a.
To verify that g(t) is indeed a probability density function, we need to check for two conditions:
Condition 1: g(t) is non-negative for all values of t in the support of U.
Condition 2: The integral of g(t) integrated over the entire support of U integrates to 1.
Condition 1
Case I: When -1 ≤ t ≤ 0:
=> -1 ≤ t
=> 0 ≤ 1 + t
=> 0 ≤ g(t)
=> g(t) is non-negative when -1 ≤ t ≤ 0
Case II: When 0 ≤ t ≤ 1:
=> t ≤ 1
=> 0 ≤ 1 - t
=> 0 ≤ g(t)
=> g(t) is non-negative when 0 ≤ t ≤ 1
Thus, g(t) is non-negative for all values of t in the support of U.
Condition 2
We find:
Thus, the integral of g(t) integrated over the entire support of U equals to 1.
Thus, g(t) is a valid probability density function.
b.
The expected value of U is given by:
Now, before finding the variance of U, we find:
Thus, the variance of U is given by:
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