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Binomial Distribution

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DISTRIBUSI BINOMIAL

Distribusi binomial berasal dari percobaan binomial yaitu suatu proses Bernoulli yang diulang sebanyak n kali dan saling bebas. Distribusi Binomial merupakan distribusi peubah acak diskrit. Secara langsung, percobaan binomial memiliki ciri-ciri sebagai berikut: – – – – percobaan tersebut dilakukan berulang-ulang sebanyak n kali setiap percobaan menghasilkan keluaran yang dapat dikategorikan sebagai gagal dan sukses probabilitas sukses p tetap konstan dari satu percobaan ke percobaan lain percobaan yang berulang adalah saling bebas

Ruang sampel A untuk percobaan E yang terdiri dari himpunan tak hingga tetapi masih terhitung dari titik – titik sampel: Jika S = Sukses dan G = Gagal E1 : S (sukses pada percobaan pertama) E2 : GS (gagal pada percobaan pertama dan sukses pada percobaan kedua) E3 : SG (sukses pada percobaan pertama, gagal pada percobaan kedua) E4 : GGS (gagal pada percobaan 1 dan 2, sukses pada percobaan ketiga) E5 : GSG (gagal pada percobaan 1 dan 3, sukses pada percobaan kedua) E6 : SGG (gagal pada percobaan 2 dan 3, sukses pada percobaan pertama)

En : SSS ...S GGG...G (sukses sebanyak x kali, gagal sebanyak n – x kali) x n x

Jika peluang sukses dinotasikan dengan p maka peluang gagal adalah q = 1 – p. Peubah acak X menyatakan banyaknya sukses dari n percobaan yang saling bebas. Maka peluang X pada masing – masing percobaan E adalah:

P(X)  p P(X)  qp  pq P(X)  pq
2 2

untuk E1 untuk E2 untuk E3

P(X)  q p  pq untuk E4 P(X)  qpq  pq 2 untuk E5 P(X)  pqq  pq 2 untuk E6 P(X)  p x q n  x untuk En

 2 Dapat dilihat bahwa E2 dan E3 memberikan hasil yang sama. Jumlahnya   , yaitu 1 

jumlah semua titik sampel yang mungkin menghasilkan x = 1 yang sukses dan n – x = 2 – 1 = 1 yang gagal dari 2 percobaan. Begitupun untuk E4 , E5 , dan E6 juga
 3 memberikan hasil yang sama. Jumlahnya   , yaitu jumlah semua titik sampel yang 1 

mungkin yang menghasilkan x = 1 yang sukses dan n – x = 3 – 1 = 2 yang gagal dari 3 percobaan. Secara umum, jumlah titik sampel yang mungkin untuk menghasilkan x sukses dan n – x gagal dalam n percobaan adalah banyaknya cara yang berbeda dalam
n mendistribusikan x sukses dalam barisan n percobaan, sehingga terdapat   cara.  x

Dan distribusi peluang atau Probability Mass Function (PMF) X dinyatakan pada definisi berikut:
n  n P(X  x)  f ( x)    p x q n x    p x (1  p)n  x  x  x

untuk x  1, 2,..., n dan 0  p  1 . Pembuktian distribusi Binomial merupakan suatu PMF. Bukti: Untuk membuktikan suatu peubah acak adalah PMF, maka harus ditunjukan: 1. 2. f ( x)  0

 f ( x)  1 x Akan ditunjukkan distribusi binomial memenuhi kedua syarat di atas: 1. f ( x)  0

Karena 0  p  1 dan nilai kombinasi pasti positif maka f(x) pasti positif. 2.

 f ( x)  1 x Menggunakan persamaan binomial Newton pada

 f ( x) , akan diperoleh: x  f ( x)    x  p (1  p) x x x 0

n

n  

n x

 ( p  (1  p)) n  1n  1

(1)

Dari 1 dan 2 dapat dikatakan bahwa distribusi binomial merupakan PMF.

Mean Jika X
B(n, p) ( X variabel random berdistribusi Binomial), maka nilai ekspektasi

dari X adalah
E ( X )   x.P( x) x   x.P( X  x) x 0 n

n

n   x.   p x (1  p) n  x x 0  x n n   x.   p x (1  p) n  x x 1  x n n!   x. p x (1  p) n  x x !(n  x)! x 1   x. x 1 n

n.(n  1)! . p. p x 1 (1  p) n  x x.( x  1)!(n  x)!

 np 

(n  1)! p x 1 (1  p) n  x x 1 ( x  1)!( n  x )! n Misalkan m = n – 1 dan s = x – 1, maka persamaan di atas menjadi
E ( X )  np m! p s (1  p)m s s  0 s !(m  s )! m Berdasarkan (1),

 s !(m  s)! p (1  p) s s 0

m

m!

m s

 1 , maka

E ( X )  np  s 0

m

m! p s (1  p) m  s s !(m  s )!

 np.1  np

Sehingga didapat mean dari X, E ( X )  np

Variansi
Var ( X )  E ( X 2 )  ( E( X ))2

Dalam meencari Var(X), kita harus tahu nilai ekspektasi dari X2 :

E ( X 2 )   x 2 .P( X  x)   x 2 . x 0 x 0

n

n

n! p x (1  p) n  x x !(n  x)!

