Fungsi Kepadatan Peluang

A.  Pengertian Fungsi Kepadatan Peluang

Kita telah mengenal dan memahami pengertian distribusi suatu peubah acak. Dimana, distribusi peubah acak merupakan kumpulan pasangan nilai-nilai dari variabel acak X dengan probabilitas nilai-nilai variabel acak X, yaitu P(X = x). Distribusi X dapat dituliskan dalam bentuk tabel atau dalam bentuk pasangan terurut. Variabel acak merupakan suatu fungsi acak X yang bernilai riil di mana nilai-nilainya ditentukan oleh titik sampel-titik sampel S dengan S merupakan ruang sampel dari suatu percobaan statistik.  Berdasarkan materi distribusi peubah acak, peubah acak terbagi dua jenis, yaitu: variabel acak diskrit dan variabel acak kontinu. Dimana variabel acak diskrit adalah variabel acak yang mempunyai nilai-nilai terhingga atau tak terhingga tetapi terbilang. Sedangkan variabel acak kontinu adalah variabel acak yang mempunyai nilai-nilai tak terhingga dan tak terbilang. Melalui pengertian-pengertian diatas kita dapat dengan mudah menghitung peluang dari suatu peristiwa. Cukup dengan mengamati tabel distribusi peluang. Pengertian tersebut dapat diperluas pada peubah-peubah acak kontinu melalui konsep fungsi kepadatan peluang (f.k.p). Dimana jika X adalah variabel acak dan P(X = x) adalah distribusi probabilitas dari X, maka fungsi f(x) = P (X = x) disebut fungsi padat peluang.

B.  Fungsi Kepadatan Peluang Dari Peuabah Acak Diskrit

Misalkan e ruang dari peubah acak diskrit X. Jadi e terbilang. Misalkan f adalah fungsi dari e ke dalam R,  fungsi  f tersebut dinamakan  fungsi  kepadatan  peluang  jika fungsi f memenuhi sifat-sifat berikut ini:

·         f (x)≥ 0 untuk setiap x di e

·        \(\sum\limits_{x\begin{array}{*{20}{c}}{}\end{array}di\begin{array}{*{20}{c}}{}\end{array}e} {f\left( x \right) = 1} \)

Jika peubah acak X diskrit dengan fungsi kepadatan peluang f(x), maka peluang suatu peristiwa A diberikan oleh:

Contoh 1:

Misalkan e = { 0, 1, 2, 3, 4} ruang dari X, dan f adalah fungsi dari e ke dalam R yang didefinisikan oleh: \(f\left( x \right) = \frac{{4!}}{{\left( {4 - x} \right)!x!}}{\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4};x\begin{array}{*{20}{c}}{}\end{array}di\begin{array}{*{20}{c}}{}\end{array}e\) Buktikan bahwa f suatu fungsi kepadatan peluang. Hitunglah P( X ≤ 1).

Penyelesaian:

Fungsi  \(f\left( x \right) = \frac{{4!}}{{\left( {4 - x} \right)!x!}}{\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4};x\begin{array}{*{20}{c}}{}\end{array}di\begin{array}{*{20}{c}}{}\end{array}e\) merupakan suatu fungsi kepadatan peluang jika memenuhi dua sifat f.k.p yaitu

·      f (x)≥ 0 untuk setiap x di e jelas bahwa f(x) ≥ 0 untuk setiap x di e karena e = { 0, 1, 2, 3, 4}

·      \(\sum\limits_{x\begin{array}{*{20}{c}}{}\end{array}di\begin{array}{*{20}{c}}{}\end{array}e} {f\left( x \right) = 1} \)

Bukti :

\(\begin{array}{l}\sum\limits_{x\begin{array}{*{20}{c}}{}\end{array}di\begin{array}{*{20}{c}}{}\end{array}e} {f\left( x \right) = 1} \\\sum\nolimits_{x = 0}^4 {f(x) = } {\sum\nolimits_{x = 0}^4 {\frac{{4!}}{{\left( {4 - x} \right)!x!}}\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]} ^4}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4}\sum\nolimits_x^4 {C_x^4} \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4}\left( {\frac{{4!}}{{\left( {4 - 0} \right)!0!}} + \frac{{4!}}{{\left( {4 - 1} \right)!1!}} + \frac{{4!}}{{\left( {4 - 2} \right)!2!}} + \frac{{4!}}{{\left( {4 - 3} \right)!3!}} + \frac{{4!}}{{\left( {4 - 4} \right)!4!}}} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4}\left( {\frac{{4!}}{{\left( 4 \right)!0!}} + \frac{{4!}}{{\left( 3 \right)!1!}} + \frac{{4!}}{{\left( 2 \right)!2!}} + \frac{{4!}}{{\left( 1 \right)!3!}} + \frac{{4!}}{{\left( 0 \right)!4!}}} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4}\left( {1 + 4 + 6 + 4 + 1} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{{16}}\left( {16} \right) = 1\end{array}\)

