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
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).
5 answers
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)
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)
3. Official answer
1/(2*(2-p))
1/(2*(2-p))
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)
Can you please show me how did you do?
(answer 3)
@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)
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)