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Math Help - Binomial distribution problem

  1. #1
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    Red face Binomial distribution problem

    Show that the variance of a binomial distribution with parameters n and p cannot exceed \frac{n}{4}
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  2. #2
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    Quote Originally Posted by zorro View Post
    Show that the variance of a binomial distribution with parameters n and p cannot exceed \frac{n}{4}
    The graph of Var(X) = np(1-p) versus p is a parabola over the domain 0 \leq p \leq 1. Do you know how to find the maximum turning point of a parabola ....?
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  3. #3
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    You need to show that pq never exceeds 1/4.

    Show that pq = -p^2 + p and find the the stationary point of this parabola (Which is a maximum).

    You should get p=1/2 produces the maximum.
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    no

    Quote Originally Posted by mr fantastic View Post
    The graph of Var(X) = np(1-p) versus p is a parabola over the domain 0 \leq p \leq 1. Do you know how to find the maximum turning point of a parabola ....?
    No
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  5. #5
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    difficulty understanding

    Quote Originally Posted by skamoni View Post
    You need to show that pq never exceeds 1/4.

    Show that pq = -p^2 + p and find the the stationary point of this parabola (Which is a maximum).

    You should get p=1/2 produces the maximum.

    From where did q come in to this question
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  6. #6
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    Differentiate the given Parabola (In this case -p^2 + p ) and put it equal to zero, to obtain the value of p for which the parabola is maximised.

    I let q=1-p (I think this is standard notation).

    pq=p(1-p) = p - p^2
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    Quote Originally Posted by mr fantastic View Post
    The graph of Var(X) = np(1-p) versus p is a parabola over the domain 0 \leq p \leq 1. Do you know how to find the maximum turning point of a parabola ....?
    Quote Originally Posted by zorro View Post
    No
    The turning point lies between the two intercepts. The intercepts are p = 0 and p = 1. So the (maximum) turning point occurs at p = 1/2. Substitute = 1/2 into Var(X) = np(1-p) to get the required result.
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  8. #8
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    Quote Originally Posted by skamoni View Post
    You need to show that pq never exceeds 1/4.

    Show that pq = -p^2 + p and find the the stationary point of this parabola (Which is a maximum).

    You should get p=1/2 produces the maximum.
    Quote Originally Posted by zorro View Post
    From where did q come in to this question
    q= 1 - p. It's a very standard notation. The variance of a binomial distribution is often written npq.
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  9. #9
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    Quote Originally Posted by zorro View Post
    From where did q come in to this question
    If you dont know Analytical geometry, there are other ways to see it as well:

    You can try:

    1) Completing the squares.

    np(1-p) = n(p - p^2) = n\left(\frac14 - \left( p - \frac12\right)^2\right). But \left( p - \frac12\right)^2 \geq 0, so what can you conclude?


    2) AM-GM inequality,

    You can apply it(since p and q are non-negative), to get \frac{p+q}2 \geq \sqrt{pq} \Rightarrow \sqrt{pq} \leq \frac12 \Rightarrow npq \leq \frac{n}4
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