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Thread: Continuous random variables - corrected

  1. #1
    Senior Member dokrbb's Avatar
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    Continuous random variables - corrected

    [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
    $f(x) = 5x^{4},$ for $0 \leq x \leq 1$.

    $\textbf{While about the independence of the side lengths, in the problem says nothing; I assumed they are independent since we are given a pdf for the side length.}$

    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.
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  2. #2
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    Re: Continuous random variables - corrected

    are the boxes cubes or rectangular prisms, with 3 independent side lengths?

    If the latter the problem is a bit complicated.
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  3. #3
    Senior Member dokrbb's Avatar
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    Re: Continuous random variables - corrected

    let's consider them as cubes, that's why I defiend the random value for volume being $X^{3}$
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    Re: Continuous random variables - corrected

    Quote Originally Posted by dokrbb View Post
    let's consider them as cubes, that's why I defiend the random value for volume being $X^{3}$
    ok

    The simple way to do this is just

    $E[V] = E[X^3] = \displaystyle{\int_0^1}~x^3 (5x^4)~dx = \dfrac 5 8$

    As a check on this we can go through the more involved route of finding the actual PDF of $V$ and finding the expected value directly from that.

    $F_V(v) = P[V<v] = P[X^3 < v] = P[X < \sqrt[3]{v}] = F_X(\sqrt[3]{v})$

    $F_X(x) = \int~f_X(x)~dx = x^5,~x\in [0,1]$

    $F_V(v) = F_X(\sqrt[3]{v}) = v^{5/3},~v \in [0,1]$

    $f_V(v) = \dfrac{d}{dv} F_V(v) = \dfrac 5 3 v^{2/3},~v \in [0,1]$

    almost there

    $E[V] = \displaystyle{\int_0^1}~v f_V(v)~dv = \displaystyle{\int_0^1}~v \dfrac 5 3 v^{2/3}~dv = \displaystyle{\int_0^1}~\dfrac 5 3 v^{5/3}~dv = \dfrac 5 8$
    Last edited by romsek; Mar 10th 2017 at 06:33 AM.
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    Senior Member dokrbb's Avatar
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    Re: Continuous random variables - corrected

    OH, the solution is awesome, Thank you very much!
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  6. #6
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    Re: Continuous random variables - corrected

    Romsek's first easy solution is certainly correct. It appears that you are'nt aware of the general fact: Given any r.v. X with density f(x) (continuous or discrete), and any function g from R to R, the following is true:
    $$E(g(X))=\int_{-\infty}^{\infty}g(x)f(x)\,dx$$
    If X is discrete, the above integral is just a sum, maybe an infinite sum.
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