Download 04Solution

University of Zurich
Institute of Mathematics
MAT924
Reinhard Furrer, Emma Hovhannisyan
Exercises 4
Problem 8
Suppose that arrival of customers at a service counter follows a Poisson process with arrival rate λ. Let X1 be
the time of arrival of the first customer, X2 the time between the arrival of the first and second customer. Hence,
X1 and X2 are independent and identically distributed exponential random variables with parameter λ. Here we
use different methods for showing that T = X1 + X2 is a sufficient statistic for λ.
(a) Show the sufficiency condition using the Factorization theorem.
Hint: T is sufficient iff there exist g1 , g2 functions such that f (x1:2 ; λ) = g1 (t, λ)g2 (x1:2 )
Solution:
f (x1 , x2 ; λ) = f (x1 ; λ)f (x2 ; λ) = λ2 exp(−λ(x1 + x2 )) = λ2 exp(−λt). Hence, we can take g1 (t, λ) =
λ2 exp(−λt) and g2 (x1:2 ) = 1 and apply the Factorization theorem..
(b) Show the sufficiency condition using the definition of sufficient statistic and Bayes formula.
Hint: Use that X1 + X2 ∼ Γ(2, λ).
Solution:
By definition the statistic is sufficient for λ, if f (x1:2 |t) does not depend on λ.
f (x1:2 |t) =
f (t) =
f (t|x1:2 )f (x1:2 )
f (t)
λ2 t exp(−λt)
Γ(2)
f (x1:2 )
f (t)
2
= λλ2 texp(−λt)
= Γ(2)
exp(−λt)
t , where we used that X1 + X2 ∼ Γ(2, λ), hence,
Γ(2)
1
if t = x1 + x2
and f (t|x1:2 ) =
0
otherwise
=
(c) Show the sufficiency condition using the result below by applying the following steps.
Result: When X1 + X2 = t whatever be λ, (X1 , X2 ) is conditionally distributed as (X, Y ), where X is
uniform on (0, t) and Y = t − X. Thus, X1 + X2 is sufficient.
X1
X1 +X2
(1)
and X1 + X2 are independent.
1
), derive the joint density function f (y1 , y2 ) using the
Hint: Let (y1 , y2 ) = g(x1 , x2 ) = (x1 + x2 , x1x+x
2
change of variables formula.
Solution:
(
y1 := x1 + x2
λ2 t exp(−λt)
As before, f (t) =
. We denote
.
Γ(2)
1
y2 := x1x+x
2
1
Then, (y1 , y2 ) = g(x1 , x2 ) = (x1 + x2 , x1x+x
). Then g is one-to-one on S = {(x1 , x2 ) : x1 > 0, x2 > 0}
2
and its range is S1 = {(y1 , y 2 ) : y1 > 0,0 < y2 < 1}. We note that on S1 , g −1 (y1 , y2 ) = (y1 y2 , y1 −y1 y2 ).
y 1 − y2 = −y1 . Then we obtain fY ,Y (y1 , y2 ) = fX ,X (g −1 (y1 , y2 )) ·
Therefore, Jg−1 (y1 , y2 ) = 2
1
2
1
2
y1
−y1 2
|Jg−1 (y1 , y2 )| = fX1 ,X2 (y1 y2 , y1 − y1 y2 ) · | − y1 | = λ y1 exp(−λy1 ). Hence, the density functions of Y1
and Y2 are gλ (y1 ) = λ2 y1 exp(−λy1 ) ⇒ Γ(2, λ), as in Hint (b), and h(y2 ) = 1, respectively.
1
and X1 + X2 are independent.
We conclude that X1X+X
2
(2)
X1
X1 +X2
is uniformly distributed on (0, 1).
Solution:
In part a) we saw that the density function of
on (0, 1).
X1
X1 +X2
is equal to 1. Hence, it is uniformly distributed
1
(3) The conditional distribution of X1X+X
given X1 + X2 = t is uniformly distributed on (0, 1) for all t.
2
Solution:
1
1
As X1X+X
and X1 + X2 are independent, the conditional distribution of X1X+X
given X1 + X2 = t is
2
2
X1
equal to the distribution of X1 +X2 . Thus, is uniformly distributed on (0, 1) for all t.
(4) Given X1 + X2 = t, X1 =
X1
X1 +X2
(X1 + X2 ) and
X1 t
X1 +X2
are the same.
Hence, the conditional distribution of X1 , given X1 + X2 = t is uniformly distributed on (0, t).
Solution:
P(X1 ≤ a|X1 +X2 = t) = P
a
t.
X1
X1 +X2
a
1
(X1 + X2 ) ≤ a|X1 + X2 = t = P X1X+X
≤
|X
+
X
=
t
=
1
2
t
2
Problem 9
Let X1 , . . . , Xn be independent random variables with pmf f (x; π) = (1 − π)x π and let Y1 , . . . , Yn be independent
random variables with pmf f (y; p) = (1 − p)y−1 p.
(a) What is the sample space of X1 and Y1 ? Try to give a probabilistic interpretation of such sample spaces!
Hint: for example, a Bernoulli random variable can be used to model a coin tossing with probability of
success p ∈ ]0, 1[.
Solution:
X = # failures until first success, ΩX = {0, 1, 2 . . . }.
Y = # trials until first success (inclusive), ΩY ∈ {1, 2, 3, . . . }.
(b) Calculate π
bML and pbML .
Solution:
1
π
bML =
,
1 + x̄
pbML =
1
.
ȳ
(c) Factorize the density f (x1:n ; π) = g1 (T (x1:n ); π)g2 (x1:n ) and identify a sufficient statistic.
Solution:
g1 (T (x), π) = π n (1 − π)
P
i
xi
,
g2 (x) = 1,
http://www.math.uzh.ch/mat924.1|04exercise.Rnw
T (x) =
P
i
xi .
2014-10-07|2014-10-15