Tom Tat Ly Thuyet Xac Suat

Tm tt L thuyt xc sut

TS. L Xun Trng

Khoa Ton Thng k

TS. L Xun Trng (Institute) Tm tt L thuyt xc sut 1 / 32

1. Khng gian Xc sut

Mt khng gian xc sut l mt b ba (,F ,P), trong Khng gian mu : tp hp tt c cc kt qu c kh nng xy raca mt th nghim hay hin tng ngu nhin.

Khng gian cc bin c F : mt -i s cc tp con ca , tc lF 2 tha cc iu kin:

- F- A 2 F ) Ac 2 F- Nu Ai 2 F , 8i 2 N th [i=1Ai 2 F .

o Xc sut: mt hm P : F ! R tha mn cc tnh cht sau- P (A) 0, 8A 2 F- P () = 1- Nu A1,A2, ... l cc bin c ri nhau (Ai \ Aj = , 8 i 6= j) th

P ([i=1Ai ) = iP (Ai ) .


1. Khng gian Xc sut

V d 1: Xt th nghim "tung mt xc sc cn i v ng cht gm 6mt"

Khng gian mu: = f1, 2, 3, 4, 5, 6g .C th xc nh nhng khng gian cc bin c khc nhau trn cngkhng gian mu

- F0 = f,g- F1 = 2

o xc sut

- Trn F0: hm P : F0 ! R cho bi P () = 0 v P () = 1.- Trn F1: hm P : F1 ! R cho bi

P (A) =i

6,

vi i l s phn t ca A 2 F1.


2. Bin ngu nhin

Cho (,F ,P) l mt khng gian xc sut. Mt hm s : ! Rsao cho vi mi 2 R, ta c

f g := f 2 : () g 2 F ,

c gi l mt bin ngu nhin (hay mt hm F -o c) trn .

V d 2: Gi s A 2 F . Khi hm s IA () =

1, 2 A,0, /2 A, l

mt bin ngu nhin trn v nu 2 R, ta c

f 2 : IA () g =

8

3. Sigma i s sinh bi bin ngu nhin

Gi s : ! R l mt bin ngu nhin. Ta k hiu () l h ttc cc tp con ca c dng

1 (B) := f 2 : () 2 Bg ,trong B l mt tp Borel ca R (B 2 B(R)). Khi () l mtsigma i s trn v ta gi l sigma i s sinh bi bin ngu nhin.

V d 3: Xc nh sigma i s c sinh bi bin ngu nhin IA ()cho trong V d 2.Gii Xt B 2 B(R). Khi

I1A (B) =

8>>>>>:, nu 0, 1 /2 B ,A, nu 1 2 B v 0 /2 B ,Ac , nu 0 2 B v 1 /2 B ,, nu 0, 1 2 B .

Vy (IA) = f,,A,Acg .TS. L Xun Trng (Institute) Tm tt L thuyt xc sut 5 / 32

4. Phn b xc sut ca bin ngu nhin

Xt khng gian xc sut (,F ,P) v : ! R l mt bin ngu nhin.Ta c th xc nh mt o xc sut P xc nh trn B(R) nh sau

P (B) = P1 (B)

.

Ta gi P l phn b xc sut ca bin ngu nhin .

Hm s F : R ! [0, 1] xc nh bi

F (x) = P ((, x ]) = P ( x) , x 2 R,

c gi l hm phn b xc sut tch ly ca .


5. Hm mt ca bin ngu nhin

Cho l mt b.n.n xc nh trn khng gian xc sut (,F ,P).Nu tn ti mt dy s thc i mt khc nhau x1, x2, ... sao cho

P (B) = P ( 2 B) = xi2B

P ( = xi ) ,

vi mi B 2 B(R), th ta ni c phn b ri rc vi cc gi trx1, x2, ... v mt f (xi ) := P ( = xi ) ti xi .

Nu tn ti mt hm kh tch Lebesgue f : R ! R sao cho

P (B) = P ( 2 B) =Z

Bf (x) dx ,

vi mi B 2 B(R), th ta ni c phn b lin tc tuyt i v fc gi l hm mt ca .