𝑥=0

Jadi Terbukti   \(\sum\nolimits_{x = 0}^4 {f(x) = } 1\) Ini berarti bahwa f adalah fungsi kepadatan peluang dari peubah acak diskrit atau f.k.p. dari X. Karena f merupakan f.k.p dari X, maka

\(\begin{array}{l}P(A) = P\left( {X \le 1} \right) = {\sum\nolimits_{x = 0}^1 {\frac{{4!}}{{\left( {4 - x} \right)!x!}}\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]} ^4}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4}\sum\nolimits_x^1 {C_x^1} \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4}\left( {\frac{{4!}}{{\left( {4 - 0} \right)!0!}} + \frac{{4!}}{{\left( {4 - 1} \right)!1!}}} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4}\left( {\frac{{4!}}{{\left( 4 \right)!0!}} + \frac{{4!}}{{\left( 3 \right)!1!}}} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {\left[ {\begin{array}{*{20}{c}}1\\2\end{array}} \right]^4}\left( {1 + 4} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{{16}}\left( 5 \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{5}{{16}}\end{array}\)

Contoh 2:

Misalkan e = { x | x = 1, 2, 3........} adalah ruang dari peubah acak X. Misalkan f adalah

fungsi dari e ke dalam R yang didefinisikan oleh \(f(x) = {\left[ {\frac{1}{2}} \right]^x}\)  untuk setiap x di e. Buktikan bahwa f suatu fungsi kepadatan peluang. Hitunglah P(A) dimana A = { x | x = 1, 3, 5................................. }.

a.       Jelas f(x ) ≥ 0 untuk setiap x di e. Akan ditunjukkan bahwa :

       \(\sum\nolimits_{x = 1}^\infty  {f(x) = } 1\)

     \(\begin{array}{l}\sum\nolimits_{x = 1}^\infty  {f(x) = } \sum\nolimits_{x = 1}^\infty  {{{\left[ {\frac{1}{2}} \right]}^4} = } \frac{1}{2} + {\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^3} + ...\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{2}(1 + \frac{1}{2} + {\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^3} + ...\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{2}\left( {1 + \sum\nolimits_{x = 1}^\infty  {f(x)} } \right)\\\sum\nolimits_{x = 1}^\infty  {f(x) = } 1 + \sum\nolimits_{x = 1}^\infty  {f(x)} \\\sum\nolimits_{x = 1}^\infty  {f(x) - } \sum\nolimits_{x = 1}^\infty  {f(x)}  = 1\\\sum\nolimits_{x = 1}^\infty  {f(x)}  = 1\end{array}\)

Ini berarti bahwa f adalah f.k.p. dari X.

\(\begin{array}{l}b.P(A) = \sum\nolimits_{x\begin{array}{*{20}{c}}{}\end{array}ganjil\begin{array}{*{20}{c}}{}\end{array}x = 1}^\infty  {f(x)} \\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{2} + {\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^5} + ...\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{2}(1 + {\left( {\frac{1}{2}} \right)^2} + {\left( {\frac{1}{2}} \right)^4} + ...)\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{2}\left( {1 + P\left( {{A^C}} \right)} \right)\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{2}\left\{ {1 + \left( {1 - P\left( A \right)} \right)} \right\}\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{2}\left( {2 - P\left( A \right)} \right)\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array}P(A) = 1 - \frac{1}{2}\left( {P\left( A \right)} \right)\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array}P(A) + \frac{1}{2}\left( {P\left( A \right)} \right) = 1\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array}\frac{3}{2}P\left( A \right) = 1\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array}P(A) = \frac{2}{3}\end{array}\)

C.   Fungsi Kepadatan Peluang dari Peubah Acak

Misalkan e ruang dari peubah acak kontinu X. Jadi e tak terbilang. Misalkan f adalah fungsi dari  e  ke   dalam   R,   fungsi  f tersebut dinamakan   fungsi   kepadatan   peluang   jika fungsi f memenuhi sifat-sifat berikut ini:

·         f (x)≥ 0 untuk setiap x di e

·         \(\int_A {f\left( x \right)dx = 1} \)

Jika peubah acak X kontinu memiliki fungsi kepadatan peluang f(x), maka peluang suatu peristiwa A diberikan oleh:

Contoh 1 : 

Misalkan A = { x | 0 < x < ∞}ruang peubah acak kontinu X, dan f adalah fungsi dari e ke dalam R yang didefinisikan oleh \(f(x) = {e^{ - x}}\)  untuk setiap x di e. Buktikanlah bahwa f merupakan f.k.p. Hitunglah P(X ≤ 1).

