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Math Help - Have the following continuous random variable....

  1. #1
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    Have the following continuous random variable....

    Hi I know this is a relatively easy question but im a bit stuck, previous one i have calculated i have found the cumulative function and plug in the upper limit and equal it to 1 and from the find the constant. Now with this having no upper limit im not to sure. The rest of the question i know how to calculate, but you need to find the constant in order to work it out, the question is as follows:
    f(x)=c/(x^3) for x greater or equal to 1
    0 otherwise.

    1) find the value of the constant c.

    Many thanks in advance.
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  2. #2
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    Quote Originally Posted by breitling View Post
    Hi I know this is a relatively easy question but im a bit stuck, previous one i have calculated i have found the cumulative function and plug in the upper limit and equal it to 1 and from the find the constant. Now with this having no upper limit im not to sure. The rest of the question i know how to calculate, but you need to find the constant in order to work it out, the question is as follows:
    f(x)=c/(x^3) for x greater or equal to 1
    0 otherwise.

    1) find the value of the constant c.

    Many thanks in advance.
    "for x greater or equal to 1" means 1 \leq x < + \infty.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    "for x greater or equal to 1" means 1 \leq x < + \infty.
    Exactly what i thought, from this would you find the continuous function, then use x=1?
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  4. #4
    MHF Contributor harish21's Avatar
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    Remember the properties of a probability density function..

    and since the random variable is continuous,

    \displaystyle \int_1^{\infty} \dfrac{c}{x^3}\;dx\;=\;1

    find c
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  5. #5
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    Quote Originally Posted by harish21 View Post
    Remember the properties of a probability density function..

    and since the random variable is continuous,

    \displaystyle \int_1^{\infty} \dfrac{c}{x^3}\;dx\;=\;1

    find c
    Yes, and since it's an improper integral care must be taken in evaluating it (using limits etc.).

    Quote Originally Posted by breitling View Post
    Exactly what i thought, from this would you find the continuous function, then use x=1?
    You ought to have examples in your textbook or class notes on this type of situation.
    Last edited by mr fantastic; January 13th 2011 at 12:56 PM. Reason: Merged posts.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Yes, and since it's an improper integral care must be taken in evaluating it (using limits etc.).


    You ought to have examples in your textbook or class notes on this type of situation.
    Unfortunately to my surprise i havent, we only have them them for the kind i have said find the cumulative and then use the upper limit.
    So honestly haven't got a clue, any help would be most appreciated.
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  7. #7
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    Would i be correct to think that the integral between infinity and 0 is:
    -c/(2x^2)=1
    ?
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  8. #8
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    Quote Originally Posted by breitling View Post
    Would i be correct to think that the integral between infinity and 0 is:
    -c/(2x^2)=1
    ?
    No.

    \displaystyle 1 = \int_1^{+\infty} \frac{c}{x^3} \, dx = \lim_{\alpha \to +\infty} \int_1^{\alpha} \frac{c}{x^3} \, dx = ....
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