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Minimum Degree of a Random Graph (Probabilistic Method)

joypav

Active member
Mar 21, 2017
151
Problem:
Suppose that the function $p : N \rightarrow [0, 1]$ satisfies $p >> n^{-1}ln(n)$ (i.e. $n^{-1}ln(n) = o(p)$).

(a) Prove that as $n \rightarrow \infty$, the random graph $G(n, p)$ has minimum degree at least $\frac{np}{2}$ almost surely.
Idea: Look at the degree of each individual vertex $v \in V(G)$. Try to find a bound on the probability of this fixed vertex having a small degree (using Chernoff's Inequality?).

I am not getting to where I think I should with this idea.
Let $X_{u,v}$ be $1$ is $uv \in E(G)$ and $0$ if $uv \notin E(G)$.
Then, $X_v = \sum_{u \in V(G), u \neq v} X_{u,v}$ will give us the degree of $v$.
Also, the $X_{u,v}$'s are uniform/independent (as $G$ is a random graph and edges are picked independently with probability $p$).
Now we may use Chernoff's? How do I use the first thing we assumed... that $p >> n^-1ln(n)$?


(b)
Conclude that for any fixed positive integer k, the random graph $G(n, p)$ has minimum
degree at least k almost surely as $n \rightarrow \infty$.
 
Last edited:

joypav

Active member
Mar 21, 2017
151
For completeness, I will post a "proof" I came up with. As you can see, at the end I skipped over some much needed justification that I couldn't seem to figure out. Some help with this would be appreciated!

However, I do think that my general idea is correct.

Proof: (a) Let $G \sim G(n,p)$. Fix $v \in V(G)$. Let $A_{v,u}$ be the event that $vu \in E(G)$ and let $X_{v,u}$ be its indicator. Then, $X_{v,u}=1$ with probability $p$ and $X_{v,u}=0$ with probability $1-p$. Because each possible edge is inserted independently, the $X_{v,u}$'s are independent. Let $X_v = \sum_{u \in V(G), u \neq v} X_{u,v}$, (giving the degree of $v$). Now we may use Chernoff's Inequality with:
$$\sigma^2 = Var[X_v] = np(1-p)$$
$$E[X_v] = \sum_{u \in V(G), u \neq v} E[X_{u,v}]=p(n-1)$$
Then,
$P[$degree of v is less than $\frac{np}{2}] = P[X_v \leq E[X_v] - \frac{np-2p}{2}] = P[X_v \leq \frac{np}{2}]$
$$ \leq e^{\frac{-(\frac{np-2p}{2})^2}{2(np+\frac{\frac{np-2p}{2}}{3})}} = e^{-\frac{3n^2p-12np+12}{28n-8}}$$
Let $X$: existence of vertices with degree less than $\frac{np}{2}$. Then,
$$P[X]= \sum_{v \in V(G)}P[X_v \leq \frac{np}{2}] \leq n\cdot e^{-\frac{3n^2p-12np+12}{28n-8}} = e^{ln(n)-\frac{3n^2p-12np+12}{28n-8}}$$
By assumption $p > > n^{-1}ln(n)$.
$$\Rightarrow ln(n)-\frac{3n^2p-12np+12}{28n-8} \rightarrow -\infty $$
$$\Rightarrow P[X]\rightarrow 0$$
Then every vertex of G must have degree at least $\frac{np}{2}$.

(b) Apply part (a). So we have $\delta(G) \geq \frac{np}{2}$ a.s. as $n \rightarrow \infty$. As long as $n \geq \frac{2k}{p}$, $\delta(G) \geq k$.