Question

Let X, Y, and Z be independent random variables, where X is Bernoulli with parameter 1/3, Y is exponential with parameter 2, and Z is Poisson with parameter 3.

(a) Consider the new random variable U = XY + (1 - X)Z. Find the transform associated with U.
(b) Find the transform associated with 2Z + 3.
(c) Find the transform associated with Y + Z.

Answer 1

To find the transforms associated with the random variables U, 2Z + 3, and Y + Z, we need to use the moment generating functions (MGFs) of the respective random variables. By applying the properties of MGFs for independent random variables and substituting the appropriate values, we can find the transforms.

Step-by-step explanation:

a. Transform associated with U:

The transform of a random variable is its moment generating function (MGF). To find the MGF of U, we can use the properties of MGFs for independent random variables.

Since X, Y, and Z are independent, the transform associated with U is the product of their MGFs:

MU(t) = MX(t) * MY(t) * MZ(t)

For a Bernoulli random variable with parameter p, the MGF is given by:

MX(t) = p * et + (1-p)

For an exponential random variable with parameter λ, the MGF is given by:

MY(t) = 1 / (1 - 2 * λ * t)

For a Poisson random variable with parameter λ, the MGF is given by:

MZ(t) = exp(λ * (et-1))

Substituting the values of p, λ, and t into the respective MGFs, we can find the transform associated with U.

b. Transform associated with 2Z + 3:

Similar to part (a), we can find the transform associated with 2Z + 3 by substituting the value of t into the MGF of Z, which is 2 * MZ(t).

c. Transform associated with Y + Z:

To find the transform associated with Y + Z, we can use the property that the transform of a sum of independent random variables is the product of their transforms.

MX(t) * MY(t)

MZ(t)

Answer 2

To find the transforms associated with the new random variables U, W, and V, we need to consider the different cases based on the values of X. The transform associated with U is (1-X)e^(3t) + X(2/(2-t)). The transform associated with W is e^(7t). The transform associated with V is 2/(2-t) + e^(3t).

Step-by-step explanation:

a) To find the transform associated with U, we need to find the probability distribution function (PDF) of U. We know that X, Y, and Z are independent random variables. The PDF of U can be found by considering the two possible cases when X = 0 and X = 1:

Case 1: X = 0
In this case, U = Z. The PDF of Z is Poisson with parameter 3, so the transform associated with U is e^(3t).

Case 2: X = 1
In this case, U = Y. The PDF of Y is exponential with parameter 2, so the transform associated with U is 2/(2-t).

Combining the two cases, the transform associated with U is (1-X)e^(3t) + X(2/(2-t)).

b) If we let W = 2Z + 3, the transform associated with W can be found by taking the transform of 2Z and adding to it the transform of 3. The transform of 2Z is (e^(2t))^2 = e^(4t), and the transform of 3 is e^(3t). Therefore, the transform associated with 2Z + 3 is e^(4t) * e^(3t) = e^(7t).

c) If we let V = Y + Z, the transform associated with V can be found by taking the transform of Y and adding to it the transform of Z. The transform of Y is 2/(2-t), and the transform of Z is e^(3t). Therefore, the transform associated with Y + Z is 2/(2-t) + e^(3t).