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Math Help - valid probability density function?

  1. #1
    Member Jason Bourne's Avatar
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    valid probability density function?

    1. For each of the following functions how do you determine whether it is a valid probabilty density function??????????????

    (a) f(x) = \frac{x}{2} ,  ~~~~~~~~     0\leq x\leq 2
    (b) f(x) = 4e^{-4x}    ,  ~~~~~~~~~     x\geq 0
    (c) f(x) = e^{x} + 1   ,   ~~~~~~~~~    0\leq x \leq 1

    2.

    (a) Find the value of a that makes f(x) = ax(1-x) for 0\leq x\leq 1 a valid probability density function
    (b) Evaluate the mean of this distribution
    (c) Evaluate the variance of this distribution

    Thanks for help
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  2. #2
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    Quote Originally Posted by Jason Bourne View Post

    (a) f(x) = \frac{x}{2} ,  ~~~~~~~~     0\leq x\leq 2
    (b) f(x) = 4e^{-4x}    ,  ~~~~~~~~~     x\geq 0
    (c) f(x) = e^{x} + 1   ,   ~~~~~~~~~    0\leq x \leq 1
    A function \geq 0 is a probability density function when its integral is 1. So just get whether the integrals are 1. Just be careful on (b) the integral has limits 0 to +oo.

    2.

    (a) Find the value of a that makes f(x) = ax(1-x) for 0\leq x\leq 1 a valid probability density function
    (b) Evaluate the mean of this distribution
    (c) Evaluate the variance of this distribution
    (a) Solve the equation \int_0^1 ax(1-x) ~ dx = 1.
    (b) Once you find 'a' compute E[X]=\int_0^1 ax^2(1-x) ~ dx
    (c) Find \int_0^1 ax^2(1-x^2)~dx - E[X].

    EDIT: Mistake, but too lazy to fix it.
    Last edited by ThePerfectHacker; February 22nd 2008 at 10:33 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    A function \geq 0 is a probability density function when its integral is 1. So just get whether the integrals are 1. Just be careful on (b) the integral has limits 0 to +oo.


    (a) Solve the equation \int_0^1 ax(1-x) ~ dx = 1.
    (b) Once you find 'a' compute E[X]=\int_0^1 ax^2(1-x) ~ dx
    (c) Find \int_0^1 ax^2(1-x^2)~dx - E[X].
    Err.. hn..

    (c) Evaluate the variance of this distribution

    \sigma^2=E((X-\mu)^2)=E(X^2)-\mu^2

    or in your notation:

    \sigma^2=\int_0^1 ax^3(1-x)~dx - (E[X])^2

    RonL
    Last edited by CaptainBlack; February 23rd 2008 at 03:44 AM.
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  4. #4
    Member Jason Bourne's Avatar
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    Quote Originally Posted by CaptainBlack View Post

    \sigma^2=\int_0^1 ax^3(1-x^2)~dx - (E[X])^2

    RonL
    Sorry, shouldn't this be \sigma^2=\int_0^1 ax^3(1-x)~dx - (E[X])^2

    where does the  x^2 come from ?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Jason Bourne View Post
    Sorry, shouldn't this be \sigma^2=\int_0^1 ax^3(1-x)~dx - (E[X])^2

    where does the  x^2 come from ?
    Of course it should, just goes to show that you should not just copy someone else's LaTeX

    The x^2 comes from E(X^2)=\int_0^1 x^2 ~p(x) dx, where p(x) is the pdf of X

    RonL
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