Exercise Solution 3.29
- By [3.94], M ~ N(8, 36).
- By [3.105] and [3.106], L ~
.
- By [3.101] and [3.102], E ~ Λ(e5, e16 – e10).
- We cannot say. We are not told if N1 and N2 are joint-normal.
- Because N1 and N2 are independent, they are joint-normal. Hence, M is normal. By [3.27] and [3.28], M ~ N(2, 6).
- H ~ χ2(1, 0)
- H ~ χ2(2, 25)
- This is a nonstandard distribution. It is clearly not normal because C is strictly positive.