Max matching size in a random graph

Consider the random graph $G(n,\frac{1}{n})$. I'm trying to estimate the size of the maximum matching in $G$.

If we look at one vertex, the expected value of its degree is $\frac{n-1}{n}$ so it seems like with high prob it should be 1.

So if I can show that with high probability half of the vertices has degree $1$, then with high probability the size of maximum match in $G$ would be of size $\frac{n}{4}$, but I couldn't prove it, and I'm looking for a hint on how to show that or something similar to that claim.

First, just because the expected value of a fixed vertex is $\frac{n-1}{n}$ does not mean w.h.p. its degree is 1. In fact, $$\text{Pr(deg(}v\text{)=1)} = {n-1 \choose 1} \frac{1}{n} \left(1-\frac{1}{n}\right)^{n-2} \approx 1 \cdot e^{-1}.$$ Now if $X$ is the random number of vertices with degree exactly 1, then $E[X] \approx e^{-1} n$.

Let $Y$ be the number of isolated edges. Note that there is clearly a matching that is at least $Y$. Now $$E[Y] = {n \choose 2} \frac{1}{n} \left(1-\frac{1}{n}\right)^{2(n-2)} \approx \frac{n}{2e^2}.$$ Now at that remains is to show that $Y$ is concentrated around its mean.

• I didn't thought of that direction, but that's probably will solve my problem, I'll try and see what I get. Thanks! Jan 8 '16 at 18:24
• Okay using Chernoff bound I showed that X is concentrated around it's mean, but I cant convince myself that I do have a matching of that size. Could you guild me how to show that? Jan 9 '16 at 15:12
• Whoops! I changed the post. I was not thinking right when thinking that there is necessarily a matching of size $X/2$. You change still show that $Y$ is concentrated about its mean (try Chebyshev's inequality). I am not sure how you are using Chernoff's bound to show that $X$ is concentrated about its mean, because $X$ is not binomially distributed (the trials are not independent). Jan 10 '16 at 21:07
• Yeah I already got that and used chebyshev.. Thanks Jan 10 '16 at 22:48

I would like to expand on the answer of D Poole, specifically to show the concentration of $$Y$$.

It is actually really easy if we use the method of bounded differences:

Theorem Suppose that $$X_1$$, $$\cdots$$, $$X_n \in \mathcal{X}$$ are independent, and $$f:\mathcal{X}^n\to \mathbb{R}$$. Let $$c_1,\cdots,c_n$$ satisfy $$\sup_{x_1,\cdots,x_n,x_i'} |f(x_1,\cdots, x_{i-1},x_i,x_{i+1},\cdots, x_n) - f(x_1,\cdots, x_{i-1},x'_i,x_{i+1},\cdots, x_n)|\le c_i$$ for $$i=1,\cdots, n$$. Then $$\mathrm{P}\{f-\mathbb{E}f\ge t\}\le \exp\left( \frac{-2t^2}{\sum_{i=1}^n c_i^2}\right).$$

Here we have $$n \choose 2$$ random variables, namely 0-1 variables saying which edges are present in the graph. When we add/remove one edge, the number of isolated edges changes by at most two. Therefore $$Y$$ is exponentially concentrated around the mean.

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