Suppose two random permutations of the numbers 1 to n placed side by side.
a) Calculate the expectation number of fixed points for $n = 5$.
b) Find the value of expectation in the amount of fixed points for n tending to infinity
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What have you tried so far ? Did you start by, say, trying to count the number of permutation with exactly one fix point ? – yago May 06 '14 at 13:51
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I am trying to calculate the probability of occurring a fixed point. $P[X_{i} = i] = 1/n$ but I don't know if it is correct – user50098 May 06 '14 at 14:00
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The term probability can be confusing here. You are in a finite universe. So you basically need to count, for $n$ fixed, how much permutations have 1 fix point, 2 fix points, and so on...if you call $\Gamma_i$ the number of permutations with exactly $i$ fixed points, then the expectation is $E[\Gamma]=\frac{\sum_{i=0}^{n} i\Gamma_i}{n!}$ since there are $n!$ permutations. Try to first count the number of permutations such that $\sigma(n)=n$, id est with exactly one fix point which is $n$. Then try to generalize when this fixpoint is $1 \leq i < n$, and finally when there are several arbitrary. – yago May 06 '14 at 14:07
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Let random variable $X_i$ be defined by $X_i=1$ if $i$ is a fixed point of the permutation, and by $X_i=0$ otherwise. Then the number $W$ of fixed points is given by $W=X_1+\cdots+X_n$.
By the linearity of expectation we have $E(W)=E(X_1+\cdots+X_n)=E(X_1)+\cdots+E(X_n)$.
The probability that $X_i=1$ is $\frac{1}{n}$. Thus the expected number of fixed points is $n\cdot\frac{1}{n}$, that is, $1$.
Remarks: $1$. If we want to find the expected number of common fixed points of two random permutations, the same technique works. Let $Y_i=1$ if $i$ is a common fixed point of the two permutations. Then the number of common fixed points is $Y_1+\cdots +Y_n$. We have $\Pr(Y_i=1)=\frac{1}{n^2}$, so the expected number of common fixed points is $\frac{1}{n}$.
$2$. The random variables $X_i$ of the answer are not independent. However, expectation is linear in all cases. The above method of indicator random variables bypasses the problem of finding the distribution of $W$.
André Nicolas
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I believe this is the expectation for the case of a unic permutation. I understand the exercise and I have two permutations when comparing the two fixed points are equal. For example:$$(2,1,3,5,4)$$ and $$(5, 4, 3, 1, 2)$$ Here was a fixed point, want to know the probability of that happening. The exercise states that the amount fixed points follows the Poisson distribution – user50098 May 06 '14 at 15:27
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1The distribution of the number of fixed points is another matter, the question asked about the mean. If your problem asks about the number of common fixed points of two random permutations, define $Y_i=1$ if the two random permutations have $i$ as a common fixed point. Then $\pr(Y_i=1)=\frac{1}{n^2}$, so the expected number is $n\cdot\frac{1}{n^2}$. – André Nicolas May 06 '14 at 15:50
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You are welcome. The distribution of $W$ could be found, and then, with some pain, the expectation. However, there are quite a few instances where the distribution of a random variable is not practically accessible, but the method of indicator random variables works smoothly. – André Nicolas May 06 '14 at 15:59
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okay. I can say that as N grows, the expected number tends to zero ? – user50098 May 06 '14 at 17:45
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If we ar looking at the expected number of common fixed points, then yes, but one can say much more, there is an explicit formula $1/n$. – André Nicolas May 06 '14 at 18:14
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4A famous mathematician --- I don't remember who, probably Alfréd Rényi --- has once said that (not word for word, but something like this) one useful fact in probability theory that is so often overlooked by teachers and students is that expectations commute with additions even when the random variables are dependent. This answer is a nice example to his remark. – user1551 May 08 '14 at 07:18
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@user1551: A related question was asked by apurv about $9$ hours ago (sorry, my attempt to get a link didn't work). There expectation was used to solve a counting problem. – André Nicolas May 08 '14 at 07:30
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