# Thread: Random Variable Distribution Function

1. ## 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

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.

2. Originally Posted by dexiangyu
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

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.