Let X,Y,Z be three independent (i.e. mutually independent) random variables, each uniformly distributed on the interval [0,1].
1. Find the mean and variance of 1/(Z+1).
E[1/(Z+1)]=
var(1/(Z+1))=
2. Find the mean of XY/(Z+1).
Hint: Use your answer to the previous part, together with the independence assumption.
E[XY/(Z+1)]=
3. Find the probability that XY/Z≤1. Enter a numerical answer.
P(XY/Z≤1)=
1.
E[1/(Z+1)] = 1/2
var(1/(Z+1)) = 0.0195
2.
E[XY/(Z+1)] = 0.1733
3.
P(XY/Z≤1) = 0.75
1. ln(2)
2. 0.5+ln(2)^2
3. 0.5^2*ln(2)
Anyone care to share the rest of the answers to the test here?
To find the mean and variance of 1/(Z+1), we need to find the expected value (mean) and variance of the random variable 1/(Z+1).
1. Mean of 1/(Z+1):
The mean of a random variable can be found by taking the integral of the random variable multiplied by its probability density function (PDF) over its range. Since Z is uniformly distributed on the interval [0,1], its PDF is constant over this interval. So we have:
E[1/(Z+1)] = ∫[0,1] (1/(z+1)) * (1) dz
Evaluating this integral, we get:
E[1/(Z+1)] = ln(2)
2. Variance of 1/(Z+1):
The variance of a random variable can be found using the formula:
var(1/(Z+1)) = E[(1/(Z+1))^2] - [E(1/(Z+1))]^2
To find E[(1/(Z+1))^2], we integrate the square of 1/(Z+1) over the range of Z:
E[(1/(Z+1))^2] = ∫[0,1] (1/(z+1))^2 * (1) dz
Evaluating this integral, we get:
E[(1/(Z+1))^2] = 2 - ln(2)
Now we can calculate the variance:
var(1/(Z+1)) = E[(1/(Z+1))^2] - [E(1/(Z+1))]^2
= (2 - ln(2)) - (ln(2))^2
3. To find the mean of XY/(Z+1), we can use the independence assumption and the linearity of expectations. Since X, Y, and Z are mutually independent, we have:
E[XY/(Z+1)] = E[X/(Z+1)] * E[Y/(Z+1)]
Using the answer from the first part, we know that E[1/(Z+1)] = ln(2). So:
E[XY/(Z+1)] = ln(2) * ln(2)
= (ln(2))^2
4. To find the probability that XY/Z ≤ 1, we can use the fact that X, Y, and Z are independent, thus allowing us to write the joint probability as the product of individual probabilities. Since X, Y, and Z are uniformly distributed on the interval [0,1], the probability density functions (PDFs) of X, Y, and Z are all constant over this interval.
P(XY/Z ≤ 1) = ∫[0,1] ∫[0,1] ∫[0,xy] 1 dx dy dz
Evaluating this triple integral, we get:
P(XY/Z ≤ 1) = 1/3
So the probability that XY/Z ≤ 1 is 1/3.