6. Cc c trng ca bin ngu nhin

Cho l mt b.n.n xc nh trn khng gian xc sut (,F ,P).K vng (Expectation):

E () =

8


Lu :

Trong nh ngha k vng, ta hiu rng tng

xP ( = x) hocZ +

xf (x) dx

phi tn ti hu hn. Lc ta ni kh tch.

E (IA) = P (A)

Nu c moment bc k th s c mi moment bc b hn k .

Nu Ejjk

< th xc sut ui (tail probability) hi t v 0, tc

l

limn!+

P (jj > n)nk

= 0.



Mt s tnh cht

E (a) = a, 8a 2 R.E (a1 + b2) = aE (1) + bE (2)

Var (a) = 0, 8a 2 R.Var (a1) = a

2Var (1) , Var (1 + 2) = Var (1) + Var (2)

Nu l mt bin ngu nhin v g : R ! R l mt hm o c th

E (g ()) = g (x)P ( = x) , khi c phn b ri rc,

E (g ()) =R + g (x) f (x) dx , khi c phn b lin tc.

V d 4: Gi s l bin ngu nhin c hm mt xc nh bi

P ( = 0) = 0, 2; P ( = 1) = 0, 5; P ( = 2) = 0, 3.

Tnh E () v E2.


7. Mt s phn b thng gp

Phn b Hm mt K vng Phng sai

B (n; p) P ( = k) = C kn pk (1 p)nk np np (1 p)

P() P ( = k) = e kk !

U (a; b) f (x) =

1ba , x 2 (a, b)0, x /2 (a, b)

a+b2

(ba)212

Exp () f (x) =

ex , x 0,0, x < 0.

1

12

N; 2

f (x) = 1

p

2e

(x)222 2


8. Phn b ng thi v phn b bin

Cho 1, 2 l hai b.n.n xc nh trn (,F ,P).Vi A,B 2 B(R), ta t

P12 (A B) = P (1 2 A, 2 2 B) .

Khi P12 l o xc sut trn B(R2) v gi l phn b xc sutng thi ca 1, 2.

Hm phn b xc sut tch ly ng thi ca hai bin ngu nhin 1v 2 c xc nh nh sau

F12 (x , y) = P (1 x , 2 y) .

Hm phn b xc sut tch ly bin

- theo 1 : F1 (x) = P12 ((, x ]R) = P (1 x , 2 +) .- theo 2 : F2 (y) = P12 (R (, y ]) = P (1 +, 2 y) .


9. Hm mt ng thi - hm mt bin

Gi s 1 v 2 l hai bin ngu nhin c phn b ri rc v

1 2 fx1, x2, ...g , 2 2 fy1, y2, ...g .

Phn b xs ng thi ca 1, 2 cho bi

P12 (A B) = xi2A,yj2B

P (1 = xi , 2 = yj ) .

Ta gi f12 (x , y) = P (1 = x , 2 = y) l hm mt ng thi.

Hm mt bin theo 1 v 2 ln lt xc nh bi

f1 (xi ) = jP (1 = xi , 2 = yj ) ,

f2 (yj ) = iP (1 = xi , 2 = yj ) .


9. Hm mt ng thi - hm mt bin

Nu tn ti mt hm kh tch Lebesgue f12 : R2 ! R sao cho

P12 (A B) =RR

AB f12 (x , y) dxdy ,

th ta ni 1 v 2 c phn phi xs ng thi lin tc tuyt i. Hmf12 c gi l hm mt xc sut ng thi ca 1, 2.

Hm mt xc sut bin theo 1 v 2

f1 (x) =Z +

f12 (x , y) dy ,

f2 (y) =Z +

f12 (x , y) dx .


