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Math Help - Density Function [a,b] and integration

  1. #1
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    Density Function [a,b] and integration

    Hello
    Here's a question:
    Let X have a continuous uniform distribution on the interval [a,b]. Find the density function for this dristribution and then verify that \mu = \frac{(a + b)}{2} and \sigma^2 = \frac{(b - a)^2}{12} .
    ok
    So here's what I "think" I should do ...

    Let f(x) = 1/(b-a) = 1 assuming that a=0 and b=1
    For a <= x <= b

    I know I need to integrate, but I'm not sure how to set up the problem
    \mu = E(x) = \int xf(x)dx where f(x) = 1
    then
    \int xdx = \frac{1}{2}x^2 which equals 1/2 when pluging in [a.b]
    Verifying ...
    \mu = \frac {(a+b)}{2} you get \mu = \frac {(0+1)}{2} = \frac {1}{2}

    and then we have \sigma^2

    \sigma^2 = E((x-\mu)^2) = \int (x-\mu)^2f(x)dx
    so ...
    \int (x-\frac{1}{2})^2dx = \int x^2-x+\frac{1}{4}dx = \frac{x^3}{3}-\frac{x^2}{2}+\frac{x}{4}
    When you plug in the interval [0,1] you get \frac{1}{12}
    verifying \mu^2 You also get \frac{1}{12}

    Is this correct? wow, that was long, sorry, I just tried to do it myself without any help at first. Be nice to me, it's my first post and it's also my birthday! WOOHOO
    Thanks for all the help
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TaoWine View Post
    ...
    Is this correct? wow, that was long, sorry, I just tried to do it myself without any help at first. Be nice to me, it's my first post and it's also my birthday! WOOHOO
    Thanks for all the help
    I can't help you with your problem, but...

    HAPPY BIRTHDAY!!

    Someone else will help you shortly
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by TaoWine View Post
    Hello
    Here's a question:
    Let X have a continuous uniform distribution on the interval [a,b]. Find the density function for this dristribution and then verify that \mu = \frac{(a + b)}{2} and \sigma^2 = \frac{(b - a)^2}{12} .
    ok
    So here's what I "think" I should do ...

    Let f(x) = 1/(b-a) = 1 assuming that a=0 and b=1
    For a <= x <= b

    I know I need to integrate, but I'm not sure how to set up the problem
    \mu = E(x) = \int xf(x)dx where f(x) = 1
    then
    \int xdx = \frac{1}{2}x^2 which equals 1/2 when pluging in [a.b]
    Verifying ...
    \mu = \frac {(a+b)}{2} you get \mu = \frac {(0+1)}{2} = \frac {1}{2}

    and then we have \sigma^2

    \sigma^2 = E((x-\mu)^2) = \int (x-\mu)^2f(x)dx
    so ...
    \int (x-\frac{1}{2})^2dx = \int x^2-x+\frac{1}{4}dx = \frac{x^3}{3}-\frac{x^2}{2}+\frac{x}{4}
    When you plug in the interval [0,1] you get \frac{1}{12}
    verifying \mu^2 You also get \frac{1}{12}

    Is this correct? wow, that was long, sorry, I just tried to do it myself without any help at first. Be nice to me, it's my first post and it's also my birthday! WOOHOO
    Thanks for all the help
    I was looking forward to seeing what Jhevon had to say. Unfortunately, what I saw was at variance with my expectation .

    Here is my birthday present to you, TaoWine:

    The pdf of X is f(x) = \frac{1}{b-a} for a \leq x \leq b and zero elsewhere.

    Then

    \mu = E(X) = \int_{-\infty}^{+\infty} x f(x) \, dx = \int_{a}^{b} \frac{x}{b-a} \, dx = \frac{1}{b-a} \int_{a}^{b} x \, dx


     = \frac{1}{b-a} \left ( \frac{b^2 - a^2}{2} \right) = \frac{1}{b-a} \left ( \frac{(b - a)(b+a)}{2} \right) = \frac{b+a}{2}.


    By symmetry this expectation is totally expected .


    I leave you to find:

    1. E \left( X^2 \right).

    2. \sigma^2 = Var (X) = E (X^2) - \left [ E(X) \right]^2.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I was looking forward to seeing what Jhevon had to say. Unfortunately, what I saw was at variance with my expectation .
    puns intended

    i'm not the probability guy, so i don't even try sorry to disappoint you
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by TaoWine View Post
    Hello
    Here's a question:
    Let X have a continuous uniform distribution on the interval [a,b]. Find the density function for this dristribution
    Mr Fantastic has told you what the density function is and finished the question, but you are asked to find it, so you need to observe that:

    X has a continuous uniform distribution on the interval [a,b]

    Means that the pdf of X is a constant on [0,1] and zero everywhere else. So let the pdf of X be:

    f(x)=\left\{ c\ x \in [a,b] \atop 0\ {\rm{otherwise}} \right..

    To be a pdf we require that the integral of this over \mathbb{R} be 1. So:

     <br />
\int_{-\infty}^{\infty}f(x)\;dx= \int_{a}^{b}c\;dx=cb-ca=1<br />

    Hence:

    c=\frac{1}{b-a}

    RonL
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  6. #6
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    Jhevon, thanks for the birthday message

    mr fantastic, I can't thank you enough, I was able to get the second answer to that question at 4:00am this morning while I was working.

    CaptainBlack, also a big thanks!

    You guys Rock! Take care everyone
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