I am reading Louigi's lecture note on random trees and graphs [here](http://problab.ca/louigi/notes/ssprob2021.pdf). I get stuck on part (b), Exercise 1.2.3 on page 19, which says the following:



> 
> Let $T\_n$ be uniformly drawn from $\mathcal{T}\_n$, the set of $n$-vertices labeled rooted trees, and $c(v;t)$ be the number of children of vertex $v$ in a fixed tree $t$. Then $\Pi\_n(c):=\frac{1}{n}\sharp\{u\in[n]:c(u;T\_n)=c\}$ converges in probability to $\mathbb{P}(Poisson(1)=c)$.
> 
> 
> 


In part (a) he proved that $c(1;T\_n)$ converges in distribution to Poisson(1) distribution, which is clear to me using the so-called "line-breaking" construction in the lecture note. Then the above, as far as I'm concerned, is just a WLLN, so I want to check the Markov's condition:


$$\frac{1}{n^2}Var(\sum\limits\_{u\in[n]}1(c(u;T\_n)=c))\rightarrow 0.$$


However I am having problem controlling the "crossing-terms":


$$\frac{1}{n^2}\sum\limits\_{u\not=v}[\mathbb{P}(c(u;T\_n)=c(v;T\_n)=c)-\mathbb{P}^2(c(u;T\_n)=c)].$$


By line-breaking construction I can write out the probability of the first, by counting the number of labeled rooted trees with vertices 1 and 2 each has exactly c children:


\begin{aligned}
\mathbb{P}(c(1;T\_n)=c(2;T\_n)=c) &= \frac{\sharp\{t\in\mathcal{T}\_n:c(1;t)=c(2;t)=c\}}{n^{n-1}}\\
&= \frac{\sharp\{v\in[n]^{n-1}:\text{ there are exactly }c\text{ copies of 1,2 in }v=(v\_1,\cdots,v\_{n-1})\}}{n^{n-1}}\\
&= \frac{1}{n^{n-1}}{n-1\choose c}{n-c-1\choose c}(n-2)^{n-2c-1}\\
&= \frac{1}{(c!)^2}(1-\frac{2}{n})^{n-2c-1}\mathop{\Pi}\limits\_{k=1}^{2c}(1-\frac{k}{n}),
\end{aligned}


and the second probability, without the square, is(as proved in earlier context in the lecture note)


$$\mathbb{P}(c(1;T\_n)=c)=\mathbb{P}(Bin(n-1,\frac{1}{n})=c)={n-1\choose c}\frac{1}{n^c}(1-\frac{1}{n})^{n-1-c}.$$


Then I have no idea how to bound the difference. Any help is appreciated.



Empirical degree distribution of random $n$ vertices labeled rooted tree converges to Poisson distribution
Essentially, you want to show that
$$p\_{12}-p\_1^2\to0,\tag{1}$$
where
$$p\_{12}:=P(c(1;T\_n)=c(2;T\_n)=c),\quad p\_1:=P(c(1;T\_n)=c).$$
We have
$$p\_1=\binom{n-1}c\frac1{n^c}\Big(1-\frac1n\Big)^{n-1-c}
=\Big(1-\frac1n\Big)^{n-1-c}\frac1{c!}\prod\_{j=1}^c\Big(1-\frac jn\Big)\to\frac{e^{-1}}{c!}.$$
If your calculation of $p\_{12}$ were correct, then we would have
$$p\_{12}
=\Big(1-\frac2n\Big)^{n-2c-1}\prod\_{k=1}^{2c}\Big(1-\frac kn\Big)\to e^{-2},$$
so that (1) would fail to hold.


So, apparently the factor $1/(c!)^2$ is missing in your ultimate expression for $p\_{12}$.


