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Math Help - Convolution of two independent random variables

  1. #1
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    Convolution of two independent random variables

    Consider two independent random variables
    X and Y:
    Let fX(x) = 1 -(x/2) if 0 <=x <=2 and 0 otherwise.
    Let fY (y) = 2 -2y for 0 <=y <=1 and 0 otherwise..

    otherwise. Find the probability density function of Z=
    X + Y:

    I am having problem in calculating the limits ......
    Please provide some help....
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  2. #2
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    In this case,
    0 \leq x \leq 2

    0 \leq y \leq 1

    So f_y(z-x) is only defined for 0 \leq z-x \leq 1

    \rightarrow z \geq x \geq z-1

    but in any case 0 \leq x \leq 2

    So i think the lower bound on x is max(z-1,0)
    and the upper bound is min(z,2)

    \displaystyle f_z(z) = \int_{x=max(z-1,0)}^{x=min(z,2)} f_x(x)f_y(z-x) dx

    for 0 < z < 3

    You need to evaluate the integral seperately for the cases where
    z < 1
    1< z <2
    2 < z < 3

    ....i think
    Last edited by SpringFan25; June 23rd 2010 at 02:59 PM.
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  3. #3
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    Quote Originally Posted by Pushpa View Post
    Consider two independent random variables





    X and Y:

    Let fX(x) = 1 -(x/2) if 0 <=x <=2 and 0 otherwise.
    Let fY (y) = 2 -2y for 0 <=y <=1 and 0 otherwise..
    otherwise. Find the probability density function of Z=X + Y:

    I am having problem in calculating the limits ......
    Please provide some help....
    Here is how I would do it if using convolutions:

    h(z) = \int_{- \infty}^{+ \infty} f_X(z - y) \cdot f_Y(y) \, dy = \int_0^1 f_X(z - y) (2 - 2y) \, dy


    since the support of f_Y(y) is 0 \leq y \leq 1 and where f_X(z - y) = \left\{ \begin{array}{ll} 1 - \frac{z-y}{2}, & 0 \leq z - y \leq 2 \Rightarrow -z \leq -y \leq 2 - z \Rightarrow z - 2 \leq y \leq z \\ 0 & \text{otherwise} \end{array} \right..


    Note that 0 \leq x \leq 2 and  0 \leq y \leq 1 and so it follows that the support of Z = X + Y is 0 \leq z \leq 3.


    To see the required integral limits, draw on the one set of y-z axes the lines y = 1, y = z and y = z - 2 over the domain 0 \leq z \leq 3. It follows that there are four cases (corresponding to the four regions defined by these lines) to consider:


    Case 1: 0 \leq z \leq 1. h(z) = \int_{0}^{z} \left( 1 - \frac{z - y}{2} \right) (2 - 2y) \, dy = ....


    Case 2: 1 < z \leq 2. h(z) = \int_{0}^{1} \left( 1 - \frac{z - y}{2} \right) (2 - 2y) \, dy = ....


    Case 3: 2 < z \leq 3. h(z) = \int_{z-2}^{1} \left( 1 - \frac{z - y}{2} \right) (2 - 2y) \, dy = ....


    Case 4: z < 0 or z > 3. h(z) = 0.
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