a) To find the PDF of Z, we need to find the cumulative distribution function (CDF) of Z first, and then differentiate it to obtain the PDF.
Let's start by finding the CDF of Z:
F_Z(z) = P(Z ≤ z) = P(max{X,Y} ≤ z) = P(X ≤ z and Y ≤ z)
Since X and Y are independent, the probability that both X and Y are less than or equal to z is simply the product of their individual probabilities:
F_Z(z) = P(X ≤ z) * P(Y ≤ z) = z * z = z^2
To find the PDF, we differentiate the CDF:
f_Z(z) = d/dz (F_Z(z)) = d/dz (z^2) = 2z
For 0 < z < 1, the value of f_Z(z) is 2z.
b) Let's find the CDF of Z when Z = max{2X, Y}:
F_Z(z) = P(Z ≤ z) = P(max{2X, Y} ≤ z) = P(2X ≤ z and Y ≤ z)
To find the CDF, we need to consider two cases:
Case 1: 0 ≤ z ≤ 1
In this case, since Y takes values between 0 and 1 uniformly, the probability that Y is less than or equal to z is simply z:
F_Z(z) = P(2X ≤ z and Y ≤ z) = P(2X ≤ z) * P(Y ≤ z) = P(X ≤ z/2) * z = (z/2) * z = z^2/2
Case 2: 1 < z ≤ 2
In this case, since 2X takes values between 0 and 2 uniformly, the probability that 2X is less than or equal to z is simply z/2:
F_Z(z) = P(2X ≤ z and Y ≤ z) = P(2X ≤ z) * P(Y ≤ z) = P(X ≤ z/2) * z = (z/2) * z = z^2/2
Therefore, the CDF of Z for 0 ≤ z ≤ 1 is z^2/2, and for 1 < z ≤ 2 is also z^2/2.
To find the PDF, we differentiate the CDF:
For 0 < z < 1, the value of f_Z(z) is 2z.
For 1 < z < 2, the value of f_Z(z) is also 2z.
c) Considering the assumption in b), for 1 < z < 2, the value of f_Z(z) is also 2z. The PDF remains the same as in the previous part, as the probability is not affected by the upper limit of z.
Let X and Y be independent random variables, each uniformly distributed on the interval [0,1]
a) Let Z = max{X,Y}. Find the PDF of Z. Express your answer in terms of z using standard notation. What's the value of f_Z(z) for 0 < z < 1?
b) Let Z = max{2X,Y}. Find the PDF of Z. Express your answer in terms of z using standard notation. What's the value of f_Z(z) for 0 < z < 1?
c) Considering the assumption in b). What's the value of f_Z(z) for 1 < z < 2?
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