5Continuous random variables
IA Probability
5.4 Geometric probability
Often, when doing probability problems that involve geometry, we can visualize
the outcomes with the aid of a picture.
Example. Two points
X
and
Y
are chosen independently on a line segment of
length
L
. What is the probability that
|X −Y | ≤ ℓ
? By “at random”, we mean
f(x, y) =
1
L
2
,
since each of X and Y have pdf 1/L.
We can visualize this on a graph:
A
ℓ
L
ℓ
L
Here the two axes are the values of
X
and
Y
, and
A
is the permitted region.
The total area of the white part is simply the area of a square with length
L −ℓ
.
So the area of A is L
2
− (L − ℓ)
2
= 2Lℓ − ℓ
2
. So the desired probability is
Z
A
f(x, y) dx dy =
2Lℓ − ℓ
2
L
2
.
Example (Bertrand’s paradox). Suppose we draw a random chord in a circle.
What is the probability that the length of the chord is greater than the length
of the side of an inscribed equilateral triangle?
There are three ways of “drawing a random chord”.
(i)
We randomly pick two end points over the circumference independently.
Now draw the inscribed triangle with the vertex at one end point. For the
length of the chord to be longer than a side of the triangle, the other end
point must between the two other vertices of the triangle. This happens
with probability 1/3.
(ii)
wlog the chord is horizontal, on the lower side of the circle. The mid-point is
uniformly distributed along the middle (dashed) line. Then the probability
of getting a long line is 1/2.
(iii)
The mid point of the chord is distributed uniformly across the circle. Then
you get a long line if and only if the mid-point lies in the smaller circle
shown below. This occurs with probability 1/4.
We get different answers for different notions of “random”! This is why when
we say “randomly”, we should be explicit in what we mean by that.
Example (Buffon’s needle). A needle of length
ℓ
is tossed at random onto a
floor marked with parallel lines a distance
L
apart, where
ℓ ≤ L
. Let
A
be the
event that the needle intersects a line. What is P(A)?
ℓ
X
θ
L
Suppose that X ∼ U[0, L] and θ ∼ U [0, π]. Then
f(x, θ) =
1
Lπ
.
This touches a line iff X ≤ ℓ sin θ. Hence
P(A) =
Z
π
θ=0
ℓ sin θ
L
| {z }
P(X≤ℓ sin θ)
1
π
dθ =
2ℓ
πL
.
Since the answer involves
π
, we can estimate
π
by conducting repeated exper-
iments! Suppose we have
N
hits out of
n
tosses. Then an estimator for
p
is
ˆp =
N
n
. Hence
ˆπ =
2ℓ
ˆpL
.
We will later find out that this is a really inefficient way of estimating
π
(as you
could have guessed).