1. Oct 18, 2011

### faen

In the notes we had the following example of a problem, with solution:

Example. Shooting stars. If you watch the sky form a peak of "Kékes-tet˝o" (highest
peak in Hungary) around midnight in August, you will see shooting-stars. Assume
that the amount of time between two shooting-stars, which is a random quantity, is 10
minutes on the average. If we watch the sky for 15 minutes, then how much is the
probability that we see exactly 2 shooting-stars?

Remark. It is not difficult to figure out wrong arguments to get the wrong answers:
0.5 or 0.75. The reader will enjoy to find out these wrong arguments.

Solution. The number of shooting-stars visible during 15 minutes is a random variable.
The following facts are obvious:
1. the number of meteors in the space is vary large, and
2. for each meteor the probability that it causes a shooting-star during our 10 minute
is small, and
3. the meteors cause a shooting-stars independently of each other,
These facts guarantee that the number of shooting-stars visible during 15 minutes follows
a Poisson distribution. The parameter  of this distribution is equal to the expected
value of the number of shooting-stars visible during 15 minutes. Since the amount of
time between two shooting-stars is 10 minutes in the average, the average number of
shooting-stars visible during 15 minutes is 1.5. Thus,  = 1:5. This is why the answer
to the question is
P(We see exactly 2 shooting-stars) =

So my question is, why is the expected value of visible shooting stars predicted to be 1.5 in 15 minutes? I would think it's 2.5. Here is a diagram showing why:

0*--------10*---15---20*

the * indicate the shooting stars, and numbers is the time. So it shows that the time between two stars is 10 minutes. Also at 15 minutes, we observe at average 2.5 stars... So how am i wrong, since the solution says its 1.5 stars at average during 15 minutes?

2. Oct 18, 2011

### HallsofIvy

You are counting the shooting star at t= 0 and (half of) the shooting star at 20 minutes. You cannot count both. (Imagine moving your slightly one way or the other- one of the shooting stars would move out ot the interval.)

3. Oct 18, 2011

### Stephen Tashi

The way to explain this would be to actually do the mathematics that establishes the relationship between the parmeter of a Poission distribution and the parameter of its associated exponential "waiting time distribution". As the previous post indicates, trying to reason about it by visualizing "the average case" isn't reliable.

4. Oct 18, 2011

### faen

I think i understand it now, thanks a lot to both of you :)