MATH6630 Assignment 2Applicati on of EM Algorithm

In study of HIV infecti on

b y Weiqing Fang

Multifunctional Materials | Micro-and-Nanostructuring Laboratory Slide 2

Introduction

IntroductionIntroduction

Slide 3

Sexual behavior of individuals ate risk for HIV infectionSuppose 1500 gay men were surveyed and each was asked how many risky sexual encounters he had in the previous 30 days. Let ni denote the number of respondents reporting i encounters. The reporting numbers are summarized in following table.

i 0 1 2 3 4 5 6 7 8

ni 379 299 222 145 109 95 73 59 45

i 9 10 11 12 13 14 15 16

ni 30 24 12 4 2 0 1 1

IntroductionIntroduction

Slide 4

: Probability of people representing typical behavior, who respond truthfully and their numbers of risky encounters following Poission()

: Probability of people report zero risky encounters even if this is not true.

1- -: Probability of people representing high-risk behavior, who respond truthfully and their numbers of risky encounters following Poission()

E - Step

E - step

Slide 6

Likelihood function

16

0 160

( )( | n ,..., n )

!

ini

i

Li

{i 0}( ) 1 (1 )i ii e e

Where

[ ]

E – step: Compute Q(|(t))E - step

Slide 7

( )0 (t)

0

( | )( )

tZ

( )( )

( | )( )

it

i ti

et

( )( )

(1 )( | )

( )

it

i ti

ep

Probability the i risky encounter belong to zero group:

Probability the i risky encounter belong to typical group:

Probability the i risky encounter belong to high-risk group:

M - Step

M – Step: Maximize Q(|(t)) M - Step

Slide 9

• Derive the 1st order partial derivatives of Q(|(t)) respect to

• Set 1st order partial derivatives to zero and solve (t+1) equal to maximizer of

Derive 1st order partial derivativesM - Step

Slide 10

( ) 160( ) ( )

10

( | )(1 ) ( )

( ) ( )

tii

t ti i

n nQe e

( ) 16

( )0

( | )( )

( )

ti ii

ti i

nQe e

( ) 161

( )0

( | )( )

( )

ti ii

ti i

nQi e e

( ) 161

( )0

( | )(1 )( )

( )

ti ii

ti i

nQi e e

Solve (t+1) equal to maximizer of M - Step

Slide 11

( )( 1) 0 0 ( )tt n z

N

( )16( 1)

0

( )tt i i

i

n t

N

16( )

( 1) 016

( )

0

( )

( )

ti i

t i

ti i

i

in t

n t

16( )

( 1) 016

( )

0

( )

( )

ti i

t i

ti i

i

in p

n p

Update [ ]

Results obtained using RM - Step

Slide 12

R1 0.2367108 0.5982755 3.7583188 7.9695942 0.21686172 0.18435487 0.64339931 3.83912238 7.93340085 0.012713743 0.17445448 0.64914307 3.85652696 7.78296758 0.017181914 0.17214260 0.64726307 3.85377345 7.69155469 0.010503075 0.171290563 0.644548457 3.845279737 7.642039844 0.0058313116 0.170665118 0.642305767 3.836370909 7.613279770 0.0035192357 0.170094226 0.640599481 3.828644860 7.594382538 0.0023967358 0.169588178 0.639297710 3.822346057 7.580625490 0.0017811959 0.16916086 0.63828644 3.81731303 7.57001612 0.0013849910 0.168810776 0.637489459 3.813311140 7.561629160 0.00109757811 0.168528271 0.636856228 3.810128908 7.554942270 0.000875622

Estimate standard errors

Standard ErrorEstimate standard errors

Slide 14

10 0

T1

0

0, I( )

( | ) ( | )I( ) i i

ni

N

l n l n

Use the sample covariance of the first derivate, instead of calculating the second derivative.

Slide 15

Questions ?