Let X and Y be independent random variables, each uniformly distributed on the interval [0,1].

Let Z=max{X,Y}. Find the PDF of Z. Express your answer in terms of z using standard notation .

For 0<z<1, fZ(z)= 1 - unanswered
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Let Z=max{2X,Y}. Find the PDF of Z. Express your answer in terms of z using standard notation .

For 0<z<1, fZ(z)= - unanswered

For 1<z<2, fZ(z)= - unanswered

To find the probability density function (PDF) of Z in each case, we need to determine the cumulative distribution function (CDF) first. Then, we can differentiate the CDF to obtain the PDF.

1. Let's find the CDF of Z = max{X, Y}.
For 0 < z < 1, the maximum of X and Y will be less than or equal to z only if both X and Y are less than or equal to z. Since X and Y are independent and uniformly distributed on [0,1], the probability of each being less than or equal to z is z.
Therefore, the CDF of Z is:
FZ(z) = P(Z ≤ z) = P(X ≤ z, Y ≤ z) = P(X ≤ z) * P(Y ≤ z) = z * z = z^2

2. Now, let's find the PDF of Z = max{2X, Y}.
For 0 < z < 1, the maximum of 2X and Y will be less than or equal to z only if both 2X and Y are less than or equal to z. Since Y is uniformly distributed on [0,1], the probability of Y being less than or equal to z is z.
For 0 < z < 1, the maximum of 2X and Y will be less than or equal to z if 2X is less than or equal to z. The probability of this event can be calculated as P(2X ≤ z) = P(X ≤ z/2) = z/2, since X is uniformly distributed on [0,1].
Therefore, the CDF of Z is:
FZ(z) = P(Z ≤ z) = P(max{2X,Y} ≤ z) = P(2X ≤ z, Y ≤ z) = P(2X ≤ z) * P(Y ≤ z) = (z/2) * z = z^2/2

3. Finally, let's find the PDF of Z = max{2X, Y} for 1 < z < 2.
For 1 < z < 2, the maximum of 2X and Y will be less than or equal to z only if both 2X and Y are less than or equal to z. Since Y is uniformly distributed on [0,1], the probability of Y being less than or equal to z is z.
For 1 < z < 2, the maximum of 2X and Y will be less than or equal to z if 2X is less than or equal to z. The probability of this event can be calculated as P(2X ≤ z) = P(X ≤ z/2) = z/2, since X is uniformly distributed on [0,1].
Therefore, the CDF of Z is:
FZ(z) = P(Z ≤ z) = P(max{2X,Y} ≤ z) = P(2X ≤ z, Y ≤ z) = P(2X ≤ z) * P(Y ≤ z) = (z/2) * z = z^2/2

To obtain the PDF, we can differentiate the CDF with respect to z.
Therefore, for 0 < z < 1, we have:
fZ(z) = d/dz (FZ(z)) = d/dz (z^2) = 2z

And for 1 < z < 2, we have:
fZ(z) = d/dz (FZ(z)) = d/dz (z^2/2) = z/2

So, the PDF of Z is:
For 0 < z < 1: fZ(z) = 2z
For 1 < z < 2: fZ(z) = z/2

Note that fZ(z) = 0 for z <= 0 or z >= 2, since Z is defined only on the interval [0, 2].