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Thread: probability density function

  1. #1
    Junior Member
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    probability density function

    Hey,

    Can anyone help me with the following question:

    f(x) = K(1+x)^n such that x>0
    and
    f(x) = 0 otherwise.

    for some constants K and n.

    For what values of n can f be made into a valid probability density?

    Any help would be greatly appreciated.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jedoob
    Hey,

    Can anyone help me with the following question:

    f(x) = K(1+x)^n such that x>0
    and
    f(x) = 0 otherwise.

    for some constants K and n.

    For what values of n can f be made into a valid probability density?

    Any help would be greatly appreciated.
    A function is defined as follows:

    $\displaystyle
    f(x)=\left{ \begin{array}{cc} K(1+x)^n & \mbox{,\ if}\ x>0\\
    0 & \mbox{,\ if}\ x \le 0 \end{array}\right
    $,

    for some $\displaystyle K$ and $\displaystyle n$.

    For what values of $\displaystyle n$ can $\displaystyle f$ be made into a valid probability density.

    Now $\displaystyle f$ is a probability density if and only if $\displaystyle f(x)\ge\ 0$ for all $\displaystyle x$, and

    $\displaystyle
    \int_{-\infty}^{\infty}f(x) dx = 1
    $.

    Now the integral is infinite when $\displaystyle n \ge \ 0$, and if $\displaystyle n < \ 0$:

    $\displaystyle
    \int_{-\infty}^{\infty}f(x) dx = \frac{-K}{n+1}=1
    $.

    Now $\displaystyle K > \ 0$ as otherwise $\displaystyle f(x)$ would not be be
    greater than or equal to zero. Then $\displaystyle n+1=-K$, which can only
    be made to work if $\displaystyle n <0$.

    RonL
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  3. #3
    Junior Member
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    Thanks

    Thanks a million.
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