Download I) Show that the following properties of a Cumulative Distribution

H OMEWORK 1.3
I) Show that the following properties of a Cumulative Distribution Function (CDF) follow
from the definition of the CDF and the properties of probability:
(1)
(2)
(3)
(4)
(5)
F (−∞) = 0 and F (+∞) = 1.
F is a non-decreasing function, i.e. if x1 ≤ x2 , then F (x1 ) ≤ F (x2 ).
If F (x0 ) = 0, then F (x) = 0 for every x ≤ x0 .
P {X > x} = 1 − F (x).
P ({x1 < X ≤ x2 }) = F (x2 ) − F (x1 ).
Proof. Property 1: By definition, FX (a) = P (A) where A = {X ≤ a}. Then F (= +∞) =
P (A), with A = {X ≤ +∞}. This event is the certain event, since A = {X ≤ +∞} =
{X ∈ R} = S. Therefore F (= +∞) = P (S) = 1.
Similarly, Then F (= −∞) = P (A), with A = {X ≤ −∞}. This event is the impossible
event, RV X is a real variable, so F (= −∞) = P (∅) = 0.
Proof. Property 2:
This is a consequence of the fact that probability is a nondecreasingf function, i.e. if A ⊆ B
then P (A) ≤
P (B). Indeed, recall that if A ⊆ B then the set B can be partitioned into B =
A + B ∩ Ā , so, using the additivity property of probability, P (B) = P (A) + P B ∩ Ā ≥
P (A).
If x1 ≤ x2 , then the corresponding events A = {X ≤ x1 } and B = {X ≤ x2 } have the
property that A ⊂ B. Therefore P ({X ≤ x1 }) ≤ P ({X ≤ x2 }), which by the definition of
a CDF means F (x1 ) ≤ F (x2 ).
Proof. Property 3:
This property is related to the nondecreasing property of the CDF: going backwards, to
the left, toward −∞, the CDF can only decrease or stay constant. If F (x0 ) = 0, to the left
of x0 it can only stay constant (=0) since it cannot take negative values.
For x ≤ x0 the events A = {X ≤ x} and B = {X ≤ x0 } have the property that A ⊂ B,
therefore
0 ≤ F (x) = P ({X ≤ x}) ≤ P ({X ≤ x0 }) = F (x0 ) = 0
therefore F (x) = 0.
Proof. Property 4:
This is a consequence of the properties of probability of complementary events: recall
that P Ā = 1 − P (A). SetA = {X ≤ x}; then its complement (the set of all elementary
events
which are not in A) is Ā = {X > x}, i.e. {X ≤ x} + {X > x} = S = R. As above,
P Ā = 1 − P (A), but with from the definition of the CDF, P Ā = P ({X > x}) =
1 − P ({X ≤ x}) = 1 − F (x).
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Proof. Property 5:
This property shows how to use the CDF to calculate the probability that RV X returns a
value in a given interval [x1 , x2 ]. The CDF is defined on semi-infinite intervals of the form
(−∞, x1 ] and (−∞, x2 ]. Note that the events A = {X ∈ (−∞, x1 ]}, B = {X ∈ (−∞, x2 ]}
and C = {X ∈ (x1 , x2 ]} are in the relationship
B =A+C
since A and C are disjoint and B = A ∪ C. Therefore P (B) = P (A) + P (C), or P (C) =
P (B) − P (A). Hence the property.
II) The number xu for which the CDF F (x) has the value F (xu ) = u is called the udF
is even, that is f (−x) = f (x),
percentile of F . Show that if the derivative f (x) =
dx
then
(1) F (−x) = 1 − F (x)
(2) x1−u = −xu .
(3) Draw a sketch of the CDF and its derivative. Explain on the plot the above relations.
dF
, one can write
dx
Z x
Z
F (x) =
f (s) ds, and, F (−x) =
Proof. 1) Since f (x) =
−∞
−x
f (s) ds.
