「ベータ分布の謎に迫る」第6回 プログラマのための数学勉強会 発表資料 (2016/3/19[sat]) 確率・統計を学んだことがある方向けに、ベータ分布とは何かを解説してみた記事です。特にベイズ統計学を学んでいるとベータ分布が出現しますが、いまいちどんな事象が対応している分布かわかりにくいので、その辺りに迫ります。

6. f(x) =
x↵ 1
(1 x) 1
B(↵, )
B(x, y) =
Z 1
0
tx 1
(1 t)y 1
dt
B(↵, )
0  x  1
↵
↵ +
↵
(↵ + )2(↵ + + 1)
x, y > 0,
↵, > 0,

10. (́・ω・) ん?
(́・ω・) で、どういう事象がこの分布するの?

12. 0  x  1
f(p) =
p↵ 1
(1 p) 1
B(↵, )
, 0  p  1
0  t  1
B(↵, ) =
Z 1
0
p↵ 1
(1 p) 1
dp

13. 0  x  1
f(p) =
p↵ 1
(1 p) 1
B(↵, )
, 0  p  1
0  t  1
B(↵, ) =
Z 1
0
p↵ 1
(1 p) 1
dp

14. Z 1
0
f(p) =
Z 1
0
p↵ 1
(1 p) 1
B(↵, )
dp
=
Z 1
0
p↵ 1
(1 p) 1
dp
1
B(↵, )
=
B(↵, )
B(↵, )
= 1

15. f(p) =
p↵ 1
(1 p) 1
B(↵, )
/ p↵ 1
(1 p) 1
B(↵, )

16. p) =
p↵ 1
(1 p) 1
B(↵, )
, 0  p  1
) =
p↵ 1
(1 p) 1
B(↵, )
, 0  p  1

17. (́・ω・) ん?
f(p) =
p↵ 1
(1 p) 1
B(↵, )
,

18. (́・ω・) ん?
(́・ω・) よく見るとコイン投げ？
f(p) =
p↵ 1
(1 p) 1
B(↵, )
,

19. x = 1 x = 0
f(x; p) =
8
<
:
p if x = 1,
1 p if x = 0.
f(x; p) = px
(1 p)1 x
, x = {0, 1}
x
p
p
1 p

20. f(x1, · · · , xn; p) =
nY
i=1
pxi
(1 p)1 xi
, xi = {0, 1}
f(x1, · · · , xn; p) = pa
(1 p)b a =
nX
i=1
xi, b = n a

21. f(x1, · · · , xn; p) =
nY
i=1
pxi
(1 p)1 xi
, xi = {0, 1}
f(x1, · · · , xn; p) = pa
(1 p)b a =
nX
i=1
xi, b = n a
f(p) =
p↵ 1
(1 p) 1
B(↵, )
/ p↵ 1
(1 p) 1

22. f(p) = p↵ 1
(1 p) 1
f(x1, · · · , xn; p) = pa
(1 p)b a =
nX
i=1
xi, b = n a
x1, · · · , xn
🌾

23. f(x1, · · · , xn; p) = pa
(1 p)b a =
nX
i=1
xi, b = n a
x1, · · · , xn
🌾
f(p) = p↵ 1
(1 p) 1

24. f(x1, · · · , xn; p) = pa
(1 p)b a =
nX
i=1
xi, b = n a
x1, · · · , xn
🌾
f(p) = p↵ 1
(1 p) 1

25. でもこのpを確率変数とみなしちゃう、
そう、ベイズならね (๑•`ㅁ•́๑)✧

26. < < < < < < < <
< < < < < < < <

27. … …
n = i + j 1
f(p) =
(
n!
(i 1)!(n i)! pi 1
(1 p)j 1
if 0  p  1
0 otherwise
<< < < < <
p ⇠ Unif(0, 1)

28. f(p) =
(
n!
(i 1)!(n i)! pi 1
(1 p)j 1
if 0  p  1
0 otherwise
… …<< < < < <
i-1個
p ⇠ Unif(0, 1)

29. f(p) =
(
n!
(i 1)!(n i)! pi 1
(1 p)j 1
if 0  p  1
0 otherwise
… …<< < < < <
i-1個
p ⇠ Unif(0, 1)

30. B(a, b) =
(a) (b)
(a + b)
(n + 1) = n!、
=
1
B(i, j)
n = i + j 1 ! n i = j 1
n!
(i 1)!(n i)!
=
n!
(j 1)!(i 1)!
=
(n + 1)
(i) (j)
=
(i + j)
(i) (j)

31. f(p) =
(
n!
(i 1)!(j 1)! pi 1
(1 p)j 1
if 0  p  1
0 otherwise
=
(
1
B(i,j)! pi 1
(1 p)j 1
if 0  p  1
0 otherwise
f(p)
🌾

32. f(p) =
(
n!
(i 1)!(j 1)! pi 1
(1 p)j 1
if 0  p  1
0 otherwise
=
(
1
B(i,j)! pi 1
(1 p)j 1
if 0  p  1
0 otherwise
f(p)
🌾

