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Thread: 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, $\displaystyle P(x) = \binom{10}{x}\pi^x(1 - \pi)^x$.

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

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

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

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

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

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

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

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

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

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

    $\displaystyle P'(\pi) = 0$

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

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

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

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

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

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

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

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

    $\displaystyle \pi(1 - \pi) = 0$

    $\displaystyle \pi = 0$ or $\displaystyle \pi = 1$
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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
    $\displaystyle \binom{10}{k}(\frac{d}{d\pi}[\pi^k] \times (1 - \pi)^k + \pi^k \times \frac{d}{d\pi}[(1 - \pi)^k]) = \binom{10}{k}(k\pi^{k - 1}(1 - \pi)^k + k\pi^k(1 - \pi)^{k - 1})$

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

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

    $\displaystyle P'(\pi) = 0$

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

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

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

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

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

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

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

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

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

    $\displaystyle \pi(1 - \pi) = 0$

    $\displaystyle \pi = 0$ or $\displaystyle \pi = 1$
    Hence three values for pi. These will be either maxima or minima, however, note that when pi is 0 or 1, $\displaystyle P(\pi)$ 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 $\displaystyle P(\pi)$ 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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