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Lecture 20
Agenda
1. Uniform Distribution
Since we have setup a general framework for continuous random variables,
let’s look at particular examples. These particular examples are not being
studied because they are mathematically pleasing, though that’s one reason;
these examples are of immense practical value and we use them everyday.
Let’s start with the easiest of them.
Uniform Distribution
Consider an experiment which consists of choosing a point from the interval
[a, b] such that “all points are equally likely to be chosen”. Our random
variable X is the point chosen. So if x1 and x2 are two points in [a, b] we
are not saying that P (X = x1 ) and P (X = x2 ) are equal, because obviously
they are both 0 and everybody knows that. We know that if the pdf at some
point x is high, then x lies in a high probability region and if pdf is small
then x lies in a low probability region. So here we mean that, inside [a, b]
there is no high or low probability region. So the pdf should also be the same
everywhere inside [a, b].
In other words,
if x ∈ [a, b]
otherwise
fX (x) = c
= 0
where c is some unknown constant. How do we find c ? Well, since fX is a
pdf,
Z ∞
fX (x)dx = 1
−∞
Z
⇒
b
c dx = 1
a
⇒
c =
1
1
b−a
In conclusion the density of X is
1
fX (x) =
if x ∈ [a, b]
b−a
= 0
otherwise
This density is known as the “UNIFORM DENSITY ON [a, b]” and
X is known as a “UNIFORM RANDOM VARIABLE on [a, b]”. It is
written as X ∼ U [a, b]. Now, let’s derive some properties of X.
Distribution Function
FX (x) = 0
Z
=
x
a
= 1
if x < a
1
x−a
dy =
b−a
b−a
if x > b
if a ≤ x ≤ b
Hence for a < c < d < b,
d−a c−a
d−c
−
=
b−a b−a
b−a
Notice that this depends on length of the interval [c, d] only. This property
is a characteristic of the uniform distribution.
Caution however this result holds only if [c, d] ⊂ [a, b].
P (X ∈ [c, d]) = FX (d) − FX (c) =
Mean and Variance
Z
∞
E(X) =
xfX (x)dx
−∞
Z b
=
=
=
=
=
1
dx
b−a
a
Z b
1
xdx
b−a a
1 b 2 − a2
b−a 2
1 (b + a)(b − a)
b−a
2
(b + a)
2
x
2
This result is intuitive. Since all points in [a, b] are equally likely, the
mean is the middle of the interval. Let’s find the variance,
V (X) = E[(X − E(X))2 ]
2
Z b
a+b
1
=
x−
dx
2
b−a
a
Now put z =
x− a+b
2
b−a
hence dx = (b − a)dz,
V (X) = (b − a)
2
Z
1
2
z 2 dz
− 12
= (b − a)
2
z3
3
12
− 21
(b − a)2
12
=
Example
Every 15 minutes the bus of
at a bus stop let X be the
X ∼ U [0, 15].
What’s the probability that
Find the time that there is
come by that time.
route 17 leaves RTS Downtown. If I arrive
number of minutes I have to wait.
a bus will come in 5 mins ?
only 10% chance that the bus will not
5−0
15 − 0
1
=
3
P (X < 5) =
For the second question, if a is the answer then,
P (X > a)
⇒ P (X ≤ a)
a
⇒
15
⇒
a
= 0.1
= 0.9
= 0.9
= 15 ∗ 0.9 = 13.5
3
Homework ::
1. Let X ∼ U nif orm[a, b] and for any two constants m > 0 and c
define Y = mX + c. Prove Y ∼ U nif orm(ma + c, mb + c).
2. Let X be a discrete random variable with the following distribution.
X
1.3 2.1 5.7
P (X = x) 0.25 0.3 0.2
8.9
0.25
Now suppose U ∼ U nif orm(0, 1), and based on U define Y as
follows
if 0 ≤ U < 0.25
if 0.25 ≤ U < 0.55
if 0.55 ≤ U < 0.75
if 0.75 ≤ U < 1
Y = 1.3
= 2.1
= 5.7
= 8.9
Prove that Y has same distribution as X.
3. 4.43, 4.50
4