33. @interact(a=(1,15,1),
b=(1,15,1))
def
draw_norm_dist(a=2,
b=2):
set_size
=
a+b-­‐1
trial_size
=
30000
bin_width
=
51
def
gen_orderd_unif(size):
unif
=
rd.rand(size)
unif.sort()
return
unif
result
=
[gen_orderd_unif(set_size)[a-­‐1]
for
_
in
np.arange(trial_size)]
plt.hist(result,
bins=np.linspace(0,1,bin_width))
plt.plot(p,
st.beta.pdf(p,
a,
b)*trial_size/bin_width,
c="g",
lw=3)
plt.show()

35. おわり

36. どうしても順序統計量が
気になる人のためのAPPENDIX

37. FX (x)
fX (x)
X1, X2, · · · , Xn FX (x) fX (x)
X(1), X(2), · · · , X(n) Xi
X(j), j = 1, 2, · · · , n
fX(j)
=
n!
(j 1)!(n j)!
fX(x)[FX(x)]j 1
[1 FX(x)]n j
1 FX(x)FX(x)
x
fX (x)
i
j

38. Y = #{Xj, j = 1, 2, · · · , n|Xj  x}
x
Y個
Zj =
(
1 if {Xj  x}
0 otherwise
x
Z4=1
Z3=1
Z9=1
Z8=1 Z6=1
Z2=1 Z1=0
Z5=0
Z7=0
Y =
nX
j=1
Zj

39. P(Zj = 1) = Pi = FX(x)
1
1O
FX(x)
Xi
Zi
Zi
xP(Zj = 1) = Pi = FX (x)
x
Zi
Pi

40. FX(j)
(x) = P(Y j) =
nX
k=j
✓
n
k
◆
[FX (x)]k
[1 FX(x)]n k
1
FX (x) = P6
x
Y j
Xi x

41. FX(j)
(x) fX(j)
(x)
fX(j)
(x) =
dFX(j)
(x)
dx
(f · g)0
= f0
g + fg0
=
d
dx
nX
k=j
✓
n
k
◆
[FX(x)]k
[1 FX(x)]n k
[f(g(x))]0
= f0
(g(x))g0
(x)
=
nX
k=j
✓
n
k
◆
kfX(x)[FX(x)]k 1
[1 FX(x)]n k
(n k)fX(x)[FX(x)]k
[1 FX(x)]n k 1
=
✓
n
k
◆
jfX (x)[FX (x)]j 1
[1 FX (x)]n j
+
nX
k=j+1
✓
n
k
◆
kfX (x)[FX (x)]k 1
[1 FX (x)]n k
n 1X
k=j
(n k)fX (x)[FX (x)]k
[1 FX (x)]n k 1

42. =
n!
(j 1)!(n j)!
fX (x)[FX (x)]j 1
[1 FX (x)]n j
+
n 1X
k=j
✓
n
k + 1
◆
(k + 1)fX (x)[FX (x)]k
[1 FX (x)]n k 1
n 1X
k=j
✓
n
k
◆
(n k)fX (x)[FX (x)]k
[1 FX (x)]n k 1
=
✓
n
k
◆
jfX (x)[FX (x)]j 1
[1 FX (x)]n j
+
nX
k=j+1
✓
n
k
◆
kfX (x)[FX(x)]k 1
[1 FX (x)]n k
n 1X
k=j
✓
n
k
◆
(n k)fX(x)[FX (x)]k
[1 FX (x)]n k 1

43. =
n!
(j 1)!(n j)!
fX (x)[FX(x)]j 1
[1 FX (x)]n j
+
n 1X
k=j
✓
n
k + 1
◆
(k + 1)fX(x)[FX (x)]k
[1 FX (x)]n k 1
n 1X
k=j
✓
n
k
◆
(n k)fX (x)[FX (x)]k
[1 FX(x)]n k 1
=
n!
(j 1)!(n j)!
fX(x)[FX(x)]j 1
[1 FX(x)]n j
+ fX(x)[FX(x)]k
[1 FX(x)]n k 1
0
@
n 1X
k=j
✓
n
k + 1
◆
(k + 1)
n 1X
k=j
✓
n
k
◆
(n k)
1
A
= 0
✓
n
k + 1
◆
(k + 1) =
n!
k!(n k 1)!
=
✓
n
k
◆
(n k)
=
n!
(j 1)!(n j)!
fX(x)[FX(x)]j 1
[1 FX(x)]n j

44. 【証明】(cont.)
X1, X2, · · · , Xn
fX(j)
(x) =
n!
(j 1)!(n j)!
fX (x)[FX (x)]j 1
[1 FX (x)]n j
fX(x) =
(
1 0 < x < 1
0 otherwise
FX (x) =
8
><
>:
0 x  0,
x 0 < x < 1,
1 x 1
= 1, (0 < x < 1) = x, (0 < x < 1)
fX(j)
(x) =
(
0 otherwise
n!
(j 1)!(n j)!
xj 1
(1 x)n j
, 0 < x < 1
n = j + i 1 ! n j = i 1
!
n!
(j 1)!(i 1)!
xj 1
(1 x)i 1