E ( X 2 )   x 2 .P( x) x   x 2 .P ( X  x ) x 0 n n   x 2 .   p x (1  p) n  x x 0  x n n   x 2 .   p x (1  p) n  x x 1  x n n!   x2. p x (1  p) n  x x !(n  x)! x 1

n

  x2. x 1 n

n

n.(n  1)! . p. p x 1 (1  p) n  x x.( x  1)!(n  x)! (n  1)! p x 1 (1  p) n  x ( x  1)!(n  x)!

 np  x. x 1

Misalkan m = n – 1 dan s = x – 1, maka persamaan di atas menjadi
E ( X 2 )  np ( s  1) s 0 m m  m m! p s (1  p) m s  np  ( s  1)   p s (1  p) m s s !(m  s)! s 0 s 

E ( X 2 )  np  ( s  1) s 0

m

m! p s (1  p) m  s s !(m  s )!

m m  np  ( s  1)   p s (1  p) m s s 0 s  m  m m s  m m s  np.   s.   p (1  p)  1.   p s (1  p) m s  s 0 s   s 0  s    np[ mp  1]

 np[( n  1) p  1]  np[ np  p  1]

Var ( X )  E ( X 2 )  ( E ( X )) 2  np[np  p  1]  (np )2  (np )2  np 2  np  (np )2  np  np 2  np (1  p )
Sehingga didapat variansi dari X, Var ( X )  np(1  p)

Contoh: 1. Probabilitas bahwa sejenis komponen tertentu yang lolos uji kelayakan adalah ¾. Tentukan probabilitas dimana 2 dari 4 komponen yang selanjutnya diuji akan dinyatakan layak!
n  n P(X  x)  f ( x)    p x q n x    p x (1  p)n  x  x  x

p = ¾, q = 1 – ¾ = ¼ untuk x = 2
42

 4 3   1  P( X  2)        2 4   4 
2

 6.

9 1 27 .  16 16 128

2. Berdasarkan data biro perjalanan PT Sentosa, yang khusus menangani perjalanan wisata turis mancanegara, 20% dari turis menyatakan sangat puas berkunjung ke Indonesia, 40% menyatakan puas, 25% menyatakan biasa saja, dan sisanya menyatakan kurang puas. Apabila kita bertemu dengan 5 orang dari peserta wisata turis mancanegara yang pernah menggunakan jasa biro perjalanan tersebut. Tentukan probabilitas: a. Tepat 2 diantaranya menyatakan biasa saja

b. Paling banyak 2 diantaranya menyatakan sangat puas

Jawab: n=5 a. p = 0.25, q = 1 – 0.25 = 0.75
5 2

5 1   3  P( X  2)        2 4   4 
2

 0.2637

b. p = 0.2, q = 1 – 0.2 = 0.8
P( X  2)  P( X  0)  P( X  1)  P( X  2) 5  5 5    (0.2) 0 (0.8)5    (0.2)1 (0.8)51    (0.2) 2 (0.8)52 0 1   2  0.32768  0.40960  0.20480  0.94208…...

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Binomial Trees

...Problem 1: Binomial Trees and Real Options Suppose that the spot price, 6-month forward price, 12-months forward price and 18-months forward price for wheat are 250, 260, 270.4 and 281.216 cents per bushel, respectively.[1] A farmer has a project that involves an immediate expenditure of $20,000 and a further expenditure of $90,000 in six months. The farmer collects two wheat harvests per year, and if this project is implemented, it will increase wheat that is harvested and sold by 20,000 bushels in one year and in 18 months (as compared to his typical harvest). Assume that there is no uncertainty about the harvests (all uncertainty comes from future wheat prices). The continuously compounded risk-free rate is 5% p.a. a) If the project has no flexibility (a “yes or no” decision), what is the value of the project? Notice that you do not have to use the binomial tree to answer this question. b) Suppose that the wheat price follows a binomial tree process with parameters u=1.1 and d=0.95 (determine y from forward prices and remember to take this variable into account when computing risk-neutral probabilities). Suppose that the farmer can abandon the project in six months and avoid paying the $90,000 cost at that time (in this case, the increase in harvest will be 5,000 bushels per year instead of 20,000). What is the value of this abandonment option? Is this abandonment option a put or a call? What is its strike? Explain. c) Now, assume that there is no option......

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