Let ๐ and ๐ be independently random variables, with ๐ uniformly distributed on [0,1] and ๐ uniformly distributed on [0,2] . Find the PDF ๐๐(๐ง) of ๐=max{๐,๐} .
For ๐ง<0 or ๐ง>2 :
๐๐(๐ง)=
For 0โค๐งโค1 :
๐๐(๐ง)=
For 1โค๐งโค2 :
๐๐(๐ง)=
I think for the first scenario is the PDF is 0 since the bounds are outside of the range.
if anybody got hint for the other two, thanks
Well, for the second scenario, where 0 โค ๐ง โค 1, you can think about it like this: In order for ๐ to be less than or equal to ๐ง, at least one of ๐ or ๐ must be less than or equal to ๐ง. Since both ๐ and ๐ are uniformly distributed, the probability of either of them being less than or equal to ๐ง is simply ๐ง itself. So, ๐๐(๐ง) will be equal to ๐ง for that range.
Now, let's move on to the third scenario, where 1 โค ๐ง โค 2. In this case, both ๐ and ๐ have to be less than or equal to ๐ง in order for ๐ to be equal to ๐ง. The probability of both ๐ and ๐ being less than or equal to ๐ง is the product of the probabilities of ๐ being less than or equal to ๐ง and ๐ being less than or equal to ๐ง. Since ๐ is uniformly distributed on [0,1], the probability of it being less than or equal to ๐ง is ๐ง itself. Similarly, the probability of ๐ being less than or equal to ๐ง is ๐ง/2, since ๐ is uniformly distributed on [0,2]. Therefore, ๐๐(๐ง) will be equal to (๐ง)(๐ง/2) = ๐ง^2/2 for that range.
So, to summarize:
- For ๐ง < 0 or ๐ง > 2, ๐๐(๐ง) = 0
- For 0 โค ๐ง โค 1, ๐๐(๐ง) = ๐ง
- For 1 โค ๐ง โค 2, ๐๐(๐ง) = ๐ง^2/2
For 0 โค z โค 1:
To find the probability density function (PDF) of Z, we need to find the probability that Z takes on a value in the interval (z, z + dz) for a small positive value of dz.
Let's consider the case when Z takes on a value between z and z + dz. This occurs when X takes on a value between z and z + dz and Y takes on a value greater than or equal to z.
Since X is uniformly distributed on [0,1], the probability that X takes on a value between z and z + dz is dz.
Since Y is uniformly distributed on [0,2], the probability that Y takes on a value greater than or equal to z is (2 - z)/2 = (2 - z)/2.
Therefore, the probability that Z takes on a value between z and z + dz is dz * (2 - z)/2.
Thus, the PDF of Z for 0 โค z โค 1 is:
fZ(z) = dz * (2 - z)/2
Next, let's consider the case when 1 โค z โค 2:
To find the PDF of Z in this range, we need to find the probability that Z takes on a value in the interval (z, z + dz) for a small positive value of dz.
In this case, Z takes on a value between z and z + dz when both X and Y take on values greater than or equal to z.
Since X is uniformly distributed on [0,1], the probability that X takes on a value greater than or equal to z is 1 - z.
Since Y is uniformly distributed on [0,2], the probability that Y takes on a value greater than or equal to z is (2 - z)/2 = (2 - z)/2.
Therefore, the probability that Z takes on a value between z and z + dz is (1 - z) * (2 - z)/2.
Thus, the PDF of Z for 1 โค z โค 2 is:
fZ(z) = (1 - z) * (2 - z)/2
For z < 0 or z > 2, the PDF is 0.
So, the complete PDF of Z is:
For z < 0 or z > 2:
fZ(z) = 0
For 0 โค z โค 1:
fZ(z) = dz * (2 - z)/2
For 1 โค z โค 2:
fZ(z) = (1 - z) * (2 - z)/2
To find the PDF of ๐, we need to consider the different ranges of values for ๐ง.
1. For ๐ง < 0 or ๐ง > 2:
In these cases, ๐ cannot take on any values as it is defined as the maximum of ๐ and ๐, both of which are bounded by [0,1] and [0,2] respectively. Therefore, ๐๐(๐ง) = 0 for ๐ง < 0 or ๐ง > 2.
2. For 0 โค ๐ง โค 1:
In this range, ๐ will be less than or equal to ๐ง only if both ๐ and ๐ are less than or equal to ๐ง.
The probability that ๐ โค ๐ง is ๐(๐ โค ๐ง) = ๐ง/1 = ๐ง (since ๐ is uniformly distributed on [0,1]).
Similarly, the probability that ๐ โค ๐ง is ๐(๐ โค ๐ง) = ๐ง/2 (since ๐ is uniformly distributed on [0,2]).
Now, since ๐ and ๐ are independent random variables, the probabilities multiply:
๐(๐ โค ๐ง and ๐ โค ๐ง) = ๐(๐ โค ๐ง) * ๐(๐ โค ๐ง) = ๐ง * ๐ง/2 = ๐ง^2/2
To find the PDF for this range, we need to find the derivative of the cumulative probability:
๐๐(๐ง) = d/d๐ง (๐(๐ โค ๐ง)) = d/d๐ง (๐ง^2/2) = ๐ง/2
Therefore, for 0 โค ๐ง โค 1, ๐๐(๐ง) = ๐ง/2.
3. For 1 โค ๐ง โค 2:
In this range, ๐ will be equal to ๐ง only if either ๐ > ๐ง or ๐ > ๐ง.
The probability that ๐ > ๐ง is ๐(๐ > ๐ง) = 1 - ๐(๐ โค ๐ง) = 1 - ๐ง (since ๐ is uniformly distributed on [0,1]).
Similarly, the probability that ๐ > ๐ง is ๐(๐ > ๐ง) = 1 - ๐(๐ โค ๐ง) = 1 - ๐ง/2 (since ๐ is uniformly distributed on [0,2]).
Again, since ๐ and ๐ are independent random variables, the probabilities multiply:
๐(๐ > ๐ง and ๐ > ๐ง) = ๐(๐ > ๐ง) * ๐(๐ > ๐ง) = (1 - ๐ง) * (1 - ๐ง/2)
To find the PDF for this range, we need to find the derivative of the cumulative probability:
๐๐(๐ง) = d/d๐ง (๐(๐ โค ๐ง)) = d/d๐ง (1 - (1 - ๐ง)(1 - ๐ง/2)) = d/d๐ง (1 - 3/2๐ง + ๐ง^2/2) = 1 - 3/2 + ๐ง/2 = 1/2 + ๐ง/2
Therefore, for 1 โค ๐ง โค 2, ๐๐(๐ง) = 1/2 + ๐ง/2.
In summary:
For ๐ง < 0 or ๐ง > 2: ๐๐(๐ง) = 0
For 0 โค ๐ง โค 1: ๐๐(๐ง) = ๐ง/2
For 1 โค ๐ง โค 2: ๐๐(๐ง) = 1/2 + ๐ง/2