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Math Help - Check that each of the following is a valid probability function...

  1. #1
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    Thumbs up Check that each of the following is a valid probability function...

    Check that each of the following is a valid probability function for a discrete random variable.

    (i) g(n) = (2/3) (1/3)^(n-1), n=1,2,3,...

    (ii) g(k) = (k+1) (2/3)^2 (1/3)^k, k=0,1,2,...

    (Hint: use the formula (1-q)^(-2) = sum from zero to infinity of (n+1)q^n with 0<q<1.)

    Last edited by mr fantastic; May 6th 2010 at 06:04 PM. Reason: Copied from title into main body of post.
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  2. #2
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    Quote Originally Posted by feyomi View Post
    ...for a discrete random variable.

    (i) g(n) = (2/3) (1/3)^(n-1), n=1,2,3,...

    (ii) g(k) = (k+1) (2/3)^2 (1/3)^k, k=0,1,2,...

    (Hint: use the formula (1-q)^(-2) = sum from zero to infinity of (n+1)q^n with 0<q<1.)

    hi feyomi,

    if the formula represent probabilities, then the sum of all probabilities
    must equal 1.

    These are sums to infinity.

    For (i)

    g(n)=\frac{2}{3}\ \left(\frac{1}{3}\right)^{n-1} n=1, 2, 3....

    the sum of these "probabilities" is

    \frac{2}{3}\ \left(\frac{1}{3}\right)^0+\frac{2}{3}\ \left(\frac{1}{3}\right)^1+\frac{2}{3}\ \left(\frac{1}{3}\right)^2+....

    which is a geometric series, first term 2/3, common ratio 1/3, whose sum to infinity is \frac{T_1}{1-r}=\frac{a}{1-r}

    T1=a=\frac{2}{3},\ r=\frac{1}{3}

    S_{\infty}=\frac{a}{1-r}=\frac{\frac{2}{3}}{1-\frac{1}{3}}=\frac{\left(\frac{2}{3}\right)}{\left  (\frac{2}{3}\right)}=1

    All probabilities sum to 1, hence the function g(n) is valid.


    For (ii), use the hint

    q=\frac{1}{3}

    Hence, the sum is

    \left(\frac{2}{3}\right)^2\left(1-\frac{1}{3}\right)^{-2}

    =\left(\frac{2}{3}\right)^2\left(\frac{2}{3}\right  )^{-2}

    As this is 1, this function is also valid.
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