Definition
Independence
X1,⋯,Xn are indepdendent
iff for all x1,⋯,xn∈R
fX1,⋯,Xn(x1,⋯,xn)=Πni=1fXi(xi).
iff for all x1,⋯,xn∈R
FX1,⋯,Xn(x1,⋯,xn)=Πni=1FXi(xi).
iid (independently identically distributioned)
We say X1,⋯,Xn are iid if
- X1,⋯,Xn are indepdendent
- They have the same distribution, ie, Fi(x)=F(x),fXi(x)=f(x) for all i and where Fi(x),fXi(x) is the (marginal) cdf and pdf/pmf of Xi respectively.
Implications
If X1,⋯,Xn are indepdendent then
Probability P(X1∈A1,⋯,Xn∈An)=Πni=1P(Xi∈Ai).
For any functions gi, i=1,⋯,n
E(Πni=1gi(Xi))=Πni=1E(gi(Xi)
provided the expectations on the right-hand side exist.It is much easier to work out what Y=g(X1,⋯,Xn) is (ie, find out what its cdf or pdf/pmf or mgf is) or at least some snapshots such as expectation, variance or moments of Y.
Illustrations
- Let X1,⋯,Xn∼iidBernoulli(p),p∈(0,1) then
n∑i=1Xi∼Bin(n,p). - Let X1,⋯,Xn∼iidN(μ,σ2)) then
ˉX=1nn∑i=1Xi∼N(μ,σ2n). - Even when the distribution is only known to expectation and variance, we can still get some “pictures” of it. In the sense,
if X1,⋯,Xn∼iid with μ=E(Xi),σ2=Var(Xi) then
E(ˉX)=μ,Var(ˉX)=σ2n.
Note that this is consistent with our Bernoulli and normal cases.
Homework 7
Textbook: Sec. 8.6: 8.1, 8.5, 8.6, 8.9; Sec 9.7: 9.1, 9.3, 9.5, 9.10, 9.11, 9.12. Sec 10.5: 10.2, 10.6, 10.8, 10.16. To be discussed in class 12/27.