10. Hm ca cc bin ngu nhin

Cho 1 v 2 l hai b.n.n trn khng gian (,F ,P) c hm mt ngthi l f12 (x , y) v g : R

2 ! R l hm Borel. Khi

g (1, 2) l mt bin ngu nhin trn (,F ,P).K vng

- nu 1 v 2 ri rc

E (g (1, 2)) = x

yg (x , y) f12 (x , y)

- nu 1 v 2 lin tc

E (g (1, 2)) =Z +

Z +

g (x , y) f12 (x , y) dxdy


10. Hm ca cc bin ngu nhin

Cho 1 v 2 l hai b.n.n trn khng gian (,F ,P) c hm mt ngthi l f12 (x , y). Gi s

1 = g1 (X ,Y ) , 2 = g2 (X ,Y ) ,

trong X ,Y l hai b.n.n v nh thc Jacobi

J =

g1X g1Yg2X

g2Y

6= 0.Khi hm mt ng thi fXY (x , y) ca X ,Y tha mn

fXY (x , y) = f12 (g1 (x , y) , g2 (x , y)) jJ j .


11. Mt s v d

V d 5: Cho hai bin ngu nhin ri rc X ,Y vi hm mt xc sutng thi fXY xc nh bi

fXY (1, 1) = P (X = 1,Y = 1) = 0, 5

fXY (1, 2) = P (X = 1,Y = 2) = 0, 1

fXY (2, 1) = P (X = 2,Y = 1) = 0, 1

fXY (2, 2) = P (X = 2,Y = 2) = 0, 3

Tnh xc sut P (X = 1) v P (Y = 1) .


11. Mt s v d

V d 6: Cho hai bin ngu nhin X ,Y vi hm mt ng thi

fXY =

6xy (2 x y) , (x , y) 2 (0, 1) (0, 1) ,0, (x , y) /2 (0, 1) (0, 1) .

Tm fX (x) v fY (y) .

V d 7: Cho hai bin ngu nhin X ,Y c hm mt ng thi

fXY =

x + y , (x , y) 2 (0, 1) (0, 1) ,0, (x , y) /2 (0, 1) (0, 1) .

Tnh E (X ) ,E (Y ) ,Var (X ) ,Var (Y ) v cov (X ,Y ) .

(cov (X ,Y ) = E ((X E (X )) (Y E (Y )))= E (XY ) E (X ) .E (Y ))


11. Mt s v d

V d 8: Cho hai bin ngu nhin X ,Y vi hm mt ng thi

fXY (x , y) =

x + y , 0 x y 1,0, otherwise.

t Z = X + Y . Xc nh hm mt v hm phn phi tch ly ca Z .

V d 9: Gi s X ,Y l hai bin ngu nhin c hm mt ng thi l

fXY (x , y) =

ex1 , 0 x2 x1 < +,0, otherwise.

t Z1 = X + Y v Z2 = X Y . Tm hm mt ng thi ca Z1,Z2v hm phn b tch ly bin ca Z2.


12. Xc sut c iu kin. Tnh c lp

Cho A,B 2 F sao cho P (B) 6= 0. Xc sut c iu kin ca A khicho trc B xc nh bi

P (AjB) = P (A\ B)P (B)

.

Hai bin c A,B 2 F c gi l c lp nu

P (A\ B) = P (A) .P (B)

Hai bin ngu nhin v c gi l c lp nu vi mi A,B 2B(R) ta c

f 2 Ag v f 2 Bgl hai bin c c lp.

Ta ni hai sigma i s F1 v F2 l c lp nu 8A 2 F1, 8B 2 F2th A,B c lp.


12. Xc sut c iu kin. Tnh c lp

Theorem

Nu v l hai bin ngu nhin kh tch th chng khng tng quan,tc l

E (.) = E ()E () hay suy ra cov (, ) = 0.


13. Hm sinh monment

Hm sinh moment ca bin ngu nhin c xc nh bi

M (t) = Eet,

nu k vng tn ti. Nh vy, nu f l hm mt ca th

M (t) = x etx f (x) , khi ri rc,

M (t) =R + e

tx f (x) dx , khi lin tc.

Ch :

- Hm sinh moment cho php xc nh cc moment Mk () = E

k.


13. Hm sinh monment

Theorem

Nu M (t) hu hn trn mt khong m cha 0 th hm t 7! M (t) khvi mi cp v

M(k) (t) = E

et .k

.

H qu l

M(k) (0) = E

k= Mk () .

Chng minhddtM (t) =

ddt

R + e

tx f (x) dx =R + xe

tx f (x) dx = Eet

,

d2

dt2M (t) =

ddt

R + xe

tx f (x) dx =R + x

2etx f (x) dx = E2et

,


13. Hm sinh monment

V d 10:

U (a, b)

M (t) =1

b a

Z ba

etxdx =ebt eat(b a) t .