     Penyelesaian:

Jelas f(x) ≥ 0 untuk setiap x di e. Akan tetapi ditunjukkan bahwa        

\(\begin{array}{l}\int_0^\infty  {f(x)dx = 1} \\\int_0^\infty  {f(x)dx = \int_0^\infty  {{e^{ - x}}dx} } \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} =  - {e^{ - x}}|_0^\infty \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} =  - {e^{ - \infty }} - \left( { - {e^0}} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = 0 + 1\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = 1\end{array}\)

Jadi fungsi f adalah f.k.p dari X.

\(\begin{array}{l}P\left( {X \le 1} \right) = \int_0^1 {f(x)dx = \int_0^1 {{e^{ - x}}dx} } \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} =  - {e^{ - x}}|_0^1\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} =  - {e^{ - x}} - \left( { - {e^0}} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} =  - {e^{ - 1}} + 1\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} =  - \frac{1}{e} + 1\end{array}\)

Contoh 2 

Misalkan e = { x | 0 < x < 1} adalah ruang dari peubah acak X. Jika f(x) = KX² untuk setiap x

di e, carilah harga X sehingga f merupakan f.k.p dari X. Kemudian, hitung \(P\left( {\frac{1}{4} < X \le \frac{1}{2}} \right)\)

Penyelesaian 

a.   Jelas f(x) ≥ 0 untuk setiap x di e. Agar f merupakan f.k.p.

Haruslah : 

            \(\int_A {f\left( x \right)dx = 1} \) , akan tetapi

             

             \(\begin{array}{l}\int_A {f\left( x \right)dx = \int_0^1 {k{x^2}dx} } \\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array} = \frac{k}{3}{x^3}|_0^1\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array} = \frac{k}{3}{\left( 1 \right)^3}\\\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}}\end{array}\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{}\end{array}}\end{array} = \frac{k}{3}\end{array}\)

karena haruslah \(\int_A {f\left( x \right)dx = 1} \)  dimana \(\int_A {f\left( x \right)dx = \int_0^1 {k{x^2}dx}  = \frac{k}{3}} \) maka, \(\int_A {f\left( x \right)dx = \frac{k}{3}} \) sehingga : 

          \(\frac{k}{3} = 1 \Rightarrow k = 3\)

b. Karena k = 3 maka :

           \(\begin{array}{l}P\left( {\frac{1}{4} < X \le \frac{1}{2}} \right) = \int_{1/4}^{1/2} {3{x^2}} \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {x^3}|_{1/4}^{1/2}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {\left( {\frac{1}{2}} \right)^3} - {\left( {\frac{1}{4}} \right)^3}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \left( {\frac{1}{8}} \right) - \left( {\frac{1}{{64}}} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{7}{{64}}\end{array}\)

Jadi, \(P\left( {\frac{1}{4} < X \le \frac{1}{2}} \right) = \frac{7}{{64}}\)

C.   Fungsi Kepadatan Peluang Bersama dari Beberapa Peubah Acak

Misalkan e ruang bersama cari \({X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}\) dalam hal ini \({X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}\) semuanya diskrit yang berarti e terbilang maka fungsi f dari e ke dalam r yang bersifat :

·      \(f\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right) \ge 0\) untuk setiap \(\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right)\)di e

·      \(\sum\nolimits_{{x_1}} {\sum\nolimits_{{x_2}} {\sum\nolimits_{{x_3}} {......\begin{array}{*{20}{c}}{}\end{array}.......\begin{array}{*{20}{c}}{}\end{array}.......\sum {{x_n}\begin{array}{*{20}{c}}{}\end{array}f\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right)\begin{array}{*{20}{c}}{}\end{array}di\begin{array}{*{20}{c}}{}\end{array}e} } } } \)

Dinamakan f.k.p bersama dari  \({X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}\) dalam hal ini jika \(A \subseteq e\) maka :

\(\begin{array}{l}P\left( A \right) = P\left[ {\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right)di\begin{array}{*{20}{c}}{}\end{array}e} \right]\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \sum\nolimits_{{x_1}} {\sum\nolimits_{{x_2}} {\sum\nolimits_{{x_3}} {......\begin{array}{*{20}{c}}{}\end{array}.......\begin{array}{*{20}{c}}{}\end{array}.......\sum {{x_n}\begin{array}{*{20}{c}}{}\end{array}f\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right)} } } } \end{array}\)

Misalkan e ruang bersama cari \({X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}\) dalam hal ini

\({X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}\) semuanya kontinu berarti e tidak terbilang, maka fungsi f dari e ke dalam R yang memenuhi sifat :

·      \(f\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right) \ge \) untuk setiap \(\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right)\) di e

