1
$\begingroup$

Let $G=(V,E)$ be an undirected random graph such that

  • $V$ is the set of nodes, and $E$ is the set of edges
  • Assume the ground graph $G$ is sparse enough, for example, $\frac{|E|}{|V|}= c \in [10, 40]$ or some not large number ?
  • each edge $uv\in E$ is associated with a probability $p_{uv}$, i.e., $uv$ is kept with probability $p_{uv}$, otherwise removed.

Let $a,b,k$ be three nodes of the graph.

Let $a-b$ be the event that $a$ is reachable from $b$. Of course, because the graph is undirected, $b$ is also reachable from $a$.

My question is

Can we estimate the conditional probability $\mathbf{P}(a-k|a - b)$?

When the graph only has $a,b,k$, and $p_{ab}=1$, then its value is $p_{ak}+(1-p_{ak})p_{bk}$. But this case impossible when the graph is big enough.

When $a=b$, then its value is $\mathbf{P}(a-k,a-k|a - a)=\mathbf{P}(a-k)$.

What about the general case?

$\endgroup$
  • 1
    $\begingroup$ Do you know anything about the values of the $p_{uv}$? This is likely to be extremely difficult in full generality, but straightforward for many particular examples. For instance, if the $p_{uv}$ are constant and $|G|$ is large then all of your expressions are approximately equal to $1$ as $G$ is very likely to be connected. $\endgroup$ – Ben Barber Aug 17 '15 at 15:46
  • $\begingroup$ @BenBarber OK, we assume that $p_{uv}$ are small enough, for example, $\sup_{uv} p_{uv} \leq 0.05$ or at least most probabilities are small enough. And assume $G$ is large enough which has thousands or millions of nodes. $\endgroup$ – Wieshawn Aug 17 '15 at 16:30
3
+50
$\begingroup$

If you're willing to take $n$ large then we can use some known results about the "giant component," since the asymptotic situation is a good approximation to reality. Here is a paper by Erdos and Renyi, and here is a nice expository write-up (the break-down on pages 3 and 4 is excellent). If $p = c\log(n)/n$ then the random graph is connected, so your $P(a-b|a-k) = 1$ for any $a,b,k$. See also this MO question. If $p \sim c/n$ for a constant $c>1$ then there is a giant component and some smaller components. If you look into the Erdos-Renyi paper you can write down the probability a vertex lands in the giant component as a function of $n$. Let's call this probability $q$. Since there's no relationship between b and k, the probability you seek is bounded below by the probability that each of a,b,k are in the giant component, which is approximately $q^3$. You can get even more precise if you want, using the distribution of smaller components to discuss the situation where a and b are in the same component but not the giant component. In that case your $P(a-k|a-b)$ is just $P(k$ is in that component too).

If $p$ is very small compared to $n$, like $o(1/n)$ then $G(n,p)$ is a disjoint union of trees, and Erdos-Renyi analyze the sizes of these trees. You can use their formulas to compute for each tree the probability that $k$ is in the tree. Then take the weighted sum over all trees (weighted by the size of the tree, which is proportional to the probability $a$ is in the tree) and you can compute $P(a-k)$. I think here the extra information that $a-b$ is not going to help much, since $b$ and $k$ have no relationship. To me, knowing $a-b$ occurs just tells me that either $a=b$ or that $a$ is not isolated. The first is very unlikely if $a,b,k$ are chosen at random, while the second can be factored into your computation by ruling out trees of size 1, again using the formulas from Erdos-Renyi to see how likely those trees are. You can conduct a similar analysis for $p\sim c/n$ for $c<1$ or other cases, but I think those in the first paragraph sound more likely to be helpful based on your question. The lecture notes are a good reference.

$\endgroup$
  • 1
    $\begingroup$ Thanks. Erdos and Renyi assume the graph are generated from the ground complete graph with $C_n^2$ edges. But I intended the ground graph sparse enough, for example $\frac{|E|}{|V|} = c \in [10, 40]$? Under such assumption, I think the probability would not be close to $1$? $\endgroup$ – Wieshawn Aug 18 '15 at 2:00
  • $\begingroup$ I'm having trouble parsing your comment. You ask if the probability is not close to 1. Could you clarify by telling me which probability you're talking about? $\endgroup$ – David White Aug 18 '15 at 11:41
  • $\begingroup$ Also, just to clarify, Erdos-Renyi write $C_n^N$ to mean (n choose 2) choose N. But when you write $C_n^2$ you must mean n choose 2 (i.e. all possible options). I would not say they start with a ground complete graph. I think they start with an empty graph on $n$ vertices and add edges randomly (I suppose equivalently they could start with a complete graph and discard edges randomly). In your new edit I guess you mean some edges are already there and you randomly add more edges? $\endgroup$ – David White Aug 18 '15 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.