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Random variable sequence?

  1. Sep 13, 2012 #1
    Hi all,
    I am really confused about the random variables
    Toss a coin three times, so the set of possible outcomes is

    Ω={HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

    Define the random variables

    X = Total number of heads, Y = Total number of tails

    In symbol,

    X(HHH)=3
    X(HTT)=X(HTH)=X(THH)=2
    X(HTT)=X(THT)=X(TTH)=1
    X(TTT)=0

    Y(TTT)=3
    Y(TTH)=Y(THT)=Y(HTT)=2
    Y(THH)=Y(HTH)=Y(HHT)=1
    Y(HHH)=0

    The probability of head on each toss is 1/2 and the probability of each element in Ω is 1/8, then:

    P{ω∈Ω; X(ω)=0}=P{TTT}=1/8

    P{ω∈Ω; X(ω)=1}=P{HTT,THT,THH}=3/8

    P{ω∈Ω; X(ω)=2}=P{HHT, HTH,THH}=3/8

    P{ω∈Ω; X(ω)=3}=P{HHH}=1/8


    P{ω∈Ω; Y(ω)=0}=P{HHH}=1/8

    P{ω∈Ω; Y(ω)=1}=P{THH,HTH,HHT}=3/8

    P{ω∈Ω; Y(ω)=2}=P{TTH,THT,HTT}=3/8

    P{ω∈Ω; Y(ω)=3}=P{TTT}=1/8

    I have taken this example from text, now my question is that what is a sequence of random variable? The text says that the sequence of random variable is: X_1,X_2,X_3,......X_n. So in the above example, can we say that there are two sequence of variables which are,
    X(HHH)=3 is X_1
    X(HTT)=X(HTH)=X(THH)=2 is X_2
    X(HTT)=X(THT)=X(TTH)=1 is X_3
    X(TTT)=0 is X_4

    Y(TTT)=3 is Y_1
    Y(TTH)=Y(THT)=Y(HTT)=2 is Y_2
    Y(THH)=Y(HTH)=Y(HHT)=1 is Y_3
    Y(HHH)=0 is Y_4

    OR

    X is just one variable but taking different values so in the following
    X(HHH)=3
    X(HTT)=X(HTH)=X(THH)=2
    X(HTT)=X(THT)=X(TTH)=1
    X(TTT)=0
    there is no sequence

    Similarly Y is just one variable but taking different values so in the following
    Y(TTT)=3
    Y(TTH)=Y(THT)=Y(HTT)=2
    Y(THH)=Y(HTH)=Y(HHT)=1
    Y(HHH)=0

    there is no sequence

    or X,Y together forms a sequence?

    I will really appreciate if someone can help me.

    Thanks in advance.
     
  2. jcsd
  3. Sep 13, 2012 #2

    Simon Bridge

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    It is a sequence of numbers that come from an idealized random process - this means it is an abstract concept. How you tell is a particular sequence of numbers is random is a hard problem.

    But I think I see your problem ...

    Say I repeat the triple-coin-toss 5 times ... I get a sequence of 5 random numbers ... something like this perhaps: [itex]\{X_1, X_2, X_3, X_4, X_5\} = \{3, 0, 1, 1, 2\}[/itex] ... this is to say that [itex]X_n[/itex] is the result of the nth coin toss. If I did the experiment 20 times, I'd have [itex]X_1, X_2, X_3 \cdots[/itex] all the way to [itex]X_{20}[/itex] each one capable of having one of four distrete values. We can write [itex]X_n = x_n \in \{0,1,2,3\}[/itex] because, strictly speaking, each "X" (cap X) is a symbol that represents the act of tossing three coins and counting up the heads. The number of heads is usually represented by a lower-case "x".

    That help?
     
    Last edited: Sep 13, 2012
  4. Sep 14, 2012 #3
    Sorry, I didn't get it :(((((
     
  5. Sep 14, 2012 #4

    lavinia

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    A sequence of random variables is just a set of sequences in which the n'th number is chosen from the n'th random variable. The sequence itself is a new random variable.
     
  6. Sep 14, 2012 #5

    Simon Bridge

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    Then I did not understand the question ... try restating it.
    Let me give you a language to do that with:


    Let ##Z## be the event that I toss one coin (for simplicity) and count the number of "heads" that result.

    Then the result of that toss can be ##z=1## for "heads" or ##z=0## for "not heads". So I can say that ##z \in \{0,1\}##.

    If I toss the coin more than once, the ##Z_1## will be the first time I do it, ##Z_2## the second time, and so on.
    Then the first result will be ##z_1## and the second result will be ##z_2## and so on.

    If I toss the coin N times I get a sequence of random numbers.
    Each member in the sequence can be a 1 or a 0.
    The entire sequence will be the set of numbers:
    ##\{z_1, z_2, z_3, \cdots , z_{N-1}, z_N\}##

    All I am doing here is defining a notation.

    In this notation:
    Z is a random process - a random number generator (if you will).
    z is a random variable.
    ... some texts get a bit casual about the distinction.

    Do you follow this so far?
     
    Last edited: Sep 14, 2012
  7. Sep 15, 2012 #6
    yes now I do

    Thank you sir
     
  8. Sep 15, 2012 #7

    Simon Bridge

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    Um ... OK. No worries then.
     
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