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Math Help - Statistics - PDF, E(U), Probability

  1. #1
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    Statistics - PDF, E(U), Probability

    Gasoline is to be stocked in a bulk tank once at the beginning of
    each week and then sold the individual customers. Let X denote the
    proportion of the capacity of the bulk tank that is stocked at the
    start of the week. Let Y denote the proportion of the capacity of
    the bulk tank that is sold during the week. Because X and Y are both
    proportions, both variables take on values between 0 and 1. Further,
    the amount sold, Y, cannot exceed the amount available, X. Suppose
    that a joint pdf model for X and Y is given by:

    f(x,y) = 3x for 0 <= y <= x <= 1, 0 for elsewhere

    Find the pdf for U = X - Y, the proportional amount of gasoline
    remaining at the end of the week. Use the pdf of U to find the
    expected value of U, E(U). Use the pdf of U to find the probability
    that the proportional amount of gasoline remaining at the end of
    the week, U, is between 0.25 and 0.5.

    If anyone can help me get the problem started or guide me in the right direction, it would be greatly appreciated. I'm not sure how to start this problem.
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  2. #2
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    Re: Statistics - PDF, E(U), Probability

    $f_{XY}(x,y)=3x~~0 \leq y \leq x \leq 1$

    $F_U(u) = Pr[U < u] = Pr[X-Y<u] = Pr[X-u<Y]$

    $F_U(u)=\int_0^1 \int_{\max(x-u,0)}^x f_{XY}(x,y)~dy
    ~ dx = \int_0^1 \int_{\max(x-u,0)}^x 3x ~dy ~dx$ for $0 \leq u \leq 1$

    $f_U(u)=\frac{\partial}{\partial u}F_U(u)$

    this should get you started
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  3. #3
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    Re: Statistics - PDF, E(U), Probability

    Thank you so much for your help! That really helped a lot!!

    I got f(u) = 3/2 for 0 ≦ u ≦ 1, 0 elsewhere.

    Now using that information to get the E(u) is confusing to me and I'm lost once again.
    Last edited by AwesomeHedgehog; March 3rd 2014 at 11:46 AM.
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    Re: Statistics - PDF, E(U), Probability

    Would E(u) just equal the integral from 0 to 1 of f(u)*u?
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    Re: Statistics - PDF, E(U), Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    Thank you so much for your help! That really helped a lot!!

    I got f(u) = 3/2 for 0 ≦ u ≦ 1, 0 elsewhere.

    Now using that information to get the E(u) is confusing to me and I'm lost once again.
    That's not correct. Check your integration and be careful using the max function as a integration limit.

    and yes, once you have the proper $f_U(u)$ for $a \leq u \leq b$ then

    $E(U) = \int_a^b u f_U(u) du$
    Last edited by romsek; March 3rd 2014 at 02:45 PM.
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    Re: Statistics - PDF, E(U), Probability

    Wait, for the p.d.f. wouldn't the integral be: The integral from 0 to 1, then the integral 0 to x-u?

    Since f(x,y) is between 0 ≦ y ≦ x ≦ 1, the lower bound would be 0 on the integrals and the highest it goes up to would be y, which y = x-u
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    Re: Statistics - PDF, E(U), Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    Wait, for the p.d.f. wouldn't the integral be: The integral from 0 to 1, then the integral 0 to x-u?

    Since f(x,y) is between 0 ≦ y ≦ x ≦ 1, the lower bound would be 0 on the integrals and the highest it goes up to would be y, which y = x-u
    $X-Y<u\Rightarrow Y>X-u$

    $0\leq Y \leq 1$ so the lower limit of integration is $\max(x-u,0)$
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    Re: Statistics - PDF, E(U), Probability

    I'm not quite understanding it with the whole max(x-u, 0). Like, how would I even calculate that in the integral?
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    Re: Statistics - PDF, E(U), Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    I'm not quite understanding it with the whole max(x-u, 0). Like, how would I even calculate that in the integral?
    Ok let's rework it slightly differently.

    $X-Y<u$ and $0\leq Y \leq 1$ so for a given u

    $X \leq u$ and as always $Y \leq X$

    so this leads to a slightly different but equivalent integral of (see if you can work it yourself)

    Spoiler:
    $F_U(u)=Pr[U < u]=Pr[X-Y<u]=\displaystyle{\int_0^u \int_0^x} 3x ~dx=u^3, ~~ \mbox{for }0\leq u \leq 1$

    $f_U(u)=\dfrac{\partial}{\partial u} F_U(u) = 3u^2, ~~ \mbox{for }0\leq u \leq 1$

    given this you should be able to complete the problem.
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    Re: Statistics - PDF, E(U), Probability

    Alright, then for E(u) I got:

    E(u) = the integral from 0 to 1 (3u^2)*u du = 3/4

    P[ 0.25 < u < 0.5 ] = the integral from 0.25 to 0.5 (3/4) du = 0.1875
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  11. #11
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    Re: Statistics - PDF, E(U), Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    Alright, then for E(u) I got:

    E(u) = the integral from 0 to 1 (3u^2)*u du = 3/4
    correct

    P[ 0.25 < u < 0.5 ] = the integral from 0.25 to 0.5 (3/4) du = 0.1875
    No.

    to find the probability you integrate the pdf, not the expectation. The expectation is just a number.

    you want $\displaystyle{\int_{\frac{1}{4}}^{\frac{1}{2}}} 3 u^2~du$
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  12. #12
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    Re: Statistics - PDF, E(U), Probability

    Oh, I had it right the first time I did it on my paper, then changed it. Thank you so much for your help!
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