Let X be a random variable with distribution function mXx de

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 .

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

Solution

a)

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)

mX(1) = mz(1^2) = mz (1) = 1/5,
mX(0) = mz(0^2) = mz(0) = 1/5,
mX(1) = mz(1^2) = mz(1) = 2/5,
mX(2) = mz(2^2) = mz(4) = 1/5

mz(1^2) = mz(1^2) mz(1) = 1/5 + 2/5 = 3/5

Answer:
mz(0) = 1/5,

mz(1) = 3/5,

mz(4) = 1/5.