5.1 Exercises
Exercise 99. Let 0 X 1 be a continuous random variable such that fX (x) = c for 0 x 1. Find c, E[X],
V ar[X], M ed[X] and E[etX ] and P (0.25 X 0.75).
Solution: Using denition 26,
fX (x)dx
= 1. Observe that:
1
cdx = cx|1 = c
0
fX (x)dx =
0
Observe that, in the special case in which k = 1, the Gamma distribution corresponds to the Exponential
distribution.
Lemma 58. For every a > 0, E[X a ] =
(k+a)a
(k) .
If X Gamma(k, ), then E[X] = k and V ar[X] = k2 .
Proof.
x
1
xk1 e dx
k
(k)
0
x
1
=
xk+
5.8 Beta Distribution
Denition 34. We say that X follows Beta distribution with parameters (, ) and denote by X Beta(, ) if
the density of X is
fX (x) =
(+)
()()
x1 (1 x)1
0
, if 0 < x < 1
, otherwise
Lemma 59. The density in Denition 34 is a valid pdf.
Proof.
P (X > t + s X > t)
P (X > t)
P (X > t + s)
=
P (X > t)
P (X > t + s|X > t) =
=
e
Conditional Probability (Denition 12)
t+s
Exponencial cdf (Lemma 53)
t
e
s
= e = P (X > s)
Exponencial cdf (Lemma 53)
5.5 Exercises
Exercise 109. Let X Exp(1). Find
Proof. Let Z N (0, 1).
etz fZ (z)dz
MZ (t) = E[etZ ] =
z2
1
etz e 2 dz
2
z 2 +2tz+t2
t2
1
2
e
=e2
dz
2
(z+t)2
t2
1
e 2 dz
=e2
2
=
t2
t2
=e2 1=e2
(Denition 35)
If X N (, 2 ), then
E[etX ] = E et(
X
+
= et E et
)
X
= et M X (t)
t
= e MZ (t)
= et e
(t)2
2
5.2 Uniform Distribution
Denition 30. We say that X follows the uniform distribution on the interval (a, b) and denote by X U(a, b)
if the density of X is such that:
fX (x) =
1
ba
, if a x b
0
, otherwise
Lemma 51. If X U (a, b), then for every set (c, d)
5.10 Normal Distribution
Denition 35. We say that X follows the Normal distribution with parameters (, 2 ) and denote by X N(, 2 )
if the density of X is
(x)2
1
fX (x) = e 22
2
If = 0, and = 1, we say that X follows the standard normal distribution.
Deni
5.4 Exponential Distribution
Denition 31. We say that a random variable X follows an exponential distribution with parameter and denote
by X Exp() if the density of X is:
x
1 e
fX (x) =
0
, if x 0
, otherwise
x
x
Lemma 53. If X Exp(), then FX (x) = 1 e .
5.6 Gamma Distribution
Denition 32. : R+ R+ is called the Gamma function and is such that:
(z) =
tz1 et dt
0
Lemma 56. The function satises the following properties:
1. For a 1, (a) = (a 1)(a 1)
2. If n N, (n) = (n 1)!
Proof.
1.
(a) =
ta1 et dt
0
= ta1 et