If I recall, this example forms a graph with (earthquake, burglary) being the two parents of a child (alarm), and the statement is that the two parents are conditionally independent only when child is unknown.

The counter-example is, the alarm goes on (child is known), and there was no burglar (one parent is known), therefore we conclude that the alarm must be caused by earthquake.

However I think the statement should be modified to say the two parents are conditionally independent only when child is unknown and the known parent is in a particular state.

I think the example requires that the burglar variable be known and false?

The question: What about the case where the burglar is known and true? In this case, the alarm goes on, and we know that there has been a burglar. I believe in this case the (earthquake, burglar) are still conditionally independent. Is that correct? Knowing that there was a burglar should not have any effect on when an earthquake happens!

  • $\begingroup$ You know an earthquake didn't happen, because the probability of the alarm going off at exactly the same time for an earthquake and a burglary is zero. Therefore, either one or the other, but not both, happened. $\endgroup$
    – jbowman
    Commented Aug 21, 2022 at 23:23
  • $\begingroup$ @jbowman That leads you towards these events having probability $0$, which rather changes statements about independence. $\endgroup$
    – Henry
    Commented Aug 22, 2022 at 0:56
  • $\begingroup$ @Henry - if I think of, say, an exponential time between earthquakes and another exponential time between burglaries, then, when the alarm goes off, we know one occurred, but we also know that only one occurred because, as you observe, the probability of the other occurring at exactly the same time is zero. If we don't know whether the alarm has gone off, then whether or not a burglary has occurred before "now" is independent of whether an earthquake has occurred before "now"... what am I overlooking? $\endgroup$
    – jbowman
    Commented Aug 22, 2022 at 1:25
  • $\begingroup$ @jbowman - I think you should choose a description which allows the possibility of both a burglary and an earthquake so that consideration of conditional or other independence is meaningful. Restricting to an instant in the conditional case but an interval in the unconditional case seems to be avoiding the question. For example in the conditional case you could know that the alarm has gone off during some time interval, and find that whether or not a burglary has occurred in that interval is not independent of whether an earthquake has occurred in that interval. $\endgroup$
    – Henry
    Commented Aug 22, 2022 at 1:46

1 Answer 1


Let's suppose a burglary is event $B$ and an earthquake is event $Q$, each with probability strictly between $0$ and $1$.

Let's also suppose that if and only if either (or both) happen then the alarm goes off as event $A$. You have $A=B \cup Q$. Clearly $\mathbb P(A) = \mathbb P(B \cup Q)$ is a higher probability than either $\mathbb P(B)$ or $\mathbb P(Q)$, but still less than $1$.

Burglaries and earthquakes are independent, i.e. $$\mathbb P(B \cap Q) = \mathbb P(B)\,\mathbb P(Q).$$

Conditioned on the alarm going off, $\mathbb P(B \mid A) = \frac{\mathbb P(B \cap A)}{\mathbb P(A)}= \frac{\mathbb P(B)}{\mathbb P(A)}$ and similarly $\mathbb P(Q \mid A) = \frac{\mathbb P(Q)}{\mathbb P(A)}$ and $\mathbb P(B \cap Q \mid A) = \frac{\mathbb P(B \cap Q)}{\mathbb P(A)}= \frac{\mathbb P(B)\,\mathbb P(Q)}{\mathbb P(A)}$. So

$$\mathbb P(B \cap Q\mid A) = \mathbb P(A)\, \mathbb P(B\mid A)\,\mathbb P(Q \mid A) \not =\mathbb P(B\mid A)\,\mathbb P(Q \mid A)$$

and you do not have conditional independence given the alarm going off.

I think your second question about conditional independence given the alarm going off and there being a burglary is correct but only in a trivial sense, because when you have two events and one has probability $1$ then the other is always independent of it. You get

$$\mathbb P(B \cap Q\mid A \cap B) = \mathbb P(B \cap Q\mid B) = \mathbb P(Q\mid B) \\ = \mathbb P(B\mid B) \, \mathbb P(Q\mid B) = \mathbb P(B\mid A \cap B) \, \mathbb P(Q\mid A \cap B)$$

using $\mathbb P(B\mid B)=1$ and $A \cap B=B$.


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