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Thread: 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,
    $\displaystyle 0 \leq x \leq 2$

    $\displaystyle 0 \leq y \leq 1$

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

    $\displaystyle \rightarrow z \geq x \geq z-1$

    but in any case $\displaystyle 0 \leq x \leq 2$

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

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

    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; Jun 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:

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


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


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


    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 $\displaystyle 0 \leq z \leq 3$. It follows that there are four cases (corresponding to the four regions defined by these lines) to consider:


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


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


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


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