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I am currently studying the concept of convergence in distribution in probability theory.

We have the permutation $\tau : \{1,\dots,n\} \to \{1,\dots,n\}$. Let $\tau_n$ be a random permutation chosen uniformly random from all $n!$ possible permutations.

We let $X_n$ be the number of fixed points of $\tau_n$.

I want to show that $\lim_{n\to\infty} F_{X_n} = F_X$ where $X$ ~ $\text{Poi}(1)$.

I am given a hint that I can explain that it is enough to show that $P(X_n = j) \to \frac{1}{ej!}$ for any fixed $j$. After that I can use the fact that $P(X_n = 0) \to 1/e$ as $n \to \infty$ (I have already proven this stament) to estimate the number of permutations with exactly j fixed points.

The thing I cannot get my head around is the part where I need to explain that it is enough to show that $P(X_n = j) \to \frac{1}{ej!}$ for any fixed $j$. Can someone clarify this a little for me?

I started with writing out that

$F_{X_n} = P(X_n \leq j) = P(X_n = 0) + P(X_n = 1) + \dots + P(X_n = j)$

But everything I tried after that didn't really make things clearer for me..

Any ideas?

Thanks in advance!

hello
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1 Answers1

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If you want to calculate $P\left(X_n=j\right)$, you need to calculate the amount of permutations $\tau_n:\{1,...,n\}\longrightarrow \{1,...,n\}$ with j fixed points.

If we were to construct such a permutation we'd have $binomial(n,j)=\frac{n!}{\left(n-j\right)! j!}$ possibilities to choose j fixed points. Now we have n-j ponints left, to construct permutations without fixed points. Following the wikipedia aricle about derangement, there are \begin{align} \left(n-j\right)! \sum_{k=0}^{n-j} \frac{\left(-1\right)^k}{k!} \end{align} permutations $\tau_{n-j}:\{1,...,n-j\}\longrightarrow \{1,...,n-j\}$ without fixed points.

Thus there are $\frac{n!}{\left(n-j\right)! j!} \left(n-j\right)! \sum_{k=0}^{n-j} \frac{\left(-1\right)^k}{k!} $ permutations with exactly $j$ fixed points.

Overall there are $n!$ permutations, therefore \begin{align} P\left(X_n=j\right)&= \frac{\frac{n!}{\left(n-j\right)! j!} \left(n-j\right)! \sum_{k=0}^{n-j} \frac{\left(-1\right)^k}{k!} }{n!} \\ &= \frac{1}{j!} \sum_{k=0}^{n-j}\frac{\left(-1\right)^k}{k!} \\ &\underset{n\rightarrow \infty}{\longrightarrow} \frac{1}{j!} \frac{1}{e}. \end{align}

In conclusion we see, that \begin{align} P\left(X_n=j\right) \underset{n\rightarrow \infty}{\longrightarrow} \frac{1}{j! e} =\frac{1^j e^{-1}}{j!}, \end{align} which ist the probability density of the Poisson distribution with parameter $\lambda=1$. Then $\text{lim}_{n\rightarrow \infty }F_{X_n}= F_X$, wehere $X \sim \text{Poi}\left(1\right)$.

RobH
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  • Thank you for your response! I understand your reasoning, however I do not really understand why it is sufficient enough to show that $P(X_n = j) \to \frac{1}{ej!}$. Can you maybe corroborate this a bit? – hello Jun 11 '19 at 09:23
  • I edited the answer :-) – RobH Jun 11 '19 at 09:41
  • Yeah, I get that haha :D but, I mean this: $F_{X_n} = P(X_n \leq j) = P(X_n = 0) + P(X_n = 1) +...+ P(X_n = j)$. So I would say that we need to show that $\lim_{n \to \infty}F_{X_n} = \lim_{n \to \infty}( P(X_n = 0) + P(X_n = 1) +...+ P(X_n = j))$.

    Why is showing that the limit of $P(X_n = j)$ (i.e. only the last term) converges to a Poisson distribution enough?

     – hello Jun 11 '19 at 09:48
  • It holds $P\left( X_n = t \right)\underset{n\rightarrow \infty}{\longrightarrow} \frac{1}{t!}\frac{1}{e} \forall t\in{1,...,n}$. – RobH Jun 11 '19 at 16:01
  • But if that relation holds for all $t$, that would imply that $\lim_{n\to\infty} P(X_n \leq j) = \lim_{n\to\infty}(P(X_n=0)+P(X_n=1)+ \dots +(X_n=j)) = \lim_{n\to\infty}P(X_n = 0) + \lim_{n\to\infty}P(X_n = 1) + \dots \lim_{n\to\infty}P(X_n = j) = \frac{1}{e0!} + \frac{1}{e1!} + \dots + \frac{1}{ej!}$

    which is different from $ \frac{1}{ej!}$. Right?

     – hello Jun 11 '19 at 18:51
  • Sorry, somehow I could get the fomat right: the word "which" should be on the last line before "is different from". – hello Jun 11 '19 at 18:57
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    That's correct! So you have $lim_{n\rightarrow \infty}P\left(X_n\leq j\right)= \frac{1}{e 0!}+\frac{1}{e 1!}+...+\frac{1}{e j!}= P\left(X\leq j\right)=F_X\left(j\right)$, where $X\sim \text{Poi}\left(1\right)$. As you wanted. – RobH Jun 12 '19 at 07:14
  • Thankyou!! I was thinking stupid, of course the CDF is the sum of the PMF's. – hello Jun 12 '19 at 08:46