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 = X^2. Find the distribution function mZ (z) of Z.

my = (1/5, 1/5, 2/5, 1/5) for Y = (2,3,4,5)

mz = (1/5, 3/5, 1/5) for Z = (0,1,4)

Step-by-step explanation:

for Y

for X = (-1), Y = - 1+3=2, my=mx=1/5 → my (Y=2) = 1/5

for X=0, Y = 0+3=3, my=mx=1/5 → my (Y=3) = 1/5

for X=1, Y = 1+3=4, my=mx=2/5 → my (Y=4) = 2/5

for X=2, Y = 2+3=5, my=mx=1/5 → my (Y=5) = 1/5

for Z

for X = (-1), Z = (-1) ²=1, mz [X = (-1) ]=mx=1/5

for X=0, Z = 0²=0, mz=mx=1/5 → mz (Z=0) = 1/5

for X=1, Z = 1²=1, mz [X=1]=mx=2/5 → mz (Z=1) = 1/5 + 2/5 = 3/5

for X=2, Z = 2²=4, mz=mx=1/5 → mz (Z=4) = 1/5