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Thread: [SOLVED] P.D.F: Continuous random variables

  1. #1
    Super Member Quacky's Avatar
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    [SOLVED] P.D.F: Continuous random variables

    Firstly, I must apologise for the poor notation. I tried to use the \array command but I couldn't get it to work.

    1) The random variable Q has P.D.F given by:

    $\displaystyle f(q)=kq^3$ if $\displaystyle 0\leq q\leq 2$

    $\displaystyle f(q)=0$ otherwise

    Find k (answer: $\displaystyle \frac{1}{4}$)

    2)The random variable J has probability density function $\displaystyle f(j)=3j^k$ if $\displaystyle 0\leq j\leq 1 $

    $\displaystyle f(j)=0 $ otherwise

    Find k (answer: 2)

    I honestly don't know where to start because I missed a lesson and am trying to catch up, but reading and retaining information can be a lot harder than being taught it.
    Last edited by Quacky; Mar 15th 2010 at 02:05 PM.
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  2. #2
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    Quote Originally Posted by Quacky View Post
    Firstly, I must apologise for the poor notation. I tried to use the \array command but I couldn't get it to work.

    1) The random variable Q has P.D.F given by:

    $\displaystyle f(q)=kq^3$ if $\displaystyle 0\leq q\leq 2$

    $\displaystyle f(q)=0$ otherwise

    Find k (answer: $\displaystyle \frac{1}{4}$)

    2)The random variable J has probability density function $\displaystyle f(j)=3j^k$ if $\displaystyle 0\leq j\leq 1 $

    $\displaystyle f(j)=0 $ otherwise

    Find j (answer: 2)
    Hint $\displaystyle \int_0^2 {q^3 dq} = 4$
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Quacky View Post
    Firstly, I must apologise for the poor notation. I tried to use the \array command but I couldn't get it to work.

    1) The random variable Q has P.D.F given by:

    $\displaystyle f(q)=kq^3$ if $\displaystyle 0\leq q\leq 2$


    $\displaystyle f(q)=0$ otherwise


    Find k (answer: $\displaystyle \frac{1}{4}$)


    2)The random variable J has probability density function $\displaystyle f(j)=3j^k$ if $\displaystyle 0\leq j\leq 1 $

    $\displaystyle f(j)=0 $ otherwise

    Find j (answer: 2)

    I honestly don't know where to start because I missed a lesson and am trying to catch up, but reading and retaining information can be a lot harder than being taught it.
    1) If $\displaystyle f(q)=kq^3$ is the pdf of the random variable, then we know that

    $\displaystyle \int_0^2 {kq^3 dq} = 1$


    or, $\displaystyle k [\frac{q^4}{4}]_0^2 = 1$


    or, $\displaystyle k [\frac{2^4}{4} - \frac{0}{4}] = 1$

    or, $\displaystyle k \frac{16}{4} = 1$

    0r, $\displaystyle k = \frac{1}{4}$

    Try doing the same for your second problem. Show If you come up with any problems. By the way, are you supposed to find j or k in your second question? I assume you are supposed to find k.
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  4. #4
    Super Member Quacky's Avatar
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    Thankyou Plato and Harish, I understand now. I shall attempt 2) by myself.
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  5. #5
    Super Member Quacky's Avatar
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    Smile

    After a minor integration error, I've solved part 2). Thanks again
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