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Math Help - help with discrete random variables

  1. #1
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    Exclamation help with discrete random variables

    1. The problem statement, all variables and given/known data

    1.
    Suppose u flip a coin
    Z = 1 if the coin is heads
    Z = 3 if the coin is tails
    W = Z^2 + Z
    a)
    what is the probability function of Z?
    b)
    what is the probability function of W?




    2.
    Let Z ~ Geometric (theta). Compute P(5<=Z<=9).

    2. Relevant equations



    3. The attempt at a solution

    1.
    what i think
    a)
    Pz(Z) = 1 if head
    3 if tail
    0 otherwise
    b)
    Pw(W) = 2 if head
    12 if tail
    0 otherwise

    I don't see how the probability part comes into this, where do i include the 1/2 chance part...


    or maybe its:
    a)
    Pz(Z) = 1/2 if Z is 1 or 3
    0 otherwise
    b)
    Pw(W) = 1/2 if W is 2 or 12
    0 otherwise

    2.
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  2. #2
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    Z can be equal to 1 or 3.
    The probability distribution function of Z will tell you what is the probability that Z takes each of these 2 values.

    So P(Z=1) =P(outcome is head)=1/2
    and P(Z=2)=P(outcome is tail)=1/2

    Similar reasoning applies to W.
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  3. #3
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    I dont understand this question.

    Let Z ~ Geometric(theta). Compute P(5<=Z<=9).

    How do i do this?
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  4. #4
    MHF Contributor harish21's Avatar
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    X ~ \mbox{Geometric}(p) has the probability Distribution:

    P(X=x) = p^{x-1} (1 - p)^{x}

    You should now be able to compute P(5<=Z<=9)!!
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  5. #5
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    I am still very confused..

    I dont understand what values p should be and what values x should be...
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  6. #6
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    p is the parameter of the distribution (same as the theta in your question).
    X is the variable we look at (Z in your question)
    So P ( 5<=Z<=9) = P(Z=5)+P(Z=6)+P(Z=7)+P(Z=8)+P(Z=9)


    Z ~ Geo(\theta)

    P(Z=z)= {(1-\theta)}^{(z-1)} * \theta

    P(Z=5)= {(1-\theta)}^{(5-1)} * \theta

    P(Z=6)= {(1-\theta)}^{(6-1)} * \theta

    P(Z=7)= {(1-\theta)}^{(7-1)} * \theta

    P(Z=8)= {(1-\theta)}^{(8-1)} * \theta

    P(Z=9)= {(1-\theta)}^{(9-1)} * \theta
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  7. #7
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    from your previous post, isn't it supposed to be


    P(Z=z)=( theta^(z-1) ) * (1 - theta)^z
    where z is 5,6,7,8,9


    or is this exactly what you have but simplified?
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  8. #8
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    Quote Originally Posted by Sneaky View Post
    from your previous post, isn't it supposed to be


    P(Z=z)=( theta^(z-1) ) * (1 - theta)^z
    where z is 5,6,7,8,9


    or is this exactly what you have but simplified?
    It is:
    P(Z=z)={(1-\theta)}^{(z-1)}*\theta
    exactly what I wrote in my post.

    Check your notes for the way geometric distribution is defined: Z ~ Geo ( \theta), then P(Z=z) = probability of obtaining a success after z-1 failures.
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