−∞
Changing the variable in the last integral to t = −s, with
s = −t;
ds = −dt;
s = −x 7→ t = x;
and using the fact that f (−t) = f (t) yields
Z −x
Z x
Z ∞
Z
f (s) ds =
f (−t)(−dt) =
f (−t) dt =
−∞
∞
x
s = −∞ 7→ t = ∞;
∞
f (t) dt = F (∞) − F (x) = 1 − F (x).
x
2)Using the above definition of the percentile, F (x1−u ) = 1 − u and F (xu ) = u. Obviously,
this means that F (x1−u ) = 1 − F (xu ). From point 1), this implies that x1−u = −xu .
3) A sketch is shown below
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III) Show that if X is an RV that can take values only in a finite interval, a ≤ X ≤ b, then
F (x) = 1 for every x ≥ b and F (x) = 0 for every x ≤ a.
Proof. If a ≤ X ≤ b, from the definition of a CDF, F (a) = 0 and F (b) = 1. Since F is
nondecreasing with values in the interval [0, 1], from x ≥ b follows that
1 = F (b) ≤ F (x) ≤ 1,
therefore F (x) = 1 for any x ≥ b. A similar reasoning works for the second statement: if
x ≤ a,
0 ≤ F (x) ≤ F (a) = 0.
IV) Show that if X and Y are two RVs such that X ≤ Y at every trial, then FY (a) ≤ FX (a)
for every a ∈ R.
Proof. From the hypothesis, X and Y are two RVs such that X ≤ Y at every trial, i.e. for
every elementary event,
X(ζ) ≤ Y (ζ).
Let AX be the set of elementary events ζ such that the observed value X(ζ) ≤ a. In set
algebra notation AX = {ζ : X(ζ) ≤ a} . Denote similarly, AY = {ζ : Y (ζ) ≤ a}.
By the definition of CDF,
FX (a) = P (AX ) ;
FY (a) = P (AY ) .
If AY ⊆ AX , then using the “monotonicity” property of probability, P (AY ) ≤ P (AX ),
and consequently FY (a) ≤ FX (a) .
Indeed, let ζ ∈ AY . This means that Y (ζ) ≤ a, but also since X(ζ) ≤ Y (ζ) ≤ a, so ζ also
belongs to AX , ζ ∈ AX .
V) RV X takes only positive values. Show that if the conditional probability
P ({s < X ≤ s + s1 } |{X > s}) = P ({X ≤ s1 })
for any s, s1 ≥ 0, then the Cumulative Distribution Function (CDF) FX (t) = 1 − e−ct , where
c is a constant.
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Proof. Denote A = {s < X ≤ s + s1 }, B = {X > s} and C = {X ≤ s1 }; with the definition
of conditional probability, the relationship can be written
P (A ∩ B)
= P (C) .
P (B)
Note also that A ⊂= B, which means that A ∩ B = A = {s < X ≤ s + s1 }, so in fact the
relation is
P (A)
= P (C) .
P (B)
Denote by FX (t) the CDF of RV X. For the sake of clarity, let’s rename s1 = ∆s. Then
P (A |B ) =
P (A) = F (s + ∆s) − F (s); P (B) = 1 − F (s); P (C) = F (∆s).
The relation becomes
F (s + ∆s) − F (s)
= F (∆s).
1 − F (s)
valid for any s, ∆s ≥ 0. For a fixed s, let ∆s → ds. Then
dF
ds,
F (∆s) → F (ds) = F (0) +
ds 0
dF
F (s + ∆s) − F (s) → F (s + ds) − F (s) =
ds.
ds
Note also that F (0) = 0.(Why?) The relation then becomes
1
dF
dF
ds =
ds
1 − F (s) ds
ds 0
or equivalently
F0
dF
.
= c, where c =
1−F
ds 0
This is a differential equation; integrating from s = 0:
Z F
Z F
dF
d(1 − F )
dF
= c ds,
= cs; −
= cs.
1−F
1−F
0 1−F
0
Z F
−
d log(1 − F ) = cs; log (1 − F (s)) = −cs,
so
0
F (s) = 1 − e−cs .
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