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Math Help - Random Variable Distribution Function

  1. #1
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    Random Variable Distribution Function

    Let X be a random variable with distribution function mX(x) defined by

    mX(−1) = 1/5, mX(0) = 1/5, mX(1) = 2/5, mX(2) = 1/5 .


    (a) Let Y be the random variable defined by the equation Y = X + 3. Find the distribution function mY (y) of Y .

    Solution:
    This is what I put down, but I don't think it's right.

    mX(−1) = my(−1+3) = my(2) = 1/5,
    mX(0) = my(0+3) = my(3) = 1/5,
    mX(1) = my(1+3) = my(4) = 2/5,
    mX(2) = my(2+3) = my(5) = 1/5.

    Answer: my(2) = 1/5, my(3) = 1/5, my(4) = 2/5, my(5) = 1/5.


    (b) Let Z be the random variable defined by the equation Z = X2. Find the distribution function mZ(z) of Z.

    Solution:
    mX(−1) = mz(−12) = mz (1) = 1/5,
    mX(0) = mz(02) = mz(0) = 1/5,
    mX(1) = mz(12) = mz(1) = 2/5,
    mX(2) = mz(22) = mz(4) = 1/5

    mz(−12) = mz(12) → mz(1) = 1/5 + 2/5 = 3/5

    Answer:
    mz(0) = 1/5, mz(1) = 3/5, mz(4) = 1/5.


    I'm completely confused on how the equation Y = X + 3 relates back to the distribution function. Thanks.
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  2. #2
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    Quote Originally Posted by dexiangyu View Post
    Let X be a random variable with distribution function mX(x) defined by

    mX(−1) = 1/5, mX(0) = 1/5, mX(1) = 2/5, mX(2) = 1/5 .

    (a) Let Y be the random variable defined by the equation Y = X + 3. Find the distribution function mY (y) of Y .

    Solution:
    This is what I put down, but I don't think it's right.

    mX(−1) = my(−1+3) = my(2) = 1/5,
    mX(0) = my(0+3) = my(3) = 1/5,
    mX(1) = my(1+3) = my(4) = 2/5,
    mX(2) = my(2+3) = my(5) = 1/5.

    Answer: my(2) = 1/5, my(3) = 1/5, my(4) = 2/5, my(5) = 1/5.


    (b) Let Z be the random variable defined by the equation Z = X2. Find the distribution function mZ(z) of Z.

    Solution:
    mX(−1) = mz(−12) = mz (1) = 1/5,
    mX(0) = mz(02) = mz(0) = 1/5,
    mX(1) = mz(12) = mz(1) = 2/5,
    mX(2) = mz(22) = mz(4) = 1/5

    mz(−12) = mz(12) → mz(1) = 1/5 + 2/5 = 3/5

    Answer:
    mz(0) = 1/5, mz(1) = 3/5, mz(4) = 1/5.


    I'm completely confused on how the equation Y = X + 3 relates back to the distribution function. Thanks.
    It all looks OK.
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