Distribution of ages of 3 children in a family
Please consider Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 30, p 88.
- A family has 3 children, creatively named A,B, and C.
(a) Discuss intuitively (but clearly) whether the event “A is older than B” is independent of the event “A is older than C”.
(b) Find the probability that A is older than B, given that A is older than C.
Let X, Y, Z be the ages of A, B, C respectively. I assumed X, Y, Z are i.i.d. random variables with each following uniform distribution U(0,18). (18 can be changed to any constant, obviously it doesn't affect the probabilities in question here.)
By this logic the event (X>Y) and (X>Z) are independent events, each having probability 0.5. $P(X>[Y \cap Z])$ is just $P((X>Y)∩(X>Z))=P(X>Y)\times P(X>Z)=1/4$.
However the author's solution takes a discrete approach, assuming the 6 possible birth orders to be equally likely events with probability 1/6 each. If we make this assumption we can see that A is older than B means A is not the youngest, which takes out 2 of these 6 possibilities. Making the event "A is the oldest" 50% likely instead of 1/3 likely originally. Hence they're not independent. Given (A>B) increases the probability of (A>C).
This is completely logical and understandable, but my question is why are the two approaches not producing the same conclusions? What am I missing? I can spot one obvious difference that the birth order approach ignores the possibilities X=Y, Y=Z etc., but even under the uniform distribution assumption these events have zero probability (by standard continuous distribution properties).
Solution 1:
The ages of $A,B,$ and $C$ may be independent†, but the events of pairwise orders are not. If you are told that $A$ is one of the two oldest children (because $A$ is older than $C$) it should raise your anticipation that $A$ is also older than $B$.
(† though, actually, they are not independent if they have the same mother; but that is beside the point.)
Assuming that even in the case of twins and triplets the children will still be described as older and younger, there are $6$ equi-plausible arrangements of age. $$\begin{array}{|c|c|} & C<A & B<A & \text{Both} \\ A<B<C \\A<C<B\\B<A<C & & \checkmark \\B<C<A & \checkmark& \checkmark& \checkmark\\C<A<B & \checkmark\\C<B<A & \checkmark& \checkmark& \checkmark\end{array}$$
The events $C<A$ and $B<A$ are not independent because they are codependent on the age of $A$.
Solution 2:
The problem is that $P((X\gt Y)\cap(X\gt Z))=P(X\gt Y)P(X\gt Z)=\frac14$ isn't right. The probability that $X$ is greater than both $Y$ and $Z$ is just $\frac13$, the probability of one of three equivalent children to be the eldest. Thus the events aren't independent.
Solution 3:
Let the children $A, B, C$ have ages $X, Y, Z$ respectively. For each child pair, $\{A,B\}, \{A,C\}, \& \{B,C\}$ there are three age possibilities. For the pair $\{A,B\}$, the possibilities are as follows:$$1.\ \ \ \ \ X \gt Y$$ $$2.\ \ \ \ \ X = Y$$ $$3.\ \ \ \ \ X<Y$$ Note that the distribution holds for the other children pairs. Now to answer your first question
(a) Discuss intuitively (but clearly) whether the event “A is olderthan B” is independent of the event “A is older than C”.
Consider the event
- $Y \gt Z$ and $X \lt Z$
In this case, $X \gt Y$ is not feasible because of the relationship in 1. If instead the relationship was
- $Y \gt Z$ and $X \gt Z$
Then, we know for a fact that $X \gt Y$.
From 1 and 2, we can conclude that the event “A is older than B” is NOT INDEPENDENT of the event “A is older than C”
(b) Find the probability that A is older than B, given that A is older than C.
Given that A is older than B implies we do not have scenario 1 above. In this situation, there are 3 possibilities:
(i). $X \gt Y$
(ii). $X \lt Y$
(iii). $X=Y$
And each of them are just as likely with a probability of $\frac{1}{3}$.
What you calculated when you wrote $P((X>Y)∩(X>Z))=P(X>Y)*P(X>Z)=1/4$ was "The probability that $X \gt Y$ AND $X>Z$". This is wrong because $X>Z$ has been given and should not be computed again.
Moreover, you did not consider the possibility of the children having the same age. (Whatever happened to twins, triplets and step siblings?)
Note also that the pair distribution scheme I used here doesn't rule out the possibility of triplets. Bonus points if you tell me what that probability is.