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Conditional probability with marginal and joint density

  1. Dec 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Determine ##P(X<Y|x>0)##
    2. Relevant equations
    X and Y are random variables with the joint density function
    $$
    f_{XY}(x,y)=
    \begin{cases}
    4|xy|,-y<x<y,0<y<1\\
    0,elsewhere
    \end{cases}$$
    The marginal densities are given by
    $$
    f_X(x)=2x\\
    f_Y(y)=4y^3
    $$
    3. The attempt at a solution
    The formula for conditional probability is
    $$
    P(B|A)=\frac{P(B\cap A)}{P(A)}
    $$
    In this case we have
    $$
    P(X<Y|x>0)=\frac{P(X<Y\cap x>0)}{P(x>0)}=\frac{F_{XY}(x,y)}{F_X(x)}=\frac{\int_{}4|xy|dxdy}{\int_0^{\infty}2xdx}
    $$
    This is where I get stuck, I do not know what boundaries I should put for the joint density function, can someone please help me?
     
  2. jcsd
  3. Dec 12, 2015 #2

    LCKurtz

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    Draw the lines ##x=y##, ##x=-y##, ##y=0##, and ##y=1##. Shade the part where x and y are between the appropriate lines. The limits are just like in any double integral in calculus. Also, I didn't check your work but surely if the marginal density is ##2x##, it wouldn't be valid clear to ##\infty##.
     
  4. Dec 12, 2015 #3

    Ray Vickson

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    You claimed ##f_X(x) = 2x## cannot possibly be correct. In the sample space, ##x## is allowed to be both ##< 0## and ##>0##, and when ##x < 0## your formula delivers a negative probability.
     
  5. Dec 14, 2015 #4
    The shaded area is a inverted triangle in the upper half of the plane with y vertical and x horizontal, so we have that
    $$
    P(X<Y|x>0)=\frac{\int_0^1\int_{-y}^y4|xy|dxdy}{\int_0^12xdx}=\frac{1}{1}=1
    $$
    I guess this concludes this topic, thanks :)
     
  6. Dec 14, 2015 #5

    Ray Vickson

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    For the record: your final answer is correct, but you have computed ##f_X(x)## incorrectly. You should have gotten
    [tex] f_X(x) = \frac{1}{2} |x| ( 1 - x^2),\; 0 \leq x \leq 1 [/tex]
    Unlike your answer, this has ##f_X > 0## for both ##-1 < x < 0## and ##0 < x < 1##.
     
  7. Dec 14, 2015 #6

    LCKurtz

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    Did you mean $$2*|x| ( 1 - x^2),\; 0 \leq x \leq 1\text{ ?}$$
     
  8. Dec 14, 2015 #7

    Ray Vickson

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    Yes. I accidentally omitted the factor '4' in front, so got 1/2 instead of 4/2 = 2.
     
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