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Math Help - Is my answer sheet wrong?

  1. #1
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    Is my answer sheet wrong?

    pdf:

    f(x,y)=\left\{<br />
\begin{array}{lr}<br />
\frac{2x+y}{4}&0\le x\le 1;0\le y\le 2\\<br />
0&otherwise<br />
\end{array}<br />
\right.

    Question: Find cdf (remember to consider all cases).

    Answer:

     x \le 0 \, , F(x,y)=0
     y \le 0 \, , F(x,y)=0

    x > 1\, , y > 2, \, F(x,y)=1

    0 \le x \le 1, 0 \le y \le 2 :

    F(x,y)=\int_{0}^{x}\int_{0}^{y}\frac{2u+v}{4}\,dvd  u = \frac{xy}{8}(2x+y)

    Shouldn't that be it? What's with these next two cases if x and y are never greater than 1 and 2 respectively?


    0 \le x \le 1 \, , y \ge 2 \, : F(x,y)=\frac{x}{2}(x+1)

    0 \le y \le 2 \, , x \ge 1 \, , F(x,y)=\frac{y}{8}(2+y)
    Last edited by MathSucker; December 10th 2009 at 04:07 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Do you have a typo?
    It doesn't make sense that x is bounded above by 1
    and x is larger than y
    and yet you have 0<y<2
    clearly y cannot pass 1.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    Do you have a typo?
    It doesn't make sense that x is bounded above by 1
    and x is larger than y
    and yet you have 0<y<2
    clearly y cannot pass 1.
    You beat me to it. Actually I think there are typos ....
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  4. #4
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    oops. fixed.
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  5. #5
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    Quote Originally Posted by MathSucker View Post
    0 \le x \le 1 \, , y \ge 2 \, : F(x,y)=\frac{x}{2}(x+1)

    0 \le y \le 2 \, , x \ge 1 \, , F(x,y)=\frac{y}{8}(2+y)
    Even with the correction, what is the point of these two cases. Why would I be finding equations for cases where x and y are greater than 1 and 2 if x only takes values between 0 and 1 and y only takes values between 0 and 2?

    Do I fundamentally misunderstand this problem?
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  6. #6
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    Quote Originally Posted by MathSucker View Post
    Even with the correction, what is the point of these two cases. Why would I be finding equations for cases where x and y are greater than 1 and 2 if x only takes values between 0 and 1 and y only takes values between 0 and 2?

    Do I fundamentally misunderstand this problem?
    The pdf is defined for x > 1 and y > 2 (just as it's defined for x < 0 and y < 0). The question said all cases. Therefore ....
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  7. #7
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    So from the beginning do I just integrate f(x,y)
    to get F(x,y). And then find:

    F(0,y)=F(x,0)=F(0,0)=0 for x>0 and  y>0

    F(2,1)=1 for x>1 and  y>2

    F(x,y)=F(x,y) for 0 \le x \le 1 ; 0 \le y \le 2

    F(1,y)= (given above) for x>1 ; 0 \le y \le 2

    F(x,2)= (given above) for y>2 ; 0 \le x \le 1

    Correct?
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