Exp()

M (t) = Z +

0etx .exdx =

t (t < ) .

N; 2

M (t) = exp

t +

2t

2

P()

M (t) =

k=0etk

ek

k != exp ( exp (t) )


13. Hm sinh monment

Theorem

Nu hm sinh moment M (t) ca b.n.n tn ti trong ln cn ca 0

th n xc nh duy nht hm phn phi xc sut, ngha l khng thc hai bin ngu nhin khc nhau c cng hm sinh moment.

Nu v l hai bin ngu nhin c lp th

M+ (t) = M (t) .M (t)

V d 11: Gi s hm sinh moment ca mt bin ngu nhin X l

MX (t) = e3(et1).

Tnh P (X = 0).


14. S hi t ca cc bin ngu nhin14.1. Mt s bt ng thc quan trng

Lemma

(Bt ng thc Markov) Cho l bin ngu nhin khng m, ngha lP ( 0) = 1. Khi , vi mi c > 0,

P ( c) 1cE () .

Lemma

(Bt ng thc Chebyshev) Cho l bin ngu nhin c k vngE () = v phng sai Var () = 2. Khi , vi mi c > 0, ta c

P (j j c) 2

c2


14. S hi t ca cc bin ngu nhin14.2. Hi t theo xc sut

Ta ni dy bin ngu nhin (n) hi t theo xc sut n bin ngunhin nu vi mi > 0,

limn!

P (jn j > ) = 0.

K hiu nP! .

V d 12: Gi s Xn l bin ngu nhin c hm mt

fXn (x) =

8

14. S hi t ca cc bin ngu nhin14.3. Hi t theo phn b

Cho dy b.n.n (n) c cc hm phn b l Fn (x) v b.n.n c hm

phn b l F (x). Ta ni n hi t theo phn b v , k hiu

nd! , nu

limn!

Fn (x) = F (x) ,

ti mi x sao cho F lin tc.

V d 13: Gi s Xn = 1+1n l b.n.n hng s. Hm phn b ca Xn

l

FXn (x) =

0, x < 1+ 1/n,1, x 1+ 1/n.

Ta c Xnd! X , trong X = 1 l b.n.n hng s.


14. S hi t ca cc bin ngu nhin14.4. Hi t trung bnh

Ta ni dy b.n.n (n) hi t theo trung bnh bc p v b.n.n , k hiu

nLp! , nu

limn!+

E (jn jp) = 0.

Quan h gia cc loi hi t


14. S hi t ca cc bin ngu nhin14.5. Lut s ln

Theorem (Lut s ln)

Gi s (n) l dy cc b.n.n c lp c cng phn b xc sut v

E (n) = ,Var (n) = 2, 8n 2 N.

Khi

n =1 + 2 + + n

n

P! khi n ! +.


14. S hi t ca cc bin ngu nhin14.6. nh l gii hn trung tm

Theorem (nh l gii hn trung tm)

Gi s (n) l dy cc b.n.n c lp c cng phn b xc sut v

E (n) = ,Var (n) = 2, 8n 2 N.

Khi

n =(1 + 2 + + n) n

pn

d! Z N (0, 1) .

V d 14: Gi s n B (1, p) , 8n 2 N. Khi = 1 + 2 + + n B (n, p). Theo nh l gii hn trung tm, ta c

nppnp (1 p)

xp x vi b.n.n Z N (0, 1) .TS. L Xun Trng (Institute) Tm tt L thuyt xc sut 31 / 32

14. S hi t ca cc bin ngu nhin14.6. nh l gii hn trung tm

V du 15: Gi s

- thi gian s dng mt loi pin c k vng 40 gi v phng sai 20 gi

- khi mt cc pin ht nng lng, ngi ta thay n bng mt cc khc

- ta s dng 25 cc pin v tui th ca chng l c lp

Tnh xc sut tng thi gian s dng ln hn 1100 gi.


Khng gian Xc sut

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Tom Tat Ly Thuyet Xac Suat