I am new to random graph theory, and is it correct that the proofs of theorem 2.1 of Random Graphs by Bollobas are brief and miss out several steps ? I am particularly interested in the probabilistic part (ii):
Theorem 2.1 Suppose Q is a monotone increasing property and 0 $\leqslant$ p1 < p2$\leqslant$1, then $P_{p_1}(Q) \leqslant P_{p_2}(Q)$ . The associated proof is :
(ii) Put $p=(p_2-p_1)/(1-p_2)$. Chose independently $G_1 \in \mathcal{G}(n,p_1)$ and $G \in \mathcal{G}$(n,p) and set $G_2=G_1 \cup G$. Then the edges of $G_2$ have been selected independently and with probability $p_1$+p - $p_1$p= $p_2$, so $G_2$ is exactly an element of $\mathcal{G}$(n,$p_2$). As Q is monotone increasing, if $G_1$ has Q so does $G_2$. Hence $P_{p_1}(Q) \leqslant P_{p_2}(Q)$.
The reason I ask is because individual graphs $G_1 \in \mathcal{G}(n,p_1)$ can have higher probability than the same graph $G_2$ $ \in \mathcal{G}(n,p_2)$, even though $p_1$ < $p_2$. As an example take simplistically a graph potentially with 5 edges, edge present probabilities $p_1$=0.16, $p_2$=0.51. Suppose the graph $G_1$ has 1 edge present and 4 absent edges. With $p_1$=0.16 the probability of this graph $G_1$ = 0.16 * (1-0.16)$^4$ = 0.0797, and with $p_2$=0.51 the same graph $G_2$ probability is 0.51 * (1-0.51)$^4$ = 0.0294. So it seems that the "Hence" in Theorem 2.1 proof needs to address this issue - or does it already and I haven't seen the neatness of the proof ?
The issue is that your property is not an increasing one. Having an induced subgraph is not necessarily an increasing property. In particular, the $(1-p)^{\#}$ terms would not be present if we were just considering whether a specified graph is present as a (not necessarily induced) subgraph in the random graph.