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
1
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
Well, well, well. Looks like we've got some probability distribution fun here.
Let's start with finding the PDF of Z when Z = max{X, Y}. Now, since X and Y are independent random variables uniformly distributed on [0,1], we know that their PDFs are equal to 1 over the interval [0,1]. And since Z is defined as the maximum of X and Y, we can see that Z will be less than or equal to z only if both X and Y are less than or equal to z.
So, for 0 < z < 1, we have the probability that Z takes a value less than or equal to z as the probability that both X and Y are less than or equal to z. Since X and Y are independent, this probability is simply the product of their individual probabilities:
P(Z ≤ z) = P(X ≤ z) * P(Y ≤ z)
Both X and Y are uniformly distributed on [0,1], so their probabilities can be given as:
P(X ≤ z) = z
P(Y ≤ z) = z
Therefore, the PDF of Z for 0 < z < 1 can be expressed as:
fZ(z) = d(P(Z ≤ z)) / dz = d((z * z)) / dz = 2z
So, for 0 < z < 1, fZ(z) = 2z. There you have it!
Now, let's move on to finding the PDF of Z when Z = max{2X, Y}. Again, we need to consider different intervals for z.
For 0 < z < 1, we can see that Z will be less than or equal to z only if both 2X and Y are less than or equal to z. Using the same logic as before, we can find the probability that Z takes a value less than or equal to z:
P(Z ≤ z) = P(2X ≤ z) * P(Y ≤ z)
Since X and Y are both uniformly distributed on [0,1], we can express their probabilities as:
P(2X ≤ z) = P(X ≤ z/2) = z/2
P(Y ≤ z) = z
So, the PDF of Z for 0 < z < 1 can be written as:
fZ(z) = d(P(Z ≤ z)) / dz = d((z/2 * z)) / dz = (3/2)z^2
And for 1 < z < 2, we can see that Z will be less than or equal to z only if 2X is less than or equal to z and Y is greater than z. Using the same logic again, we have:
P(Z ≤ z) = P(2X ≤ z) * P(Y > z)
P(2X ≤ z) = P(X ≤ z/2) = z/2
P(Y > z) = 1 - P(Y ≤ z) = 1 - z
So, the PDF of Z for 1 < z < 2 can be expressed as:
fZ(z) = d(P(Z ≤ z)) / dz = d((z/2 * (1-z))) / dz = (1/2) - z/2
And there you have it! The PDF of Z when Z = max{2X, Y}. Keep in mind, though, that I'm just a Clown Bot, so please double-check my work before using it for anything serious.
To find the PDF of Z, we need to consider the range of values that Z can take.
For the first case, Z = max{X, Y}. Since X and Y are independent random variables, the probability that both X and Y are less than or equal to z is given by the product of their individual probabilities:
P(X ≤ z) = z
P(Y ≤ z) = z
Since Z is the maximum of X and Y, the probability that Z is less than or equal to z is given by the product of the probabilities:
P(Z ≤ z) = P(X ≤ z) * P(Y ≤ z) = z * z = z^2
To find the PDF, we differentiate the cumulative distribution function (CDF) of Z with respect to z:
fZ(z) = d/dz(P(Z ≤ z)) = d/dz(z^2) = 2z
Therefore, for 0 < z < 1, the PDF of Z is given by:
fZ(z) = 2z
For the second case, Z = max{2X, Y}. Again, we calculate the individual probabilities:
P(2X ≤ z) = P(X ≤ z/2) = z/2
P(Y ≤ z) = z
Since Z is the maximum of 2X and Y, the probability that Z is less than or equal to z is given by the product:
P(Z ≤ z) = P(2X ≤ z) * P(Y ≤ z) = (z/2) * z = z^2/2
To find the PDF, we differentiate the CDF of Z with respect to z:
fZ(z) = d/dz(P(Z ≤ z)) = d/dz(z^2/2) = z/2
Therefore, for 0 < z < 1, the PDF of Z is given by:
fZ(z) = z/2
For 1 < z < 2, Z = Y, so the PDF remains the same:
fZ(z) = z/2
To find the PDF (Probability Density Function) of Z, which is the maximum between X and Y, we need to consider the possible values that Z can take on and calculate the probability of Z being in each range.
1. For 0 < z < 1:
In this range, Z can only take on values between 0 and 1. To calculate the probability, we need to find the probability that both X and Y are less than or equal to z.
Since X and Y are independent uniform random variables on the interval [0, 1], the probability that X is less than or equal to z is just z. Similarly, the probability that Y is less than or equal to z is also z.
Since X and Y are independent, the probability that both X and Y are less than or equal to z is the product of their probabilities, i.e., z * z = z^2.
Therefore, for 0 < z < 1, the PDF of Z is fZ(z) = z^2.
2. For 1 < z < 2:
In this range, Z can take on values between 1 and 2. To calculate the probability, we need to find the probability that one of X or 2X is less than or equal to z, and the other is less than or equal to z / 2.
When X is less than or equal to z, the probability is z. And when 2X is less than or equal to z, X must be less than or equal to z / 2. So, the probability is z / 2.
Since X and Y are independent, the probability of either one happening is the sum of their probabilities, i.e., z + z / 2 = (3z) / 2.
Therefore, for 1 < z < 2, the PDF of Z is fZ(z) = (3z) / 2.
Now, for the remaining range of Z, which is 0 for values less than or equal to 0, and 1 for values greater than or equal to 1, the PDF is 0.
To summarize:
1. For 0 < z < 1, fZ(z) = z^2.
2. For 1 < z < 2, fZ(z) = (3z) / 2.
3. For all other values of z, fZ(z) = 0.