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Thread: n-th percentile

  1. #1
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    n-th percentile

    I need to find 10th percentile of random variable X with PDF defined with:

    $\displaystyle
    f(x) = \left\{\begin{array}{cc}c(x-x^3)&0< x < 1\\0&x\le 0 \vee x\ge 1 \end{array}\right.
    $

    where $\displaystyle c\in \mathbb{R}$ is const that should be found too.
    By integrating f from 0 to $\displaystyle x$ I got cumulative dist. function $\displaystyle F(x) = c \frac{2x^2 - x^4}{4}$, so $\displaystyle c=4$.

    First, is this approach OK, and second, how can I calculate required percentile?

    Thanks guys
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  2. #2
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    Your method for finding the CDF is correct(but i didn't check you did the integral properly).


    the 10 percentile $\displaystyle x_{10}$ satisfies:

    $\displaystyle F(x_{10}) = 0.1$
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  3. #3
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    I'm little bit confused with notation $\displaystyle x_{10}$, can you write complete calculation pls?
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  4. #4
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    i have just called the 10th percentile $\displaystyle x_{10}$

    The definition of a percentile is:

    $\displaystyle P(X<x_{10}) = 0.1$
    $\displaystyle F(x_{10}) = 0.1$

    So find the value of x that satisfies

    $\displaystyle 2x^2 - x^4 = 0.1$

    Can you finish from there?

    Hint: this is a quadratic in $\displaystyle x^2$
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  5. #5
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    Ok, thanks. So we take a root that is in interval $\displaystyle 0<x<1$ (we have 4 solutions, one in that interval)?
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  6. #6
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    yes, since x can not be outside that interval
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