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Math Help - method of transformation question

  1. #1
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    method of transformation question

    I'm wondering if someone can check me work to see if I in fact have the correct solution.

    f(y_1,y_2) = \left\{ \begin{array}{rcl}<br />
1 & \mbox{for} & 0\leq y_1 \leq 2, \ 0\leq y_2 \leq 1, \ 2y_2 \leq y_1 \\ 0 & \mbox{otherwise}<br />
\end{array}\right.

    U=Y_1-Y_2

    1-\int^2_u \int^{y_1-u}_0 1 \ dy_2 \ dy_1

    1-\int^2_u y_2 \bigg{|}^{y_1-u}_0  \ dy_1

    1-\int^2_u y_1-u  \ dy_1

    1-\left( \frac{{y_1}^2}{2} -uy_1\right) \bigg{|}^2_u

    1-\left( 2-2u+\frac{u^2}{2} \right)

    -1+2u-\frac{u^2}{2}

    F_U(u) = \left\{ \begin{array}{rcl}<br />
0 & \mbox{for} & u < 2-\sqrt{2}\\ <br />
-1+2u-\frac{u^2}{2} & \mbox{for} & 2-\sqrt{2} \leq u \geq 2+\sqrt{2} \\<br />
1 & \mbox{for} & u > 2+\sqrt{2}<br />
\end{array}\right.
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  2. #2
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    A remark: the density f is positive on a neighbourhood of (0,0), so that Y_1 and Y_2 can both be close to zero. So that the probability density function of U=Y_1-Y_2 must be positive near zero. This does not hold with your solution. In addition, it seems you forgot the condition 2y_2\leq y_1, didn't you?

    You integral is in fact: if 0\leq u\leq 2, P(Y_1-Y_2>u)=\int_u^2 \int_0^{\min(y_1-u, \frac{y_1}{2})}1 dy_2\,dy_1, I think.
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  3. #3
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    I was going more along the lines of the example in the book, which solves a problem like that in a similar fashion to what I have done.
    I was thinking that I can just remove the bottom red triangle which goes from u to 2 on the y_1 line and 0 to y_1-u leaving me only in the area in blue, which would be easier for me to solve, since I've never seen an integral that has a bound of min.
    Attached Thumbnails Attached Thumbnails method of transformation question-graph.jpg  
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  4. #4
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    Quote Originally Posted by lllll View Post
    I'm wondering if someone can check me work to see if I in fact have the correct solution.

    f(y_1,y_2) = \left\{ \begin{array}{rcl}<br />
1 & \mbox{for} & 0\leq y_1 \leq 2, \ 0\leq y_2 \leq 1, \ 2y_2 \leq y_1 \\ 0 & \mbox{otherwise}<br />
\end{array}\right.

    U=Y_1-Y_2

    1-\int^2_u \int^{y_1-u}_0 1 \ dy_2 \ dy_1

    1-\int^2_u y_2 \bigg{|}^{y_1-u}_0 \ dy_1

    1-\int^2_u y_1-u \ dy_1

    1-\left( \frac{{y_1}^2}{2} -uy_1\right) \bigg{|}^2_u

    1-\left( 2-2u+\frac{u^2}{2} \right)

    -1+2u-\frac{u^2}{2}

    F_U(u) = \left\{ \begin{array}{rcl}<br />
0 & \mbox{for} & u < 2-\sqrt{2}\\ <br />
-1+2u-\frac{u^2}{2} & \mbox{for} & 2-\sqrt{2} \leq u \geq 2+\sqrt{2} \\<br />
1 & \mbox{for} & u > 2+\sqrt{2}<br />
\end{array}\right.
    Take y_1 as the vertical axis and y_2 as the horizontal axis. Draw the lines y_1 = 2y_2 and y_1 = y_2 + u over 0 \leq y_1 \leq 2 and 0 \leq y_2 \leq 1 . The lines intersect at (u, 2u).

    To get F(u) = \Pr(U < u) = \Pr(Y_1 - Y_2 < u) = \Pr(Y_1 < Y_2 + u) you need to integrate the given joint pdf over the region bounded by the Y_1 axis and the lines y_1 = 2y_2 and y_1 = y_2 + u:

    F(u) = \int_{y_2 = 0}^{y_2 = u} \int_{y_1 = 2y_2}^{y_1 = y_2 + u} 1 \, dy_1 \, dy_2 for 0 \leq u \leq 1 and zero elsewhere.
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  5. #5
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    Because Y_1 can be close to 2 and Y_2 close to 0 simultaneously, we must expect that the density of U is positive near 2 (from below).

    If 1\leq u\leq 2, the lines intersect differently: the line y_1=y_2+u meets y_1=2 at y_2=2-u, so that (look at the graph): P(Y_1\leq Y_2+u)=\int_0^{2-u}\int_{2y_2}^{y_1+u}dy_1\,dy_2+\int_{2-u}^1\int_{2y_2}^2 dy_1\,dy_2.

    And P(U\leq u)=1 if u>2, P(U\leq u)=0 if u\leq 0.
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  6. #6
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    Quote Originally Posted by Laurent View Post
    Because Y_1 can be close to 2 and Y_2 close to 0 simultaneously, we must expect that the density of U is positive near 2 (from below).

    If 1\leq u\leq 2, the lines intersect differently: the line y_1=y_2+u meets y_1=2 at y_2=2-u, so that (look at the graph): P(Y_1\leq Y_2+u)=\int_0^{2-u}\int_{2y_2}^{y_1+u}dy_1\,dy_2+\int_{2-u}^1\int_{2y_2}^2 dy_1\,dy_2.

    And P(U\leq u)=1 if u>2, P(U\leq u)=0 if u\leq 0.
    It occured to me that I overlooked about 1 < u < 2. I was happy to see you'd done the work for me when I came back to rectify my oversight
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