contestada

(4.1.4) Let X and Y be Bernoulli random variables. Let Z = X + Y. a. Show that if X and Y cannot both be equal to 1, then Z is a Bernoulli random variable. b. Show that if X and Y cannot both be equal to 1, then pZ = pX + pY. c. Show that if X and Y can both be equal to 1, then Z is not a Bernoulli random variable.

Respuesta :

Olajidey

Step-by-step explanation:

Given that,

a)

X ~ Bernoulli [tex](p_x)[/tex] and Y ~ Bernoulli [tex](y_x)[/tex]

X + Y = Z

The possible value for Z are Z = 0 when X = 0 and Y = 0

and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0

If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,

Therefore Z is a Bernoulli random variable

b)

If X and Y can not be both equal to  1

then,

[tex]p_z = P(X=1[/tex] or [tex]Y=1)\\[/tex]

[tex]p_z = P(X=1)+P(Y=1)-P(=1[/tex] and [tex]Y =1)[/tex]

[tex]p_z = P(x=1)+P(Y=1)\\\\p_z=p_x+p_y[/tex]

c)

If both X = 1 and Y = 1 then Z = 2

The possible values of the random variable Z are 0, 1 and 2.

since a  Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution

Otras preguntas