Question

Consider three random variables X, Y, and Z, associated with the same experiment. The random variable X is geometric with parameter p∈(0,1). If X is even, then Y and Z are equal to zero. If X is odd, (Y,Z) is uniformly distributed on the set S={(0,0),(0,2),(2,0),(2,2)}. The figure below shows all the possible values for the triple (X,Y,Z) that have X≤8. (Note that the X axis starts at 1 and that a complete figure would extend indefinitely to the right.)



Find the joint PMF pX,Y,Z(x,y,z). Express your answers in terms of x and p using standard notation .

If x is odd and (y,z)∈{(0,0),(0,2),(2,0),(2,2)},

pX,Y,Z(x,y,z)=

- unanswered

If x is even and (y,z)=(0,0),

pX,Y,Z(x,y,z)=

- unanswered

Find pX,Y(x,2), for when x is odd. Express your answer in terms of x and p using standard notation .
If x is odd,

pX,Y(x,2)=

- unanswered

Find pY(2). Express your answer in terms of p using standard notation .

pY(2)=

- unanswered

Find var(Y+Z∣X=5).

Answer 1

1.a. 0.25*p*(1-p)^(x-1)

1.b. p*(1-p)^(x-1)

2. 0.5*p*(1-p)^(x-1)

3. ¿ 0*p ? (not sure about this one)

4. 2

Answer 2

3) 1/2*1/(2-p)

pY(2)=pY(2 | Odd(x)) * P(Odd(x)) = 1/2 * P(Odd(x))
sum(p*(1-p)^x-1) for x = 1, 3, 5, ... = 1/(2-p)

Answer 3

3. Official answer

1/(2*(2-p))

Answer 4

I think I'm doing a silly error in the trasformation of the sum to get the final answer.

Can you please show me how did you do?

(answer 3)

Answer 5

@cle
we are trying to calculate the sum of (1-p)^(x-1)=1+(1-p)^2+(1-p)^4+....+(1-p)^2k=1+sum(1-p)^2k for k=1,2,.....
Thus (using the sum of geometric series formula we get => 1+(1-p)^2/(1-(1-p)^2)=1+(1-p)^2/(2p-p^2)=1/(2p-p^2)
pY(2)=1/2p*1/(2p-p^2)=1/2*1/(2-p)