Download Homework 5 with Answers - Mathematical Statistics I | STAT 510 and more Assignments Mathematical Statistics in PDF only on Docsity!
STAT 510 HW #5 – Written: Answers
Due Friday, October 6, 2006 You can turn it in in class, or to my office (116B IH) or mailbox in 101 IH, by 4PM. There are also questions on Mallard.
- Suppose Z ∼ N (0, 1) and U ∼ U nif orm(0, 1), and Z and U are independent. Let Y = (Y 1 , Y 2 ) = g(z, u) be given by
Y 1 = ZU and Y 2 = U.
Y 1 is said to have the “slash” distribution. (a) What is the space of Y? Answer: R × (0, 1).
(b) Find g−^1 (y). Answer: z = uy 1 = y 1 y 2 , so g−^1 (y) = (y 1 y 2 , y 2 ).
(c) Find Jg− 1 (y). Answer:
Jg− 1 (y) =
∣∣ ∣∣ ∣∣ ∣∣
∂y 1 y 2 ∂y 1 ∂y 1 y 2 ∂y 2 ∂y ∂y 2 1 ∂y ∂y 2 2
∣∣ ∣∣ ∣∣ ∣∣^ =
∣∣ ∣∣ ∣
y 2 y 1 0 1
∣∣ ∣∣ ∣ =^ y^2.
(d) Find the pdf of Y. Answer: First, because Z and U are independent, the pdf of (Z, U ) is
fZ,U (z, u) = √^1 2 π e−^12 z^2 × 1.
Then
fY (y 1 , y 2 ) = fZ,U (g−^1 (y 1 , y 2 ))|Jg− 1 (y)| = fZ,U (y 1 y 2 , y 2 ) y 2 = √^12 π e−^12 (y^1 y^2 )^2 y 2.
(e) Are Y 1 and Y 2 independent? Why or why not? Answer: No, the pdf does not factor.
(f) What is the marginal space of Y 1? Show that the marginal pdf of Y 1 is
f 1 (y 1 ) = c. 1 −^ e
−y^21 / 2 y^21.
What is the constant c? Answer: The space of Y is R × (0, 1), since for any y 1 ∈ R and y 2 ∈ (0, 1), (y 1 y 2 , y 2 ) is in the space of (Z, U ). So the marginal space of Y 1 is R, and the conditional space of Y 2 given Y 1 is (0, 1). Thus
fY 1 (y 1 ) =
∫ (^1) 0 fY^ (y^1 , y^2 )dy^2 =^
√^1
2 π
∫ (^1) 0 e
− 12 (y 1 y 2 )^2 y 2 dy 2.
Make the substitution u = y^21 y^22 /2 in the integral, so that y 2 dy 2 = du/y^21. Then ∫ (^1) 0 e
− 12 (y 1 y 2 )^2 y 2 dy 2 = ∫ (^) y (^21) / 2 0 e
−udu/y (^21) =^1 −^ e−y
(^21) / 2 y 12 ,
and fY 1 (y 1 ) = √^12 π^1 −^ e
−y^21 / 2 y^21 , which means that c = 1/
2 π.
- Suppose (Y 1 ,... , Y 5 ) ∼ Dirichlet(α 1 ,... , α 6 ). Find the distribution of
(U 1 , U 2 ) = (Y 1 + Y 2 , Y 3 + Y 4 + Y 5 ).
(Do not use pdf’s, unless you really want to. Work directly with the definition of the Dirichlet based on gamma’s.) Answer: With X 1 ,... , X 6 being independent, Xk ∼ Gamma(αk, 1), we have
(Y 1 ,... , Y 5 ) =D
( X 1
X 1 + · · · + X 6 ,... ,^
X 5
X 1 + · · · + X 6
) .
That means that
(U 1 , U 2 ) = (Y 1 + Y 2 , Y 3 + Y 4 + Y 5 ) =D
( X 1 + X 2
X 1 + · · · + X 6 ,
X 3 + X 4 + X 5
X 1 + · · · + X 6
) .
Note that
V 1 ≡ X 1 + X 2 ∼ Gamma(α 1 + α 2 , 1), V 2 ≡ X 3 + X 4 + X 5 ∼ Gamma(α 3 + α 4 + α 5 , 1),
and V 3 ≡ X 6 ∼ Gamma(α 6 , 1),
