Post your solutions to the Nov 7 In-Class Exercise tot this thread.
Best,
Chris
len(d) = 250, ftd = 2, number of times star appears in the corpus = 73695, words in corpus = 450000000, Mct = 73695/450000000 = 0.00016,
Mμd(t) = 2 + μ (0.00016)/250+μ,
ftd = 0, 4.) Mμd(t) = 0.16/1250 = 0.00013
(Edited: 2018-11-07)when u = 100, M_u_d (t) = (2 + 100 * ( 73695/450000000) ) / (250 + 100) = 0.005 when u = 1000, M_u_d (t) = (2 + 1000 * ( 73695/450000000) ) / (250 + 1000) = 0.0017 when u = 10000, M_u_d (t) = (2 + 10000* ( 73695/450000000) ) / (250 + 10000) = 0.00035
if star does not appear in d, u = 1000, M_u_d = (1000 * ( 73695/450000000) ) / (250 + 1000) = 0.00013
(Edited: 2018-11-07)M_d^(\mu) (t) = (2 + (73695/450000000) * \mu) / (250 + \mu)
\mu = 100, M_d^(\mu) (t) = 0.5761%
\mu = 1000, M_d^(\mu) (t) = 0.1731%
\mu = 10000, M_d^(\mu) (t) = 0.0355%
Where star does not appear in d and \mu = 1000:
M_d^(\mu) (t) = 0.0131%
a)M_d^mu(t) = (2 + 100(73695 / 450000000)) / (250+100) = 0.0058
b)M_d^mu(t) = (2 + 1000(73695 / 450000000)) / (250+1000) = 0.0017
c)M_d^mu(t) = (2 + 10000(73695 / 450000000)) / (250+10000) = 0.0004
d)M_d^mu(t) = (0 + 1000(73695 / 450000000)) / (250+1000) = 0.0001
mu= 100M_d^100(mbox{star}) = 2+(100*0.00016)/350 = 2.016/350 = 0.005761
mu= 1000M_d^1000(mbox{star}) = 2+(1000*0.00016)/1250 = 2.1637/1250 = 0.0017
mu= 10000M_d^10000(mbox{star}) = 2+(10000*0.00016)/10250 = 3.6/10250 = 0.00035
If star doesn't appear in d
M_d^1000(mbox{star}) = 0+(1000*0.00016)/1248 = 0.000128
M_C("star") = l_t/l_C
M_C("star") = 73695 / 450000000
M_C("star") = 0.000163767
(a) mu = 100
#M_d^100("star") = (2 + (100 * 0.000163767))/(250 + 100)
#M_d^100("star") = 0.005761076
(b) mu = 1000
#M_d^1000("star") = (2 + (1000 * 0.000163767))/(250 + 1000)
#M_d^1000("star") = 0.001731014
(c) mu = 10000
#M_d^10000("star") = (2 + (10000 * 0.000163767))/(250 + 10000)
#M_d^10000("star") = 0.000354895
(d) mu = 1000 and f_(t,d) = 0
#M_d^1000("star") = (0 + (1000 * 0.000163767))/(250 + 1000)
#M_d^1000("star") = 0.000131014
<pre>
a) μ = 100; = (f + μ⋅MC(t))/ (ld+μ) = (2+100.(73695/45010^6))/(250+100) = 2.0164/350 = '''0.00576''' b) mu = 1000 => (2+(73695/45010^3))/250+1000 = 2.1637/1250 =''' 0.00173''' c) mu= 10000 => (2+(73695/450*10^2))/250+10000 = 3.637/10250 = '''0.00035'''
0+(73695/450*10^3)/250+1000 => 0.0164/1250 = '''0.000013'''
</pre>
(Edited: 2018-11-12)