Kenneth Kuttler

38.9. INDEPENDENCE AND CONDITIONAL PROBABILITY 793

are nonnegative functions xi → fi (xi) such that f (x) = ∏pi=1 fi (xi) . Note how this gives

the conclusion of the above theorem.

P(X1 ∈ I1, · · · ,Xp ∈ Ip) =∫

∏pi=1 Ii

f (x)dx

=∫

I1· · ·∫

f1 (x1) · · · fp (xp)dxp · · ·dx1

=∫

I1f1 (x1)dx1 · · ·

∫Ip

fp (xp)dxp

∏i=1

P(Xi ∈ Ii)

In fact, this is a specialization of what always happens in every situation. Note that thesame argument shows that if these components are an independent set, then if you consider

(g1 (X1) , · · · ,gp (Xp))

these would also be independent random variables. In this case,

P(g1 (X1) ∈ I1, · · · ,gp (Xp) ∈ Ip)

= P(X1 ∈ g−1

1 (I1) , · · · ,Xp ∈ g−1p (Ip)

)=∫

∏pi=1 g−1

i (Ii)f (x)dx

=∫

g−11 (I1)

· · ·∫

g−1p (Ip)

f1 (x1) · · · fp (xp)dxp · · ·dx1

=∫

g−11 (I1)

f1 (x1)dx1 · · ·∫

g−1p (Ip)

fp (xp)dxp

∏i=1

P(Xi ∈ g−1

i (Ii))=

∏i=1

P(gi (Xi) ∈ Ii)

This proves the following.

Proposition 38.9.4 Let X = (X1, · · · ,Xp) and suppose there is a density function f (x).Then the components are independent random variables if f has the following form.

f (x) =p

∏i=1

fi (xi)

If these are independent random variables, then so are {gi (Xi)}pi=1 whenever gi are con-

tinuous functions.

This is actually an equivalence so there is no loss of generality in taking it as the defini-tion of independence. However, it gets much more involved because some random variablesare neither discrete nor continuous. Nevertheless, this kind of thing will hold if appropri-ately generalized. To do this in full generality requires a much better mathematical theorythan any contemplated in this book. It will end up involving differentiation theory of some-thing called a Radon measure. Having noted this, all of the techniques discussed here weredeveloped with nothing more than the Riemann integral in the early 1900’s.

Also recall the definition of mean and variance given above.

38.9. INDEPENDENCE AND CONDITIONAL PROBABILITY 793are nonnegative functions x; — f;(x;) such that f(a) = []?_, fi(xi). Note how this givesthe conclusion of the above theorem.P(X Eh sXp Ep) =f f (a) dxMy hi[= [ flan sp today[ fi wad fh (Xp) dxpPp[[?>% €4)i=lIn fact, this is a specialization of what always happens in every situation. Note that thesame argument shows that if these components are an independent set, then if you consider(g1 (X1),-°° Sp (Xp))these would also be independent random variables. In this case,P(g (X1) El,+++ 8p (Xp) € Ip)= P(X € gy! (i) .-°° Xp € 8, (Ip))= f (a) dx. Tg) | (h) ( )in dey (x1) +*+ fp (Xp) dxp ++ dx}1a afi (01) dx} of fp (Xp) dXp1 (4) 8p (Ip)P(X; €g; | (li)) = [Pte (Xi) € Jj)i=1I=.=This proves the following.Proposition 38.9.4 Let X = (X,,---,X,) and suppose there is a density function f (x).Then the components are independent random variables if f has the following form.f(a) = [TatsIf these are independent random variables, then so are {g;(X;)}?_, whenever g; are con-tinuous functions.This is actually an equivalence so there is no loss of generality in taking it as the defini-tion of independence. However, it gets much more involved because some random variablesare neither discrete nor continuous. Nevertheless, this kind of thing will hold if appropri-ately generalized. To do this in full generality requires a much better mathematical theorythan any contemplated in this book. It will end up involving differentiation theory of some-thing called a Radon measure. Having noted this, all of the techniques discussed here weredeveloped with nothing more than the Riemann integral in the early 1900’s.Also recall the definition of mean and variance given above.