Examples #4: Probability distributions, etc.
1. A disk of radius R is dropped on a at sheet that has parallel lines separated by l > 2R drawn on it.
What is the probability that the disk will land on the sheet not touching a line?
• Let us position our view of the sheet and the landed disk such that the drawn lines stretch
horizontally and appear one above the other. Let x0 be the distance between the disk center
and the nearest line above it. Since the lines are the distance l apart, x0 can take a random
value between 0 and l. It is actually more convenient to work with the shifted random variable
x = x0 − l/2 ∈ (−l/2, l/2), which measures the oriented vertical distance from the disk center to
the nearest line. Without any bias, x is distributed uniformly on the interval (−l/2, l/2). In order
for the disk to touch the nearest line, |x| must be smaller than R - this event corresponds to the
range of values x ∈ (−R, R) whose size is 2R. The size of the full possible range of x values is 2l,
so the probability of the disk touching a line is:
P0 =
R
2R
=
2l
l
We are interested in the probability of not touching the line:
P = 1 − P0 =
l−R
l
2. A screen-saver plots N small circles of xed radius R at random positions on the computer screen
of width Lx and height Ly , making sure only that all of them completely t on the screen. Assume
Lx Ly N 2 πR2 . What is the probability that no two circles will overlap?
• The rst circle is placed randomly somewhere on the screen. The area available for the position of
it's center is (Lx − 2R)(Ly − 2R), since the circle must completely t on the screen. If the second
circle is not to overlap with it, the area available for the placement of its center is (Lx − 2R)(Ly −
2R)−π(2R)2 . Note that the minimum distance between the centers of two non-overlapping circles
is 2R. Therefore, the unbiased probability that the second circle will not overlap with the rst
one is:
p2 =
(Lx − 2R)(Ly − 2R) − π(2R)2
π(2R)2
4πR2
=1−
≈1−
(Lx − 2R)(Ly − 2R)
(Lx − 2R)(Ly − 2R)
Lx Ly
Note that this is not absolutely correct. If the rst circle is too near to a corner or an edge of the
screen, then our naive judgment of the available area for the second circle needs to be corrected
(more area is available for no overlap). However, the probability of the rst circle being so close
to the screen border is already small, and corrects p2 in a conditional sense by an amount much
smaller than R2 /Lx Ly , which is already small. We will neglect these corrections. The third circle
then has the area (Lx − 2R)(Ly − 2R) − 2π(2R)2 available for placement without overlap, if we
similarly neglect the probability that the rst two circles will be placed too close to each other.
The corresponding third-circle probability is:
p3 ≈ 1 − 2
4πR2
Lx Ly
The placements of individual circles are independent random events, so the nal probability that
all N circles will not overlap is:
P = p2 p3 p4 · · · pN ≈
N
−1 Y
k=1
4πR2
1−k
Lx Ly
≈ 1−
N
−1
X
k=1
1
4πR2
+O
Lx Ly
"
4πR2
Lx Ly
2 #
≈ 1−
N (N − 1) 4πR2
2
Lx Ly
Clearly, there could be problems if N is large: a new circle could be hard or impossible to t
without overlap in the insucient space left between circles, even when there is ample total area
for their placement (which is the only thing that this analysis considers). However, again, the
corrections that overcome this problem are negligible in the assumed limit Lx Ly N 2 πR2 .
3. What is the probability that two completely random real numbers between 0 and 10 will have a product
smaller than 50?
• Let the two random numbers be x and y . They are selected from the uniform distribution on the
interval (0, 10), so that their joint PDF is constant and equal to:
f (x, y) =
1
102
in this domain. The curve xy = 50 plotted in the xy coordinate system is a hyperbola that goes
through the points (x, y) ∈ {(5, 10), (10, 5)} on the boundary of the square domain 0 ≤ x, y ≤ 10.
All points below this curve satisfy the requirement that xy < 50. Therefore, the sought probability
is the area A below the curve (but bounded by the 10 × 10 square) divided by 102 :
ˆ10
A = 5 × 10 +
dx
50
= 50 + 50 log
x
10
5
= 50(1 + log 2)
x=5
P =
1
A
= (1 + log 2) = 0.85
102
2
4. If a quantity x is a random variable with PDF f (x), derive the general formula for the PDF F (y) of
the random variable y related to x by a given function g(x), i.e. y = g(x). Then, calculate specically:
(a) The PDF F (x2 ) if x is uniformly distributed on the interval (0, 1).
(b) The PDF F (x2 ) if x has the Gaussian distribution N (0, 1).
