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Thread: P(x ≥ 0), e[y], v[y]

  1. #1
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    P(x ≥ 0), e[y], v[y]

    If continuous random variable X follow distribution P(x ≥ 0),   e[y],    v[y]-captura0118.jpg
    P(x ≥ 0),   e[y],    v[y]-captura0119.jpg, P(X ≥ 0), E[Y], V[Y]

    I calculate C is:
    1=∫[-∞,+∞]f(x)dx
    =∫[-1,+2]cx^2dx
    =9c/3,
    C =1/3
    but no idea how to calculate P(X ≥ 0), E[Y], V[Y]
    anyone can help?
    Attached Thumbnails Attached Thumbnails P(x ≥ 0),   e[y],    v[y]-captura0119.jpg  
    Last edited by Amanda2018; Nov 1st 2018 at 01:49 PM.
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  2. #2
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    Re: P(x ≥ 0), e[y], v[y]

    I believe:

    $$P(X\ge 0) = \int_0^\infty f(x)dx =\int_0^2 \dfrac{x^2}{3}dx = \dfrac{8}{9}$$

    $$E[Y] = E[X^2] = \int_{-\infty}^\infty x^2f(x)dx = \int_{-1}^2 \dfrac{x^4}{3}dx = \left. \dfrac{x^5}{15}\right|_{-1}^2 = \dfrac{11}{5}$$

    $$V[Y] = E[Y^2] - (E[Y])^2 = E[X^4] - (E[Y])^2 = \int_{-1}^2 \dfrac{x^6}{3}dx-\dfrac{121}{25} = \dfrac{43}{7}-\dfrac{121}{25} = \dfrac{228}{175}$$

    I could be wrong, but that is how I would do it.
    Thanks from Amanda2018
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  3. #3
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    Re: P(x ≥ 0), e[y], v[y]

    I think you are correct, Thanks!
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