Expected Value and Arrival Times
Joe, Austin, and Garren are studying for their final Probability exam and agree to meet at 8.00pm. They will not arrive before the specified meeting time. Their lateness (the amount of time by which they are late) are independent and exponentially distributed with an average lateness of 120 seconds. Suppose are the lateness in seconds of the first, second, and third people to arrive, respectively. Find .
Assumptions : are defined according to the order they arrive. They are not defined as Joe's, Austin's, and Garren's in that order. For example, if it's Garren that arrives first, then is Garren's lateness. The answer is expressed in seconds.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Given that
T i ∼ 1 2 0 1 e − 1 2 0 t i , i ∈ { 1 , 2 , 3 }
and
P ( T i ∣ T j ) = P ( T i ) , i , j ∈ { 1 , 2 , 3 } , i = j
we have that
E ( T 1 ∣ T 3 = 1 2 0 ) = ∫ 0 + ∞ t 1 f T 1 ∣ T 3 ( t 1 ∣ t 3 ) d t 1 = ∫ 0 1 2 0 1 2 0 t 1 e − 1 2 0 t 1 d t 1
since f T 1 ∣ T 3 ( t 1 ∣ t 3 ) = f T 1 ( t 1 ) . Let t 1 = t . Solving the integral via integration by parts we have
E ( T 1 ∣ T 3 = 1 2 0 ) = ∫ 0 1 2 0 1 2 0 t e − 1 2 0 t d t = [ − t ⋅ e − 1 2 0 t ] 0 1 2 0 + ∫ 0 1 2 0 e − 1 2 0 t d t = − e 1 2 0 + [ − 1 2 0 e − 1 2 0 t ] 0 1 2 0 = 1 2 0 − e 2 4 0 = 3 1 . 7 0 8