HW-Sol-5-V1 - Massachusetts Institute of Technology
6.434J/16.391J Statistics for Engineers and Scientists MIT, Spring 2006
Apr 11 Handout #13
Solution 5
Problem 1: Let a > 0 be a known constant, and let > 0 be a parameter. Suppose X1, X2, . . . , Xn is a sample from a population with one of the following densities.
(a) The beta, (, 1), density: fX (x | ) = x-1, for 0 < x < 1.
(b) The Weilbull density: fX (x | ) = axa-1 e-xa , for x > 0.
(c)
The
Pareto
density:
fX (x | ) =
, a
x(+1)
for x > a.
In each case, find a real-valued sufficient statistic for .
Solution Let X (X1, X2, . . . , Xn) be a collection of i.i.d. random variables Xi's, and let x (x1, x2, . . . , xn) be a collection of observed data.
(a) For any x, the joint pdf is
fX (x | ) = n(x1x2 ? ? ? xn)-1,
0,
if i, 0 < xi < 1 otherwise;
= n(x1x2 ? ? ? xn)-1 ? I(0,1)(x1)I(0,1)(x2) ? ? ? I(0,1)(xn) .
g(T (x) | )
h(x)
Factorization theorem implies that
T (x) x1x2 ? ? ? xn
is a sufficient statistic for .
(b) For any x, the joint pdf is
?
fX (x | ) = nan(x1x2 ? ? ? xn)a-1 e-
, n
i=1
xai
0, ?
= n e-
n i=1
xai
if i, xi > 0; otherwise;
g(T (x) | )
? an(x1x2 ? ? ? xn)a-1 ?I(0,)(x1)I(0,)(x2) ? ? ? I(0,)(xn) .
h(x)
1
Factorization theorem implies that
T (x)
n
xai
i=1
is a sufficient statistic for .
(c) For any x, the joint pdf is
fX
(x
|
)
=
n an
(x1 x2 ???xn )+1
,
0,
if i, xi > a; otherwise;
=
nan (x1x2 ? ? ? xn)+1
? I(a,)(x1)I(a,)(x2) ? ? ? I(a,)(xn) .
g(T (x) | )
h(x)
Factorization theorem implies that
T (x) x1x2 ? ? ? xn
is a sufficient statistic for .
Problem 2:
a) Let X1, X2, . . . , Xn be independent random variables, each uniformly distributed on the interval [-, ], for some > 0. Find a sufficient statistic for .
b) Let X1, X2, . . . , Xn be a random sample of size n from a normal N (, ) distribution, for some > 0. Find a sufficient statistic for .
Solution
a) For any x (x1, x2, . . . , xn), the joint pdf is given by
n
1 fX (x | ) = 2
,
0,
=
1 2
n
,
0,
if i, - xi ; otherwise; if - min(x1, . . . , xn) and max(x1, . . . , xn) ; otherwise;
=
1 2
n
I[-,)(min(x1, . . . , xn))I(-,](max(x1, . . . , xn)) ?
1
.
g(T(x) | )
h(x)
2
Factorization theorem implies that
T(x) min(x1, . . . , xn), max(x1, . . . , xn)
is jointly sufficient for .
b) For any x (x1, x2, . . . , xn), the joint pdf is given by
fX (x) = = = =
1
? e n
-
1 2
n i=1
(xi
-)2
? ? 2
1
e n
-
1 2
(
n i=1
x2i -2
n i=1
xi +n2 )
2 1
e ? ? n
-
1 2
n i=1
x2i +
n i=1
xi -
n 2
? 2
1
n
e
? n
i=1
xi
1
e ? . n
-
1 2
n i=1
x2i -
n 2
2
h(x)
g(T (x) | )
Factorization theorem implies that
T (x)
n
x2i
i=1
is a sufficient statistic for .
Problem 3: Let X be the number of trials up to (and including) the first success in a sequence of Bernoulli trials with probability of success , for 0 < < 1. Then, X has a geometric distribution with the parameter :
P {X = k} = (1 - )k-1, k = 1, 2, 3, . . . .
