Kenneth Kuttler

59.15. THE CONVERGENCE OF SUMS 1907

By induction, it follows that if you have n independent random variables each having sym-metric distribution, then their sum has symmetric distribution.

Here is a simple lemma about random variables having symmetric distributions. It willdepend on Lemma 59.15.2 on Page 1906.

Lemma 59.15.4 Let X≡ (X1, · · · ,Xn) and Y be random variables defined on a probabilityspace, (Ω,F ,P) such that Xi, i = 1,2, · · · ,n and Y have values in E a separable Banachspace. Thus X has values in En. Suppose also that {X1, · · · ,Xn,Y} are independent andthat Y has symmetric distribution. Then if A ∈B (En) , it follows

([X ∈ A]∩

[∣∣∣∣∣∣∣∣∣∣ n

∑i=1

Xi +Y

∣∣∣∣∣∣∣∣∣∣< r

])

= P

([X ∈ A]∩

[∣∣∣∣∣∣∣∣∣∣ n

∑i=1

Xi−Y

∣∣∣∣∣∣∣∣∣∣< r

])

You can also change the inequalities in the obvious way, < to ≤ , > or ≥.

Proof: Denote by λ X and λY the distribution measures for X and Y respectively. Sincethe random variables are independent, the distribution for the random variable, (X,Y ) map-ping into En+1 is λ X×λY where this denotes product measure. Since the Banach space isseparable, the Borel sets are contained in the product measurable sets. Then by symmetryof the distribution of Y

([X ∈ A]∩

[∣∣∣∣∣∣∣∣∣∣ n

∑i=1

Xi +Y

∣∣∣∣∣∣∣∣∣∣< r

])

=∫

En×EXA (x)XB(0,r)

∑i=1

xi + y

)d (λ X×λY )(x,y)

=∫

∫En

XA (x)XB(0,r)

∑i=1

xi + y

)dλ XdλY

=∫

∫En

XA (x)XB(0,r)

∑i=1

xi + y

)dλ Xdλ−Y

=∫

En×EXA (x)XB(0,r)

∑i=1

xi + y

)d (λ X×λ−Y )(x,y)

= P

([X ∈ A]∩

[∣∣∣∣∣∣∣∣∣∣ n

∑i=1

Xi +(−Y )

∣∣∣∣∣∣∣∣∣∣< r

])

This proves the lemma. Other cases are similar.Now here is a really interesting lemma.

59.15. THE CONVERGENCE OF SUMS 1907By induction, it follows that if you have n independent random variables each having sym-metric distribution, then their sum has symmetric distribution.Here is a simple lemma about random variables having symmetric distributions. It willdepend on Lemma 59.15.2 on Page 1906.Lemma 59.15.4 Let X = (X1,--- ,X,) and Y be random variables defined on a probabilityspace, (Q,.F,P) such that X;,i = 1,2,---,n and Y have values in E a separable Banachspace. Thus X has values in E®. Suppose also that {X\,--- ,Xn,Y} are independent andthat Y has symmetric distribution. Then if A € B(E"), it follows+(xeai|} a}r(xenn[ex- <)You can also change the inequalities in the obvious way, < to <, > or >.Proof: Denote by Ax and Ay the distribution measures for X and Y respectively. Sincethe random variables are independent, the distribution for the random variable, (X,Y) map-ping into E”*! is Ax x Ay where this denotes product measure. Since the Banach space isseparable, the Borel sets are contained in the product measurable sets. Then by symmetryof the distribution of YP(weain <|)_ 2X4 (x) Poo.) [Zser)a (Ax x Ay) (x,y)E"xEPx[ Es 2a (X) 2B(0,r) (Ex) dAxddyi=l[ Es 2a (X) 2B00,r) (E> +s) ddxdi_yi=l= Ba (x) Zion er (Ax x A_y) (xy)o iThis proves the lemma. Other cases are similar.Now here is a really interesting lemma.