·      \(\int_{{x_1}} {\int_{{x_2}} {\int_{{x_3}} . } } ......\begin{array}{*{20}{c}}{}\end{array}........\begin{array}{*{20}{c}}{}\end{array}.......\int {_n\begin{array}{*{20}{c}}{}\end{array}} f\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right)d{x_1}d{x_2}d{x_3}....\begin{array}{*{20}{c}}{}\end{array}.....\begin{array}{*{20}{c}}{}\end{array}....d{x_n} = 1\)

Dikatakan f.k.p bersama dari  \({X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}\) dalam hal ini 

\(\begin{array}{l}P\left( A \right) = P\left[ {\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right)di\begin{array}{*{20}{c}}{}\end{array}A} \right]\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \int_{{x_1}} {\int_{{x_2}} {\int_{{x_3}} . } } ......\begin{array}{*{20}{c}}{}\end{array}........\begin{array}{*{20}{c}}{}\end{array}.......\int {_n\begin{array}{*{20}{c}}{}\end{array}} f\left( {{X_1},{X_2},{X_3},....\begin{array}{*{20}{c}}{}\end{array}......\begin{array}{*{20}{c}}{}\end{array}......{X_n}} \right)d{x_1}d{x_2}d{x_3}....\begin{array}{*{20}{c}}{}\end{array}.....\begin{array}{*{20}{c}}{}\end{array}....d{x_n}\end{array}\)

Contoh :

Misalkan e {(x,  y) | x = 1, 2, 3,  ...... dan y = 1,  2, 3,....... adalah ruang bersama dari X dan Y .Misalkan f (x, y) didefinisikan oleh:

\(f\left( {x,y} \right) = \frac{9}{{{4^{x + y}}}}\)

Buktikan bahwa f merupakan f.k.p bersama dari X dan Y

Penyelesaian :

Bukti :

Jelas \[f{\rm{ }}\left( {x,{\rm{ }}y} \right) \ge 0\] untuk setiap ( x, y ) di e, akan di tunjukan bahwa \(\sum\nolimits_{y = 1}^\infty  {\sum\nolimits_{x = 1}^\infty  {f\left( {x,y} \right) = 1} } \) untuk itu kita buat tabel distribusi bersama sebagai berikut :

 

(i)        Jumlah baris pertama adalah

  \(\begin{array}{l}{S_1} = 9\sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{9}{4}\left( {\frac{1}{4} + \sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} } \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{9}{{16}} + \frac{1}{4}{S_1}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{3}{4}{S_1}\end{array}\) 

Jadi \({S_1} = \frac{3}{4}\)

(ii)      Jumlah Baris Kedua adalah

\(\begin{array}{l}{S_2} = 9\sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = 9\left( {\frac{1}{4} + \sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} } \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{9}{4}\sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{4}{S_1}\end{array}\)

Jadi :

\(\begin{array}{l}{S_2} = \frac{1}{4}{S_1}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \left( {\frac{1}{4}} \right)\left( {\frac{3}{4}} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \left( {\frac{3}{{16}}} \right)\end{array}\)

(iii)    Jumlah baris ketiga adalah :

    \(\begin{array}{l}{S_2} = 9\sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = 9\left( {\frac{1}{4}\sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} } \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{9}{4}\sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{1}{{{4^2}}}{S_1}\end{array}\)

(iv)    Dengan cara yang sama seperti di atas maka jumlah baris ke- k adalah :

\({S_K} = \frac{1}{{{4^{K - 1}}}}{S_1}\)

Jadi :

\(\begin{array}{l}\sum\nolimits_{y = 1}^\infty  {\sum\nolimits_{x = 1}^\infty  {f\left( {x,y} \right) = {S_1}\left\{ {1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + \frac{1}{{{4^4}}} + .....\begin{array}{*{20}{c}}{}\end{array}.....} \right\}} } \\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {S_1}\left\{ {1 + \frac{1}{4} + \sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} } \right\}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {S_1}\left\{ {1 + \frac{1}{4} + \frac{1}{9} + \left( {9\sum\nolimits_{k = 2}^\infty  {\frac{1}{{4k}}} } \right)} \right\}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = {S_1}\left\{ {1 + \frac{1}{4} + \frac{1}{9} + \left( {{S_1}} \right)} \right\}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \frac{3}{4}\left\{ {1 + \frac{1}{4} + \frac{1}{9} + \left( {\frac{3}{4}} \right)} \right\}\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = \left( {\frac{3}{4}} \right)\left( {\frac{{16}}{2}} \right)\\\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array}\begin{array}{*{20}{c}}{}\end{array} = 1\end{array}\)

Ini berarti bahwa f adalah f.k.p bersama dari X dan Y