Finding a distribution function given constraints

[divcurl]

Airline A has commuter flights every 45 minutes from San Fran airport
to Fresno. A passenger who wants to take one of these flights arrives
at the airport at a random time. Suppose that X is the waiting time
for
this passenger; find the distribution function of X. Assume that
seats
are always available for these flights.

My book says the following:
Let the departure time of the last flight before the passenger arrives
be 0.
Then Y, the arrival time of the passenger, is a random number from (0,
45).
The waiting time is X = 45 - Y. We have that 0<= t <= 45,

P(X <= t) = P(45 - Y <= t) = P(Y >= 45 - t) = (45 - (45 - t))/45

Why do we equate P(Y >= 45 - t) to (45 - (45 - t))/45 ?

The reason I ask is this:

P(X >= t) would correspond with 1 - F(t-), hence, we exclude
the given t. So why do we include it above for P(Y >= 45 - t)?

If that is unclear, to state it another way: the above result of
(45 - (45 - t))/45 would seem to me to correspond to:
P(Y > 45 - t) vice P(Y >= 45 - t)

Elaboration welcomed.

[danh...@yahoo.com]

Since Y is a continuous random variable, P{Y=t}=0.

[Ray Vickson]

On Mar 19, 12:23 pm, divcurl <divc...@europe.com> wrote:
> Airline A has commuter flights every 45 minutes from San Fran airport
> to Fresno.  A passenger who wants to take one of these flights arrives
> at the airport at a random time.  Suppose that X is the waiting time
> for
> this passenger; find the distribution function of X.  Assume that
> seats
> are always available for these flights.
>
> My book says the following:
> Let the departure time of the last flight before the passenger arrives
> be 0.
> Then Y, the arrival time of the passenger, is a random number from (0,
> 45).
> The waiting time is X = 45 - Y.  We have that 0<= t <= 45,
>
> P(X <= t) = P(45 - Y <= t) = P(Y >= 45 - t) = (45 - (45 - t))/45

We have Y = 45 - X, so P{Y <= t} = P{45 - X <= t} = P{X >= 45 - t} = 1
- P{X < 45 - t}; but P{X = 45 - t} = 0, so P{X < 45 - t} = P{X <= 45 -
t} = (45 - t)/45 (since P{X <= v} = v/45 for 0 < v < 45). Thus, P{Y <=
t} = 1- (45-t)/45 = t/45. Note that this is the SAME as P{X <= t}, so
X and Y have the same distribution! This makes sense: if X is
uniformly distributed between 0 and 45, then so is 45 - X.

>
> Why do we equate P(Y >= 45 - t) to (45 - (45 - t))/45 ?
>
> The reason I ask is this:
>
> P(X >= t) would correspond with 1 - F(t-),

F(t-) = F(t) for a continuous random variable (because F(t) = P{X <=
t} = P{X = t} + P{X < t} and P{X = t} = 0).

R.G. Vickson

[divcurl]

Thanks for the clear explanation.