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Thread: P(y=1), e[y]

  1. #1
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    P(y=1), e[y]

    P(y=1), e[y]-captura111.jpg
    P(y=1), e[y]-captura112.jpg
    P(y=1), e[y]-captura113.jpg


    P(y=1), e[y]-captura114.jpg
    P(y=1), e[y]-captura115.jpg
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  2. #2
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    Re: P(y=1), e[y]

    For any probability distribution we must have total probability equal to 1 Here, that means that $\displaystyle \sum_{i= 0}^\infty \frac{\alpha}{2^i}= \alpha\sum_{i=0}^\infty \left(\frac{1}{2}\right)^i= 1$. That sum is a geometric series, $\displaystyle \sum_{i= 0}^\infty r^i$ with r= 1/2. It's sum is $\displaystyle \frac{1}{1- r}= \frac{1}{1- \frac{1}{2}}= \frac{1}{\frac{1}{2}}= 2$. So we must have $\displaystyle 2\alpha= 1$.


    If Y= X (mod 3)= 1 then X is 1, 4, 7, 10, etc so that $\displaystyle P(1)= \frac{\alpha}{2}+ \frac{\alpha}{2^4}+ \frac{\alpha}{2^7}+ \frac{\alpha}{2^{10}}+ \cdot\cdot\cdot= $$\displaystyle \frac{\alpha}{2}\left(1+ \frac{1}{2^3}+ \frac{1}{2^6}+ \frac{1}{2^9}+ \cdot\cdot\cdot\right)= $$\displaystyle \frac{\alpha}{2}\left(1+ \frac{1}{2^3}+ \frac{1}{(2^3)^2}+ \frac{1}{(2^3)^3}+ \cdot\cdot\cdot\right)$. That sum is again a geometric series with $\displaystyle r= \frac{1}{2^3}= \frac{1}{8}$.
    Last edited by HallsofIvy; Oct 26th 2018 at 06:12 AM.
    Thanks from SlipEternal and Amanda2018
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  3. #3
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    Re: P(y=1), e[y]

    Hi, thanks. But can you explain a bit? i do not full understand how i can get probability P[X],
    for geometric distribution
    1. if i want get E[X]=1/p, i have to first get probability P[X]

    2. I think P(Y=1)=a/2 = 1/4 (2a=1, a=1/2)

    E[Y] = 1/p, but i need get p[Y]
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  4. #4
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    Re: P(y=1), e[y]

    A "geometric distribution" is, by definition, a distribution of the form $\displaystyle P(X= k)= (1- p)^{k-1}p$. If you are still talking about the distribution in your first post, $\displaystyle P(X= I)= \frac{\alpha}{2^i}= \alpha \left(\frac{1}{2}\right)^i=$ then that is a geometric distribution with p= 1/2 so that 1- p= 1/2 also: $\displaystyle P(X= k)= (1/2)^{k-1}(1/2)= (1/2)^k$.
    Last edited by topsquark; Oct 27th 2018 at 08:27 AM. Reason: Tweaked LaTeX
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  5. #5
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    Re: P(y=1), e[y]

    hi, i study your advice again, here is my understand:
    first problem: a = 1/2
    p= 1/2
    E[X] = (1-p)/p = 1 (k∈{0,1,2,3,...})

    second problem: Y = X mod 3
    r = 1/8
    a/1-r = 1
    a = 7/8
    P(Y=1) = a/2 = 7/16
    E[Y] = (1-p)/p = 9/7

    do you think my understand correct?
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