Module 16
EQUALITY IN DISTRIBUTION
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Definition 1: Random variables X and Y are said to have the same
d
distribution (written as X = Y ) if they have the same d.f., i.e., if
FX (x) = FY (x), ∀ x ∈ R.
Result 1: Let X and Y be r.v.s with p.m.f.s/p.d.f.s fX (·) and fY (·),
respectively. Then
d
(a) fX (x) = fY (x), ∀ x ∈ R ⇒ X = Y ;
d
(b) For some h > 0, MX (t) = MY (t), ∀ t ∈ (−h, h) ⇒ X = Y ;
d d
(c) X = Y ⇒ h(X ) = h(Y ), for any function h : R → R;
d
(d) X = Y ⇒ E (h(X )) = E (h(Y )), for any function
h : R → R for which the expectation exists.
Proof: Proofs of (a), (c) and (d) are straightforward and hence omitted.
Proof of (b) is based on uniqueness of m.g.f.s.
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Example 1:
For p ∈ (0, 1), let Yp be a r.v. having p.m.f.
n y n−y
y p (1 − p) , if y ∈ {0, 1, . . . , n}
fYp (y ) = ,
0, otherwise
where n ∈ N (set of positive integers) is a fixed constant. Find the m.g.f.
d
of Yp and show that n − Yp = Y1−p .
Solution
MYp (t) = E (e tYp )
n
ty n
X
= e p y (1 − p)n−y
y
y =0
n
X n
= (pe t )y (1 − p)n−y
y
y =0
= (1 − p + pe t )n , t ∈ R.
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Let Zp = n − Yp , p ∈ (0, 1). Then
MZp (t) = E (e t(n−Yp ) )
= e nt E (e −tYp )
= e nt MYp (−t)
= e nt (1 − p + pe −t )n
= (p + (1 − p)e t )n
= MY1−p (t), ∀t∈R
d
⇒ Zp = Y1−p .
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Definition 2: A r.v. X is said to have a symmetric distribution about a
d
point µ ∈ R if X − µ = µ − X .
Remark 1: In Example 1
d
n − Y1 = Y1
2 2
n d n
⇒ − Y1 = Y1 − ,
2 2 2 2
implying that the distribution of Y 1 is symmetric about n2 . Clearly
2
n n
E − Y1 = E Y1 −
2 2 2 2
1
⇒ E Y1 = .
2 2
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Result 2:
Let X be a r.v. with p.m.f./p.d.f. fX (·) and d.f . FX (·). Let µ ∈ R.
(a) If fX (µ − x) = fX (µ + x), ∀ x ∈ R, then the distribution of X is
symmetric about µ;
(b) Distribution of X is symmetric about µ iff
FX (µ + x) + FX ((µ − x)−) = 1;
(c) Distribution of X is symmetric about µ iff the distribution of
Y = X − µ is symmetric about 0;
(d) If distribution of X is symmetric about µ and E (X ) exists then
µ = E (X );
1
(e) If distribution of X is symmetric about µ then FX (µ−) ≤ 2 ≤ FX (µ);
(FX (µ) = 12 , if FX (·) is continuous at µ);
(f) If distribution of X is symmetric about µ then
E ((X − µ)2m−1 ) = 0, m ∈ {1, 2, . . . }, provided the expectations exist.
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Proof.
(a) Let Y1 = X − µ and Y2 = X − µ. Then
fY1 (y ) = fX (µ + y ) = fX (µ − y ) = fY2 (y ), ∀y ∈R
d
⇒ Y1 = Y2 .
(b)
d
X −µ=µ−X
⇔ P ({X − µ ≤ x}) = P ({µ − X ≤ x}) , ∀x ∈R
⇔ P ({X ≤ µ + x}) = P ({X ≥ µ − x}) , ∀x ∈R
⇔ FX (µ + x) + FX ((µ − x)−) = 1, ∀ x ∈ R.
(c) Let Y1 = X − µ. Then
d
X −µ = µ−X
d
⇔ X − µ = −(X − µ)
d
⇔ Y1 − 0 = 0 − Y1 .
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(d)
d
X −µ = µ−X
⇒ E (X − µ) = E (µ − X )
⇔ E (X ) = µ.
(e) By (b)
FX (µ + x) + FX ((µ − x)−) = 1 ∀ x ∈ R
⇒ FX (µ) + FX (µ−) = 1
1
⇒ FX (µ−) ≤ ≤ FX (µ) (since FX (µ−) ≤ FX (µ)).
2
(f)
d
X −µ = µ−X
⇒ E ((X − µ)(2m−1) ) = E ((µ − X )(2m−1) )
⇒ E ((X − µ)(2m−1) ) = −E ((X − µ)(2m−1) )
⇒ E ((X − µ)(2m−1) ) = 0, m = 1, 2, . . . .
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Example 2:
Let X be a r.v. having the p.d.f.
1 1
fX (x) = . , −∞ < x < ∞.
π 1 + (x − µ)2
Clearly,
fX (µ + x) = fX (µ − x), ∀ x ∈ R
d
⇒ X − µ = µ − X.
However E (X ) does not exists.
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Take Home Problems
1 Let X be a r.v. having a p.d.f.
1
fX (x) = e −|x| , −∞ < x < ∞.
2
Show that the distribution of X is symmetric about zero. Hence find
E (X ) (Does it exists?).
2 Let X be a r.v. having the m.g.f.
t2
MX (t) = e 2 , −∞ < t < ∞.
d
Show that X = −X (i.e., distribution of X is symmetric about zero);
E (X 2r −1 ) = 0, r ∈ {1, 2, . . . } ,
and
(2r )!
E (X 2r ) = , r ∈ {1, 2, . . .} .
2r r !
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Abstract of Next Module
Let A ⊆ R, g = R → R and let X be a r.v. In many situations
P({X ∈ A}) or E (g (X )) can not be evaluated precisely. In such situations
some approximations of P({X ∈ A}) or E (g (X )) may be useful. Some
useful approximations can be provided in form of inequalities.
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Thank you for your patience
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