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).
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
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)
3. Official answer
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)
@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)
To find the joint PMF pX,Y,Z(x,y,z), we need to consider the given conditions for each case.
If x is odd and (y,z)∈{(0,0),(0,2),(2,0),(2,2)}, the joint PMF is defined as follows:
pX,Y,Z(x,y,z) = 1/|S|, where |S| represents the cardinality of set S.
Since |S| = 4, the joint PMF is:
pX,Y,Z(x,y,z) = 1/4, for x odd and (y,z)∈{(0,0),(0,2),(2,0),(2,2)}.
If x is even and (y,z) = (0,0), the joint PMF is defined as follows:
pX,Y,Z(x,y,z) = 1, for x even and (y,z) = (0,0).
For any other combination of x, y, and z, the joint PMF pX,Y,Z(x,y,z) is 0.
Next, to find pX,Y(x,2) when x is odd, we need to sum up the joint PMFs over all possible values of z, which are (0,0), (0,2), (2,0), and (2,2).
pX,Y(x,2) = ∑pX,Y,Z(x,2,z) for z∈{0,2}
Since pX,Y,Z(x,y,z) = 1/4 for x odd and (y,z)∈{(0,0),(0,2),(2,0),(2,2)}, substituting the values:
pX,Y(x,2) = ∑(1/4) for z∈{0,2}
pX,Y(x,2) = (1/4) + (1/4)
pX,Y(x,2) = 1/2
For pY(2), we need to sum up the joint PMFs over all possible values of x and z, where y = 2.
pY(2) = ∑pX,Y,Z(x,2,z) for x and z
Since pX,Y,Z(x,y,z) = 1/4 for x odd and (y,z)∈{(0,0),(0,2),(2,0),(2,2)}, and pX,Y,Z(x,y,z) = 1 for x even and (y,z) = (0,0), substituting the values:
pY(2) = (∑(1/4) for x, z) + (∑1 for x and z)
pY(2) = (8/4) + 0
pY(2) = 2
Finally, to find var(Y+Z|X=5), we need to calculate the conditional expectation E(Y+Z|X=5) and then use the formula for variance.
First, let's find the conditional PMF pY+Z|X(5, y+z), where x = 5:
pY+Z|X(5, y+z) = pY,Z|X(5, y, z) / pX(5)
Since pX(5) is the probability of X = 5 and X is geometric with parameter p, we have:
pX(5) = (1-p)^(5-1) * p = p^4 * (1-p)
Now, let's consider the cases:
1. If y+z = 0, then (y,z) = (0,0):
pY+Z|X(5, 0) = pY,Z|X(5, 0, 0) / pX(5)
pY+Z|X(5, 0) = (1/4) / (p^4 * (1-p))
2. If y+z = 2, then (y,z) can be either (0,2) or (2,0):
pY+Z|X(5, 2) = pY,Z|X(5, 0, 2) + pY,Z|X(5, 2, 0) / pX(5)
pY+Z|X(5, 2) = (1/4) / (p^4 * (1-p)) + (1/4) / (p^4 * (1-p))
Now, we can calculate the conditional expectation E(Y+Z|X=5):
E(Y+Z|X=5) = ∑(y+z) * pY+Z|X(5, y+z)
E(Y+Z|X=5) = 0 * pY+Z|X(5, 0) + 2 * pY+Z|X(5, 2)
E(Y+Z|X=5) = 0 * ((1/4) / (p^4 * (1-p))) + 2 * ((1/4) / (p^4 * (1-p)))
Finally, to find var(Y+Z|X=5), we use the formula:
var(Y+Z|X=5) = E((Y+Z - E(Y+Z|X=5))^2|X=5)
Note: The calculations for var(Y+Z|X=5) involve more computations which are specific to the parameters p and the conditions given. These calculations can be done using the probability laws and properties specific to the experiment and random variables involved.