Question

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.

Answer 1

Answer:

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