Show that the family of geometric distributions is a one-parameter exponential family with T (x) = x. [Hint : x = e ln x, for x > 0.] Solution Recall that the pmf of a one-parameter () exponential family is of the form
p(x | ) = h(x) e()T (x)-B(),
where x X . Rewriting the pmf of a Geometric random variable yields
P {X = x} = e(x-1) ln(1-)+ln = ex ln(1-)-(ln(1-)-ln ),
3
where x {1, 2, 3, . . . }. Thus, the geometric distribution is a one-parameter exponential family with
h(x) T (x)
X
1 x {1, 2, 3, . . . }.
() ln(1 - ) B() ln(1 - ) - ln
Problem 4: Let X1, X2, . . . , Xn be a random sample of size n from the truncated Bernoulli probability mass function (pmf),
p,
if x = 1;
P {X = x | p} =
(1 - p), if x = 0.
(a) Show that the joint pmf of X1, X2, . . . , Xn is a member of the exponential family of distribution.
(b) Find a minimal sufficient statistic for p.
Solution
(a) Let x (X1, X2, . . . Xn) denote the collection of i.i.d. Bernoulli random variables. The joint pmf is given by
? ? P {X = x | p} = px1 (1 - p)1-x1 px2 (1 - p)1-x2 ? ? ? pxn (1 - p)1-xn
=p
n i=1
xi
(1
-
p)n-
n i=1
xi
? ? ? = e(ln p)
e n
i=1
xi
[ln(1-p)][n-
n i=1
xi ]
= e[ln p-ln(1-p)]
, n
i=1
xi
+n
ln(1-p)
for x {0, 1}n. Therefore, the joint pmf is a member of the exponential family, with the mappings:
=p (p) = ln p - ln(1 - p) B(p) = -n ln(1 - p)
h(x) = 1
n
T (x) = xi
i=1
X = {0, 1}n.
(b) Let x, y {0, 1}n be given. Consider the likelihood ratio,
? ? P {X
P {X
= =
x | p} y | p}
=
e[ln p-ln(1-p)][
n i=1
xi -
. n
i=1
yi ]
4
Define a function k(x, y) h(x)/h(y) = 1, which is bounded and non-zero
for any x X and y X .
Note that x and y such that
n i=1
xi
=
n i=1
yi
are
equivalent
because
function k(x, y) satisfies the requirement of likelihood ratio partition.
Therefore, T (x)
n i=1
xi
is
a
sufficient
statistic.
Problem 5: Let X1, X2, . . . , Xm and Y1, Y2, . . . , Yn be two independent samples from N (?, 2) and N (?, 2) populations, respectively. Here, - < ? < , 2 > 0, and 2 > 0. Find a minimal sufficient statistic for (?, 2, 2).
Solution Let X (X1, X2, . . . , Xm) and Y (Y1, Y2, . . . , Yn) denote the col-
lections of random samples. The joint pdf (of Xj's and Yi's), evaluated at
x (x1, x2, . . . , xm) and y (y1, y2, . . . , yn), is given by
? ? fX,Y (x, y | ) =
1 22
m ? e-
m j=1
(xj
-?)2
22
?
? ? ? ? =
e-
1 22
m j=1
x2j -
1 2 2
n i=1
yi2 +
? 2
1 2 2
n ? e-
n i=1
(yi
-?)2
2 2
, m
j=1
xj
+
? 2
n i=1
yi-B(?,2, 2)
where B(?, 2, 2)
m 2
ln 22
+
n 2
ln 2 2
+
m?2 22
+
n?2 2 2
.
Notice that the joint pdf belongs to the exponential family, so that the
minimal statistic for is given by
T(X, Y)
m
n
m
n
Xj2, Yi2, Xj , Yi .
j=1
i=1
j=1
i=1
Note: One should not be surprised that the joint pdf belongs to the exponential family of distribution. Recall that Gaussian distribution is a member of the exponential family of distribution and that random variables, Xi's and Yj's, are mutually independent. Thus, their joint pdf belongs to the exponential family as well. Note: To derive the minimal sufficient statistic, one may alternatively consider likelihood ratio partition.
The set D0 is defined to be
D0 (x, y) Rm+n for all ?, for all 2 > 0, for all 2 > 0
fX,Y x, y | ?, 2, 2 = 0
=
(empty set).
5
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