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Math Help - Derivitive to Find Maximization

  1. #1
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    Derivitive to Find Maximization

    This question has ties to probability, but the question I am asking involves Calculus. See attachment.

    I was able to construct A)... or at least I think so. See other attachment as I'm not quite familiar with latex.
    Attached Thumbnails Attached Thumbnails Derivitive to Find Maximization-forum-question.jpg   Derivitive to Find Maximization-math-p-n-.jpg  
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  2. #2
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    As you found, P(x) = \binom{10}{x}\pi^x(1 - \pi)^x.

    If x = k, then P(x) = P(k) = \binom{10}{k}\pi^k(1 - \pi)^k.

    You are told to think of your function as a function of \pi. Therefore, let P(\pi) = \binom{10}{k}\pi^k(1 - \pi)^k.

    Absolute maxima occur at critical numbers c such that f'(c) = 0 or f'(c) is undefined. Therefore, let's find P'(\pi).

    P'(\pi) = \frac{d}{d\pi}[P(\pi)] = \frac{d}{d\pi}[\binom{10}{k}\pi^k(1 - \pi)^k] = \binom{10}{k}\frac{d}{d\pi}[\pi^k(1 - \pi)^k]

    Note \binom{10}{k} can be 'factored' out of the derivative operator because it is a constant (scalar multiple).

    \binom{10}{k}\frac{d}{d\pi}[\pi^k(1 - \pi)^k] = \binom{10}{k}(\frac{d}{d\pi}[\pi^k] \times (1 - \pi)^k + \pi^k \times \frac{d}{d\pi}[(1 - \pi)^k])

    Note the use of the product rule because \pi^k(1 - \pi)^k is a product.

    \binom{10}{k}(\frac{d}{d\pi}[\pi^k] \times (1 - \pi)^k + \pi^k \times \frac{d}{d\pi}[(1 - \pi)^k]) = \binom{10}{k}(k\pi^{k - 1}(1 - \pi)^k + k\pi^k(1 - \pi)^{k - 1})

    Therefore, P'(\pi) = \binom{10}{k}(k\pi^{k - 1}(1 - \pi)^k + k\pi^k(1 - \pi)^{k - 1}).

    As you can see, there are no critical numbers c such that P'(c) is undefined. Now we only have to determine if there are any critical numbers c such that P'(c) = 0.

    P'(\pi) = 0

    \binom{10}{k}(k\pi^{k - 1}(1 - \pi)^k + k\pi^k(1 - \pi)^{k - 1}) = 0

    Since \binom{10}{k} is a constant, we can eliminate it from the equation by dividing the equation by \binom{10}{k}.

    k\pi^{k - 1}(1 - \pi)^k + k\pi^k(1 - \pi)^{k - 1} = 0

    \pi^{k - 1}(1 - \pi)^k + \pi^k(1 - \pi)^{k - 1} = 0

    \pi^{k - 1}(1 - \pi)^{k - 1}[(1 - \pi) + \pi] = 0

    I factored \pi^{k - 1}(1 - \pi)^{k - 1} from both terms on the right-hand side of the equation. Since (1 - \pi) + \pi = 1, we have:

    \pi^{k - 1}(1 - \pi)^{k - 1} = 0

    [\pi(1 - \pi)]^{k - 1} = 0

    \pi(1 - \pi) = 0

    \pi = 0 or \pi = 1
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  3. #3
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    I am now asked the question: Does your answer remind you of a familiar statistic commonly calculated to estimate pi? Is so, which one?
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  4. #4
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    Quote Originally Posted by NOX Andrew View Post
    \binom{10}{k}(\frac{d}{d\pi}[\pi^k] \times (1 - \pi)^k + \pi^k \times \frac{d}{d\pi}[(1 - \pi)^k]) = \binom{10}{k}(k\pi^{k - 1}(1 - \pi)^k + k\pi^k(1 - \pi)^{k - 1})

    Therefore, P'(\pi) = \binom{10}{k}(k\pi^{k - 1}(1 - \pi)^k + k\pi^k(1 - \pi)^{k - 1}).
    Is that not:
    \binom{10}{k}(\frac{d}{d\pi}[\pi^k] \times (1 - \pi)^k + \pi^k \times \frac{d}{d\pi}[(1 - \pi)^k]) = \binom{10}{k}(k\pi^{k - 1}(1 - \pi)^k +(-1)k\pi^k(1 - \pi)^{k - 1}) which gives#
    P'(\pi) = \binom{10}{k}(k\pi^{k - 1}(1 - \pi)^k - k\pi^k(1 - \pi)^{k - 1})

    As you can see, there are no critical numbers c such that P'(c) is undefined. Now we only have to determine if there are any critical numbers c such that P'(c) = 0.

    P'(\pi) = 0

    \binom{10}{k}(k\pi^{k - 1}(1 - \pi)^k <b>-</b> k\pi^k(1 - \pi)^{k - 1}) = 0

    Since \binom{10}{k} is a constant, we can eliminate it from the equation by dividing the equation by \binom{10}{k}.

    k\pi^{k - 1}(1 - \pi)^k <b>-</b> k\pi^k(1 - \pi)^{k - 1} = 0

    \pi^{k - 1}(1 - \pi)^k <b>-</b> \pi^k(1 - \pi)^{k - 1} = 0

    \pi^{k - 1}(1 - \pi)^{k - 1}[(1 - \pi) <b>-</b> \pi] = 0

    I factored \pi^{k - 1}(1 - \pi)^{k - 1} from both terms on the right-hand side of the equation.
    This gives 1-2\pi=0, or \pi=\frac{1}{2}

    However, you still have to take into account if \pi^{k - 1}(1 - \pi)^{k - 1} is zero.

    \pi^{k - 1}(1 - \pi)^{k - 1} = 0

    [\pi(1 - \pi)]^{k - 1} = 0

    \pi(1 - \pi) = 0

    \pi = 0 or \pi = 1
    Hence three values for pi. These will be either maxima or minima, however, note that when pi is 0 or 1, P(\pi) is zero, regardless of what your value of k is.

    To find out which is max and which is min explicitly, you should get the second derivative of P(\pi) with respect to pi, and when it is negative, the function will be maximised.

    As you don't know what k which value of pi is a max/min may also vary with k.
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