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Thread: Continuous random variables – probability density functions

  1. #1
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    Continuous random variables – probability density functions

    The following could represent a probability density function (p.d.f.) or not. If could, find the value of k.

    $\displaystyle
    f(x) = \begin{cases}
    k(x + x^2),
    & \text {$-1 \le x \le 1,$} \\ 0, & \text {otherwise.} \end{cases}
    $


    How could I know it represent a p.d.f or not?
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    Newbie Black Kawairothlite's Avatar
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    evaluate if:

    $\displaystyle \int_{-\infty}^{\infty} f(x)dx=1$
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    Well, my problem is not to finding k. I’ve confusion to recognize is it p.d.f or not.

    If I try on this way,

    f(-0.5) = k(0.75)

    f(1) = 2k



    I didn’t understand how the above info/working or else will show

    f(x) < 0. So it could not be a p.d.f. (As my book’s answer.)
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    If $\displaystyle \int_{-1}^{1} f(x)dx=1$

    So: k=3/2.


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    Newbie Black Kawairothlite's Avatar
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    $\displaystyle \int_{-\infty}^{\infty} f(x)dx=1$

    you gotta "chop" the integral in three parts:

    $\displaystyle \int_{-\infty}^{-1} 0dx+\int_{-1}^{1} k(x + x^2)dx+\int_{1}^{\infty} 0dx=1$

    continue...
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  6. #6
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    Quote Originally Posted by geton View Post
    If $\displaystyle \int_{-1}^{1} f(x)dx=1$

    So: k=3/2.


    that's the right way can't say if that's the answer gotta do it but im gonna sleep atm
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    Answer:

    f(x) < 0.

    Therefore f(x) cannot be a probability density function.
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    Quote Originally Posted by geton View Post
    The following could represent a probability density function (p.d.f.) or not. If could, find the value of k.

    $\displaystyle
    f(x) = \begin{cases}
    k(x + x^2),
    & \text {$-1 \le x \le 1,$} \\ 0, & \text {otherwise.} \end{cases}
    $


    How could I know it represent a p.d.f or not?
    $\displaystyle f(x) = k(x + x^2)$ can only represent a pdf if it's non-zero for all values of x. This is clearly NOT the case therefore it's cannot represent a pdf.
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    Quote Originally Posted by mr fantastic View Post
    $\displaystyle f(x) = k(x + x^2)$ can only represent a pdf if it's non-zero for all values of x. This is clearly NOT the case therefore it's cannot represent a pdf.
    Why I'm trying to prove f(x) < 0? Totally time wasting...
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  10. #10
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    Quote Originally Posted by geton View Post
    The following could represent a probability density function (p.d.f.) or not. If could, find the value of k.

    $\displaystyle
    f(x) = \begin{cases}
    k(x + x^2),
    & \text {$-1 \le x \le 1,$} \\ 0, & \text {otherwise.} \end{cases}
    $


    How could I know it represent a p.d.f or not?
    Hi geton,

    One of the requirements for a pdf is that $\displaystyle f(x) \geq 0$ for all values of x. Now

    $\displaystyle x + x^2 < 0 \text{ for } -1 < x < 0$

    and

    $\displaystyle x + x^2 > 0 \text{ for } 0 < x < 1$,

    so if $\displaystyle k > 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle -1 < x < 0$, and if $\displaystyle k < 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle 0 < x < 1$, so no value of $\displaystyle k \neq 0$ will make f suitable for a pdf.

    The only remaining possibility is $\displaystyle k = 0$ but then $\displaystyle \int f(x) dx = 0$, so that won't work either.
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  11. #11
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    Quote Originally Posted by awkward View Post
    Hi geton,

    One of the requirements for a pdf is that $\displaystyle f(x) \geq 0$ for all values of x. Now

    $\displaystyle x + x^2 < 0 \text{ for } -1 < x < 0$

    and

    $\displaystyle x + x^2 > 0 \text{ for } 0 < x < 1$,

    so if $\displaystyle k > 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle -1 < x < 0$, and if $\displaystyle k < 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle 0 < x < 1$, so no value of $\displaystyle k \neq 0$ will make f suitable for a pdf.

    The only remaining possibility is $\displaystyle k = 0$ but then $\displaystyle \int f(x) dx = 0$, so that won't work either.
    Thank you so much. I really appreciate it.
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