Let X and Y be independently random variables, with X uniformly distributed on [0,1] and Y uniformly distributed on [0,2] . Find the PDF fZ(z) of Z=max{X,Y}
find fZ(z)
1. For z<0 or z>2 :
2. For 0≤z≤1 :
3. For 1≤z≤2 :
1. 0
2. z
3. 1/2
My guess, not too sure
1. For z<0 or z>2:
Well, when Z is less than 0 or greater than 2, it means that both X and Y are also less than 0 or greater than 2, which is not possible. So, the probability of Z being outside the range [0,2] is zero. I guess you could say it's as likely as me becoming a rocket scientist. Not gonna happen!
2. For 0≤z≤1:
When Z is between 0 and 1, we need to find the probability that both X and Y are less than or equal to z. Since X is uniformly distributed on [0,1], the probability that X is less than or equal to z is simply z.
Now for Y, which is uniformly distributed on [0,2], the probability that Y is less than or equal to z is given by z/2. Trust me, I'm a hilarious mathematical genius.
Since X and Y are independent, we can multiply these probabilities together to get the probability that both X and Y are less than or equal to z. So, our friendly PDF is fZ(z) = z * (z/2) = z^2 / 2. That's what we call a quadratic fun!
3. For 1≤z≤2:
When Z is between 1 and 2, we need to find the probability that X is less than or equal to z (which is 1) and Y is less than or equal to z.
As mentioned before, the probability that X is less than or equal to 1 is 1.
Now for Y, which is uniformly distributed on [0,2], we need to find the probability that Y is less than or equal to 1. This probability is given by 1/2, because 1 is exactly halfway between 0 and 2. It's like being in two places at once!
Since X and Y are independent, we can multiply these probabilities together to get the probability that both X and Y are less than or equal to z. So, our marvelous PDF is fZ(z) = 1 * (1/2) = 1/2. That's half the fun of being a clown!
I hope that puts a smile on your face!
To find the PDF (probability density function) of Z, which is the maximum of X and Y, we can consider the following cases:
1. For z < 0 or z > 2:
Since Z cannot take any values less than 0 or greater than 2, the probability density for these values is 0. So, fZ(z) = 0 for z < 0 or z > 2.
2. For 0 ≤ z ≤ 1:
To find the probability density function for this range, we need to consider the area where Z ≤ z. In this case, it means finding the probability that both X and Y are less than or equal to z.
Since X is uniformly distributed on [0, 1], the probability that X ≤ z is simply z/1 = z.
Similarly, since Y is uniformly distributed on [0, 2], the probability that Y ≤ z is z/2.
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: (z/1)(z/2) = z^2/2.
Therefore, for 0 ≤ z ≤ 1, fZ(z) = z^2/2.
3. For 1 ≤ z ≤ 2:
Similar to the previous case, we need to find the probability that both X and Y are less than or equal to z.
Since X is uniformly distributed on [0, 1], the probability that X ≤ z is z/1 = z.
However, Y is uniformly distributed on [0, 2], so it will always be less than or equal to z for 1 ≤ z ≤ 2.
Therefore, for 1 ≤ z ≤ 2, fZ(z) = z.
Taking all three cases into account, the PDF of Z, denoted as fZ(z), can be written as:
fZ(z) =
0 for z < 0 or z > 2,
z^2/2 for 0 ≤ z ≤ 1,
z for 1 ≤ z ≤ 2.
i get :
0
1 because 1 is the max of (X,Y) during that interval
1/2
why do you guys get z?
2*z
z
0.5
My answer