• Given the PDF f (x), the probability that x will take a value in the innitesimal interval (x0 , x0 +
dx) is f (x0 )dx. That event is the same as the event in which y takes a value in the interval
(y0 , y0 + dy), where y0 = g(x0 ) and dy = dg(x0 ), so it should have the same probability:
f (x0 )dx = F (y0 )dy
dx f (x0 ) F (y0 ) = f (x0 ) × = 0
dy x=x0
|g (x0 )| x0 =g−1 (y0 )
We took the derivative with the modulus (absolute value), because we want formally to have
dx, dy > 0 express merely the size of the intervals (without a sign). In some cases, there are
multiple values of x0 that correspond to the same value of y0 . For example, if y = x2 and x is
allowed to take both positive and negative values, than specifying y = y0 is equivalent to one of
√
√
two events, x0 = + y0 and x0 =√− y0 . The probability that y = y0 happens is equivalent to
the probability that either x = ± x0 happens. Most generally, if the equation y = y0 = f (x0 )
has multiple solutions x0 ∈ X(y0 ), were X is the set of such solutions, then:
F (y0 ) =
X
x0 ∈X(y0 )
2
f (x0 )
|g 0 (x0 )|
(a) Here,
, 0≤x≤1
, otherwise
1
0
f (x) =
√
y > 0, and g 0 (x) = 2x = 2 y :
, 0≤y≤1
, otherwise
Therefore, for y = g(x) = x2 we have a unique solution x =
√
f ( y)
F (y) = √ =
2 y
Check normalization:
ˆ1
ˆ∞
dy F (y) =
−∞
(b) Here,
1
√
2 y
0
√
1
1
√ 1
dy √ = × 2 y = 1
2 y
2
0
0
2
1
f (x) = √ e−x /2
2π
(∀x)
√
√
For y = g(x) = x2 > 0 we now have two acceptable solutions x = ± y , and |g 0 (x)| = |2x| = 2 y .
Therefore:
F (y) =
√
√
f ( y) f (− y)
1
1
1
= √
e−y/2 + √
e−y/2 = √
e−y/2
√ +
√
2 y
2 y
2 2πy
2 2πy
2πy
(∀y > 0)
Check normalization:
ˆ∞
ˆ∞
dy F (y) =
−∞
0
1
2
dy √
e−y/2 = √
2πy
2π
ˆ∞
√ 2
1
√
d( y)e−( y) /2 = √
2π
0
ˆ∞
−x2 /2
dx e
−∞
ˆ∞
=
dx f (x) = 1
−∞
5. If x and y are uniformly distributed random numbers on the interval (0, 1), nd the probability distribution f (z) for the random variable z = xy .
• The joint PDF for x and y is f2 (x, y) = 1 in the domain 0 ≤ x, y ≤ 1:
1 , 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
f2 (x, y) =
0 ,
otherwise
The line of points xy = z for a xed z is a hyperbola. The bounds on z are (0, 1). The event
of z being in the innitesimal interval z ∈ (z0 , z0 + dz) is supposed to have the probability
F (z0 )dz . However, there are many combinations of (x, y) that correspond to this event and their
probabilities must be added to give F (z0 )dz . These (x, y) pairs come from the area dA between
the two hyperbolas at xy = z0 + dz and xy = z0 , bounded also by 0 ≤ x, y ≤ 1:
ˆ
ˆ
F (z0 )dz =
dxdy f2 (x, y) =
dA
dxdy
dA
We can calculate dA = A(z0 + dz) − A(z0 ) = A0 (z0 )dz ,
where A(z) is the area inside the box 0 ≤ x, y ≤ 1
under the hyperbola xy = z :
ˆ1
A(z) = z +
z
3
z
dx = z + z log
x
1
= z − z log(z)
z
Therefore,
F (z0 ) = A0 (z0 ) =
i
dh
z − z log z
= 1 − (1 + log z0 ) = − log(z0 )
dz
z=z0
• Another way to solve this problem involves the Dirac delta function, which is dened by:
0
∞
δ(x) =
x 6= 0
x=0
,
,
ˆ∞
,
dx δ(x) = 1
−∞
This function vanishes for all values of its argument except zero. The innite value of the function
at zero argument is not an arbitrary innity, but a particular innity that integrates out to 1.
In other words, the Dirac delta function is an innitely sharp peak with an area beneath it still
equal to 1. This property gives rise to the following easy-to-prove theorem:
ˆ∞
dx δ(x − x0 )f (x) = f (x0 )
−∞
for any function f (x) and point x0 . Also, we can derive the rule for changing the argument of the
Dirac delta function:
ˆ∞
−∞
ˆ∞
ˆ∞
dx dx dx dx δ f (x) − f (x0 ) =
df δ f (x) − f (x0 ) = =
dx δ(x − x0 )
df
df x=x0
df
−∞
−∞
δ(x − x )
δ(x − x0 )
0 ≡
δ f (x) − f (x0 ) =
df |f 0 (x)|
dx ⇒
Actually, this theorem needs to be patched if there are multiple solutions to f (x) = f . Now, we
may write and calculate the following expression for the PDF of z = xy :
ˆ1
f (z) =
ˆ1
dy f2 (x, y)δ(xy − z)
dx
0
0
This is sensible because the delta function constrains the product xy to the given value of z , and
automatically normalizes f (z):
ˆ1
ˆ1
dz f (z) =
0
ˆ1
ˆ1
dy f2 (x, y) ×
dx
0
0
ˆ1
ˆ1
ˆ1
dz δ(xy − z) =
0
ˆ1
dx
0
dy f2 (x, y) = 1
0
Therefore,
f (z)
=
dy f2 (x, y)δ(xy − z) =
dx
0
0
ˆ1
ˆ1
=
dx
0
dx
ˆ1
dx
0
dy f2 (x, y)
δ(y − z/x)
=
x
0
ˆ1
=
ˆ1
0
ˆ1
0
1
= − log(z)
x
z
4
δ(y − z/x)
dy f2 (x, y) d(xy) dy 1 z
dx f2 x,
x
x