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Thread: Continuous random variables and putting them all together

  1. #1
    Senior Member dokrbb's Avatar
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    Continuous random variables and putting them all together

    [the data is modified so that it keeps the main principle of the problem I'm working on but it's not identical]

    I have a $k$ number of boxes manufactured and the length of each side of the box is determined by a random var with a density function
    $f(x) = \frac{5x}{3},$ for $0 \leq x \leq 1$.

    I need to find the mean of the volume of a box.

    So, consider the box is a cube and its volume is another random variable, define it $g(x) = V = x^3$

    I know that I can find the cumulative distribution function of $g(x)$ by letting $Y = g(x)$, then
    $F_y(y) = P(g(x) \leq y) = p(x \in g^{-1}(-\infty, y] )$, so I assume that the next step is
    $ = \int_{-infty}^{y} x^{1/3}dx = ...$, but it's kind of weird since the improper integral I get will not converge...

    Then the p.d.f is the derivative of c.d.f, but I am a bit confused about how in general I should deal with random values like this, when once comes from the other.

    Would greatly appreciate some help,
    Thanks
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  2. #2
    MHF Contributor
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    Re: Continuous random variables and putting them all together

    There's too many issues with this problem to begin.

    $f(x)$ does not integrate to 1 across it's range of support.

    You don't mention if the side lengths are independent of one another. You seem to think they will all be the same length.

    Fix the problem statement and repost it please.
    Thanks from topsquark
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