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Suppose train on line A arrives in time uniformly distributed between 0 and 4mins, train on line B arrives in time uniformly distributed between 0 and 6 mins, and furthermore the time interval between A and B arrival is uniformly distributed between 0 and 4.

a. What is the probability train on line A arrives first

b. What is the probability you wait less than 2 mins for one of the trains to arrive.

Here's how I approached this question,

Let $A : $ arrival time for train on line A $\sim U(0,4)$, $B : $ arrival time for train on line B $\sim U(0,6)$

and it is given that $A-B \sim U(-4,4)$

a. $P(A<B) = P(A-B<0) = \frac{1}{2}$

b. This is where I'm stuck, $P(\mathrm{min}(A,B) \leq 2) = 1 - P(A>2, B>2)$

But I'm not sure how I'd go about computing $P(A>2, B>2)$ any help or hint is appreciated, thanks.

Stefan
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Harsh
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    I believe there cannot exist any distribution with the properties you describe. – whuber Feb 23 '24 at 17:01
  • Interesting, can you elaborate more, what makes you believe that ? – Harsh Feb 23 '24 at 17:28
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    Hint: Write $B=(B-A)+A$ and check whether or not the marginal distribution of $B$ can be uniform. – Xi'an Feb 23 '24 at 18:24
  • Wait, I am not the most proficient at this, but I thought the underlying question was to find a joint distribution and a conditional distribution such that $P(A) = \frac{P(A,B)}{P(A|B)}$

    Am I understanding something wrong? I thought that the expression $B = (B - A) + A$ would not hold only under independence. but if B depends on A, is there a way it can hold?

     – Oscar Flores Feb 23 '24 at 20:54
  • I actually thought that there would be infinitely many possibilities, but still I have no clue on how to find one of those that satisfy the requirements. – Oscar Flores Feb 23 '24 at 21:07
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    B = (B-A) + A of course always holds, and @whuber is right you cannot have distributions satisfying these conditions, and it just seems to me that problem statement isn't stated very well – Harsh Feb 24 '24 at 07:45
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    In a linked question the easiest route is to say you should have $E[A-B]=E[A]-E[B]$ by linearity of expectation but here $0 \not = 2-3$ – Henry Mar 03 '24 at 22:45

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