Page 265 - Applied Probability

11. Radiation Hybrid Mapping
252
Show that [9, 14]
r(1 − r)(2 − θ i )
=
J θ i θ i
(1 − θ i r)θ i (1 − θ i + θ i r)
J θ i r
= J rθ i
(1 − 2r)(1 − θ i )
=
(1 − θ i r)(1 − θ i + θ i r)
m−1
1 1 − r r
J rr = + θ i + .
r(1 − r) r(1 − θ i r) (1 − r)(1 − θ i + θ i r)
i=1
Prove that all other entries of J are 0. Hints: Use the factorization
property of the likelihood. In the case of two loci, denote the proba-
bility Pr(X 1 = i, X 2 = j)by p ij for brevity. Then a typical entry J αβ
of J is given by
1 1
1 ∂p ij ∂p ij
= .
J αβ
p ij ∂α ∂β
i=0 j=0
has a maximum at r = 1
2
6. Continuing the last problem, prove that J θ i θ i
≤ 1/[2θ i(1 − θ i )].
when θ i is fixed. Use this fact to show that J θ i θ i
Given a known retention probability r, this inequality proves that
√
the asymptotic standard error of the estimated θ i will be at least 2
times greater than that calculated for a simple binomial experiment
with success probability θ.
7. Complete the calculation of the partial derivatives of the likelihood
for a single clone under the polyploid model by specifying the partial
derivatives ∂ t c,i , ∂ t c,i , and ∂ α 1 (j 1 ) appearing in equations (11.16)
∂θ i ∂r ∂r
and (11.17).
8. Under the polyploid model for two loci, consider the map
(θ, r) → (q 00 ,q 11 )
= Pr(X 1 =0,X 2 =0)
q 00
= Pr(X 1 =1,X 2 =1).
q 11
Show that this map from {(θ, r): θ ∈ [0, 1],r ∈ (0, 1)} is one to one
and onto the region
2
Q = {(q 00 ,q 11 ): q 00 ∈ (0, 1),q 11 ∈ (0, 1),q 00 q 11 ≥ q },
01
where q 01 =Pr(X 1 =0,X 2 = 1). Prove that θ = 0 if and only if
2
q 00 + q 11 = 1, and θ = 1 if and only if q 00 q 11 = q . The upper
01
boundary of Q is formed by the line q 00 + q 11 = 1 and the lower
2
boundary by the curve q 00 q 11 = q . Prove that the curve is generated
01
√
by the function q 11 =1 + q 00 − 2 q 00 .