Absence of evidence is not evidence of absence, but it helps.

A long time ago in a city far far away I was a grad student.

Now here at Berkeley it is qualifying exam season, and the following exchange took place in an elevator:

Professor X: “I was at A’s a qualifying exam today.”
Grad student Y: “How did A do?”
Professor X: “He was great!”

Of course, X wouldn’t have said that A did well if A had done poorly. But he probably also wouldn’t have said that A did poorly; instead he would have hemmed and hawed and avoiding saying anything at all. And that would have told Y what she wanted to know.

Similarly, I remember that when I was in grad school, the information that someone had passed their qualifying exam spread quite quickly among the students, whereas the information that someone failed spread less quickly. The reason for this is simple: passing is happy news, so the person who passed and their friends will tell everyone. But failing is sad news, so you can really only find out if someone failed by specifically asking them.

So assume that rumors spread according to a logistic model, and furthermore that the information of a success spreads twice as quickly (for small populations) as the information of a failure. That is, the proportion of the population that knows that person X succeeded at time $t$ after their exam is

P_1(t) = {1 \over 1+C_1 e^{-t}}

and the proportion that knows that person Y failed at time $t$ after their exam is

P_2(t) = {1 \over 1+C_2 e^{-t/2}}

for some different constant C_2. Furthermore P_1(0) = P_2(0) = 1/(n+1), where n+1 is the number of students; solving gives C_1 = C_2 = n.

So say it’s time t, and you haven’t heard if the person who had their exam at time 0 passed or not. Then by Bayes’ theorem, the odds in favor of their passing are that they’ve passed is

O(pass|N) = O(pass) {P(N|pass) \over P(N|fail)}

where O(\cdot) denotes the odds of \cdot N denotes the event of not having heard yet. Typical pass rates at the time I was there were perhaps 3 out of 4, so O(pass) \approx 3. The conditional odds are therefore

O(pass|N) = 3 {P(N|pass) \over P(N|fail)}.

But P(N|pass) = 1-P_1(t) and P(N|fail) = 1-P_2(t). So, after some algebra,

{P(N|pass) \over P(N|fail)} = {{1 \over n} e^{t/2} + 1 \over {1 \over n} e^t + 1}

and indeed this decays (exponentially fast!) as t \to \infty. So the longer you go without hearing the news of someone’s exam outcome, the more likely it is that it’s bad news.

Of course reality is more complicated. This doesn’t take into account the structure of the social network. For example, for any given person there’s probably a ring of people at a certain social distance who would hear very quickly if they passed but not if they failed. If you know that you are in that relationship to a person, you can probably guess with pretty near certainty that they failed if you don’t hear right away.

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