MATH 556 - ASSIGNMENT 2: SOLUTIONS
1 By direct calculation, we have by the theorem of total probability for y 2,
y-1
fY (y) = PY [Y = y] =
x1 =1
PX1 ,X2 [X1 = x1 , X2 = y - x1 ] =
x1 =1
PX1 [X1 = x1 ]PX2 [X2 = y - x1 ]
by independence. Thus
y-1
fY (y) =
MATH 556 - EXERCISES 2: SOLUTIONS
1 We have fR (r) = 6r(1 - r), for 0 < r < 1, and hence FR (r) = r2 (3 - 2r) with the usual cdf behaviour outside of this range. Circumference: Y = 2R, so Y = (0, 2), and from first principles, for y Y, FY (y) = P[ Y y ] =
MATH 556 - MID-TERM 2007 SOLUTIONS
1.
(a) (i) We have
MX (t) = EfX e =
tX
=
x=1
e fX (x) =
x=1
tx
1 x 1 e = log (1 ) x log (1 )
tx
x=1
et x
x
log 1 log (1 )
et
provided et < 1, or equivalently t < log , which ensures the required neighbourhood of zero
McGill University
Faculty of Science
Department of Mathematics and Statistics
December 2007
MATH 556
MATHEMATICAL STATISTICS I
SOLUTIONS
1.
(a) From rst principles (univariate transformation theorem also acceptable): for y R
X
1X
FY (y) = P [Y y] = Pr log
MATH 556 - EXERCISES 3 : SOLUTIONS
1 (a) By direct calculation, using a derivation as in lecture notes, CX (t) = EfX [eitX ] = 1 eitx |x| expcfw_|x| dx = 2
0 0
cos(tx)xex dx
as the pdf is an even function around zero. Integrating by parts, CX (t) = = =
MATH 556 - EXERCISES 4: SOLUTIONS
1 (a) This is not an Exponential Family distribution; the support is parameter dependent. (b) This is an EF distribution with k = 1: f (x|) = Icfw_1,2,3,. (x) -1 expcfw_x log = h(x)c() expcfw_w()t(x) x log (1 - ) Icfw_1,
MATH 556 - ASSIGNMENT 4: SOLUTIONS
1 (a) Directly from the notes: by Lyapunovs inequality
E[ |Xn X|s ]1/s E[ |Xn X|r ]1/r
so that
E[ |Xn X|s ] E[ |Xn X|r ]s/r 0 E[ |Xn X|s ] 0
as n , as s < r. Thus and Xn X.
sth
2 M ARKS For the given sequence, if r > s 1
MATH 556 - ASSIGNMENT 3: SOLUTIONS
1 (a) Suppose rst that
EfX [g(X)] EfY [g(Y )]
for any non-decreasing real function g. Proof given for continuous random variables, but proof in the discrete case follows after minor adjustment. Let gt (x) = I[t,) (x) ; t
MATH 556 - EXERCISES 5: SOLUTIONS
1 Key is to find the i.i.d random variables X1 , ., Xn such that
n
X=
i=1
Xi
and then to use the Central Limit Theorem result for large n
n
Zn =
i=1
Xi - n d - Z N ormal(0, 1) n 2
n
X=
i=1
Xi AN (n, n 2 )
where =EfX [Xi ]
McGill University Faculty of Science Department of Mathematics and Statistics December 2006
MATH 556 MATHEMATICAL STATISTICS I SOLUTIONS
1.
(a) From rst principles (univariate transformation theorem also acceptable): for y > 0 FY (y) = P [Y y] = P and the