be the number of cards that have never been chosen after (1 - ) n ln n steps. We will use a Poisson approximation and a Chernoff bound to upper bound the probability P [ X (1 - δ ) n ]. Let Y i Poi ((1 - ) ln n ), and let Y = Y 1 + Y 2 + ... + Y n be the number of cards not chosen in the Poisson case. Then we have E [ Y i ] = e - (1 - ) ln n = n - 1 and E [ Y ] = n . By the Chernoff bound from Theorem 5.4 for Poisson random variables, we have P [ Y b ] e - n ( - en ) b b b . Now, we use the Poisson approximation from Corollary 5.9 to get P [ X b ] e - n ( - en ) b b b e (1 - ) n ln n. Now, we compute a lower bound for the variation distance between the Markov chain and the stationary distribution. Let k = n / 2, and consider a subset S of the states for the Markov chain, where for each state in S , the k bottom cards are ordered in the same order that they were in the initial state. The chain will remain in set S any time that none of these k cards are chosen. So, the probability of being in set S after running the chain (1 - ) n steps is at least P [ X k ] = 1 - o (1) by the Poisson-approximation bound above. The probability of a state, chosen from the uniform stationary distribution, being in set S is 1 /k ! = o (1). Thus, the variation distance is at least the difference of these two probabilities, which is 1 - o (1). 1
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3. (MU 11.6) Theorem 11.5 is useful only if there exists a nonzero minimum entry in at least one column of the matrix P of the Markov chain. Argue that for any finite, aperiodic, irreducible Markov chain, there exists a time T such that every entry of P T is nonzero. Explain how this can be used in conjunction with Theorem 11.5. First, we will prove that there exists a time T such that every entry of P T is nonzero. For a finite, aperiodic, irreducible Markov chain, π i = 0 for all i . From Theorem 7.7, we know that lim t →∞ P t ji = π i . By the definition of a limit, for every i and j , there exists a T ij such that for all t T ij , P t ji > 0. We let T = max ij T ij . Now that we can calculate T , we can simply apply Theorem 11.5 to P T to get a bound on the variation distance after t additional steps. This is equivalent to T + t steps of the original chain.
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