Let Z=max{3X,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 PDF of Z, we need to consider the different cases based on the range of z.

For 0 < z < 1, we have two possibilities:
1. 3X > Y: In this case, Z = 3X since it is the maximum value. To find the PDF of Z, we need to find the probability that 3X is less than z. We can express this as P(3X < z). To find this probability, we need to find the cumulative distribution function (CDF) of 3X and evaluate it at z. The CDF of 3X is given by F3X(t) = P(3X ≤ t) = P(X ≤ t/3). Thus, fZ(z) = F3X(z).

2. Y > 3X: In this case, Z = Y since it is the maximum value. To find the PDF of Z, we need to find the probability that Y is less than z. We can express this as P(Y < z). To find this probability, we need to use the CDF of Y and evaluate it at z. Let's denote the CDF of Y as FY(t). Thus, fZ(z) = FY(z).

For 1 < z < 2, we also have two possibilities:
1. 3X > Y: In this case, Z = 3X since it is the maximum value. To find the PDF of Z, we need to find the probability that 3X is greater than 1 and less than z. We can express this as P(1 < 3X < z). This probability can be calculated as F3X(z) - F3X(1). Thus, fZ(z) = F3X(z) - F3X(1).

2. Y > 3X: In this case, Z = Y since it is the maximum value. To find the PDF of Z, we need to find the probability that Y is greater than 1 and less than z. We can express this as P(1 < Y < z). This probability can be calculated as FY(z) - FY(1). Thus, fZ(z) = FY(z) - FY(1).

Note that for values of z outside the range 0 < z < 1 and 1 < z < 2, the PDF of Z will be zero, since Z can only take values between 0 and 3X or Y, depending on which is greater.

Please note that I cannot provide the specific values for fZ(z) without knowing the specific distributions of X and Y. You would need to substitute the appropriate CDFs into the expressions above to find the PDF of Z.

To find the probability density function (PDF) of Z = max{3X, Y}, we need to consider the range of values that Z can take on and calculate the corresponding probabilities for each range.

1. For 0 < z < 1:
In this range, Z can only take on values less than or equal to 1. Since the maximum value of 3X and Y would be either 3X or Y itself, we have:
fZ(z) = P(Z ≤ z) = P(max{3X, Y} ≤ z) = P(3X ≤ z, Y ≤ z)
= P(3X ≤ z) * P(Y ≤ z) (since X and Y are independent)
= P(X ≤ z/3) * P(Y ≤ z) (using the fact that P(aX ≤ b) = P(X ≤ b/a) for a > 0)
= Fx(z/3) * Fy(z), where Fx(x) and Fy(y) are the respective cumulative distribution functions (CDFs) of X and Y.

2. For 1 < z < 2:
In this range, Z can take on values between 1 and 2. The maximum value among 3X and Y would be z if 3X > z and Y ≤ z. Similarly to the previous range, we have:
fZ(z) = P(Z ≤ z) = P(max{3X, Y} ≤ z) = P(3X ≤ z, Y ≤ z)
= P(3X ≤ z) * P(Y ≤ z)
= P(X ≤ z/3) * P(Y ≤ z)
= Fx(z/3) * Fy(z).

So, for 0 < z < 1, fZ(z) = Fx(z/3) * Fy(z).
And for 1 < z < 2, fZ(z) = Fx(z/3) * Fy(z).

Please note that we need to know the specific distributions of X and Y in order to calculate Fx(x) and Fy(y) and obtain numerical values for the PDF.