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?
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.
a) To find the PDF of Z, we need to consider the cumulative distribution function (CDF) of Z. Since X and Y are independent random variables, the probability that Z is less than or equal to z is equal to the probability that both X and Y are less than or equal to z.
P(Z ≤ z) = P(X ≤ z and Y ≤ z)
Since X and Y are uniformly distributed on the interval [0,1], their probability density functions (PDFs) are constant on this interval and zero elsewhere. Therefore, we can write:
P(X ≤ z and Y ≤ z) = P(X ≤ z) * P(Y ≤ z) = z * z = z^2
To find the PDF of Z, we differentiate the CDF with respect to z:
f_Z(z) = d/dz [P(Z ≤ z)] = d/dz [z^2] = 2z
For 0 < z < 1, the value of f_Z(z) is given by:
f_Z(z) = 2z
b) Now, let Z = max{2X, Y}. We need to find the PDF of Z. Similar to part a), we consider the probability that Z is less than or equal to z:
P(Z ≤ z) = P(2X ≤ z and Y ≤ z)
To find this probability, we need to consider two cases:
1. If z < 1, then both 2X ≤ z and Y ≤ z should hold. Thus, we have:
P(2X ≤ z and Y ≤ z) = P(X ≤ z/2) * P(Y ≤ z) = (z/2) * z = z^2/2
2. If z ≥ 1, then both 2X ≤ z and Y ≤ z should hold. This is equivalent to:
P(2X ≤ z and Y ≤ z) = P(X ≤ z/2) * P(Y ≤ z) = (z/2) * 1 = z/2
To find the PDF of Z, we differentiate the CDF with respect to z:
f_Z(z) = d/dz [P(Z ≤ z)]
= d/dz [(z^2/2) + (z/2)]
= z
For 0 < z < 1, the value of f_Z(z) is given by:
f_Z(z) = z
c) For 1 < z < 2, the value of f_Z(z) for the assumption made in part b) is zero. This is because for values of z between 1 and 2, the condition 2X ≤ z cannot be satisfied for any value of X in the interval [0, 1]. Therefore, the probability of Z being less than or equal to z in this range is zero, and hence the PDF f_Z(z) is also zero.