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Conditional Prob -cont random variable

  1. Jul 28, 2005 #1
    hello all

    I have been workin on some problems involving conditional probability and continuous random variables and the thing is i dont know if i get the limits correct, anyway here is the problem, check it out, any suggestions would be helpful

    [tex]f(y_1,y_2) =\left\{\begin{array}{cc}2,&\mbox{ if }
    0\le y_1\le 1, 0\le y_2\le 1, y_1+y_2\le 1\\0, & \mbox{elsewhere}\end{array}\right[/tex]


    what i want to find was
    [tex]P(Y_1\ge \frac{1}{2}|Y_2\le \frac{1}{4})[/tex]
    [tex]=\frac{\int_{0}^{\frac{1}{4}} \int_{\frac{1}{2}}^{1-y_2} 2 dy_1 dy_2}{\int_{0}^{\frac{1}{4}} \int_{0}^{1-y_2} 2 dy_1 dy_2}=\frac{3}{7}[/tex]

    also I wanted to find

    [tex]P(Y_1\ge \frac{1}{2}|Y_2=\frac{1}{4})[/tex]
    [tex]=\frac{\int_{\frac{1}{2}}^{\frac{3}{4}} 2 dy_1}{\int_{0}^{1} 2 dy_1}=\frac{1}{4}[/tex]
    about the 3/4 that is where the intersection occurs

    now have i got my limits correct? how do i know if i have the limits correct? are my answers corrrect?

    steven
     
  2. jcsd
  3. Jul 29, 2005 #2
    If you draw the triangle then the limits are simple. Your first problem becomes a ratio of areas and the second problem becomes a ratio of line segments. You will need to change the upper limit of the denominator of the second problem.
     
  4. Jul 29, 2005 #3
    hello there

    thanxs for that its an interesting way of looking at it, but I didnt get which areas and which segments should be put into ratio, so i have provided a diagram of my triangle, please point out which areas and segments are you refering to. from my understanding

    [tex]P(Y_1\ge \frac{1}{2}|Y_2\le \frac{1}{4})[/tex]

    [tex]=\frac{D}{A+B+C+D}=\frac{3}{16}[/tex]

    [tex]P(Y_1\ge \frac{1}{2}|Y_2=\frac{1}{4})[/tex]

    [tex]=\frac{ys}{xs}=1/3[/tex]

    is this what you mean?

    so was my answer for the first problem correct?
    and how do i determine my limits for the second problem?

    steven
     

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  5. Jul 30, 2005 #4
    Yes, except P(Y1>=0.5| Y2<=0.25)= D/(D+B)=3/7. The upper limit of integration in the denominator of your 2nd problem should be 3/4. xs=3/4 and ys=1/4 and
    P(Y1>=0.5|Y2=0.25)=(2*ys)/(2*xs)=1/3. Nice diagram.
     
  6. Jul 30, 2005 #5
    Actually P(Y1>=0.5| Y2<=0.25)= (2*D)/(2*(D+B))=3/7.
     
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