# 1 then px vn v0 0 and hence px v0 1 1 in words

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Unformatted text preview: c when i 0 p) and setting ⇡ (0) = c, we have ✓ p 1 p ◆i (1.29) There are now three cases to consider: P p &lt; 1/2: p/(1 p) &lt; 1. ⇡ (i) decreases exponentially fast, so i ⇡ (i) &lt; 1, and we can pick c to make ⇡ a stationary distribution. To ﬁnd the value of c to make ⇡ a probability distribution we recall 1 X ✓i = 1/(1 ✓) when ✓ &lt; 1. i=0 Taking ✓ = p/(1 p) and hence 1 ✓ = (1 2p)/(1 of the ⇡ (i) deﬁned in (⇤) is c(1 p)/(1 2p), so ⇡ (i) = 1 2p · 1p ✓ p 1 p ◆i = (1 p), we see that the sum ✓)✓i (1.30) To conﬁrm that we have succeeded in making the ⇡ (i) add up to 1, note that if we are ﬂipping a coin with a probability ✓ of Heads, then the probability of getting i Heads before we get our ﬁrst Tails is given by ⇡ (i). The reﬂecting random walk is clearly irreducible. To check that it is aperiodic note that p(0, 0) &gt; 0 implies 0 has period 1, and then Lemma 1.18 implies that all states have period 1. Using the convergence theorem, Theorem 1.19, now we see that I. When p &lt; 1/2, P (Xn = j ) ! ⇡ (j ), the stationary distribution in (1.30). Using Theorem 1.22 now, E0 T 0 = 1 1 1p = = ⇡ (0) 1...
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