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Question about Independent Random Variables and iid

  1. Jun 15, 2005 #1
    I have a question about independent random variable:
    Let say we flip a fair coin, the set of outcome is S={H,T}, P(H)=1/2, P(T)=1/2. Define random variable X:S->R by X(H)=1, X(T)=-1.
    From what I read in books, I can define X1 and X2 as independent identically distributed (iid) random variables with the same distribution as X. Then, that would mean P(X1=1,X2=1)=P(X1=1)P(X2=1).
    It can easily be seen that the event {X1=1,X2=1}={w in S:X1(w)=1,X2(w)=1}={w in S:w=H}={w in S:X1(w)=1}={X1=1}={w in S:X2(w)=1}={X2=1}, and since P({w:w=H})=P(H)=1/2, we have P(X1=1,X2=1)=1/2 and P(X1=1)P(X2=1)=1/4.
    So, I am not sure exactly what I mistake I made. Please help me clear up my confusion. Thanks.
  2. jcsd
  3. Jun 15, 2005 #2


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    The sample space (S) for two flips has four points (H,H), (H,T), (T,H), (T,T), each of which has probability 1/4. In abstract terms, the sample space for n independent random variables is the direct product of the sample spaces of each of the variables.
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