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Distribusi Bivariat Normal

Jika peubah acak \(X\) dan \(Y\) masing-masing memiliki sebaran distribusi normal \(N(\mu_x,\sigma_x^2)\) dan \(N(\mu_y,\sigma_y^2)\), maka gabungan dari kedua peubah acak tersebut akan mempunyai sebaran distribusi bivariat normal \(N(\mu_x,\mu_y,\sigma_x^2,\sigma_y^2,\rho)\) yang fungsi kepadatan peluangnya adalah \[f(x,y)=\frac{1}{2\pi\sqrt{(1-\rho^2)\sigma_x^2\sigma_y^2}}\exp{\left\{-\frac{1}{2(1-\rho^2)}\left[\left(\frac{x-\mu_x}{\sigma_x}\right)^2-2\rho\left(\frac{x-\mu_x}{\sigma_x}\right)\left(\frac{y-\mu_y}{\sigma_y}\right)+\left(\frac{y-\mu_y}{\sigma_y}\right)^2\right]\right\}}\] atau dapat disingkat menjadi \[f(x,y)=c\exp{\left\{-\frac{1}{2}Q\right\}}\] dimana \[c=\frac{1}{2\pi\sqrt{(1-\rho^2)\sigma_x^2\sigma_y^2}}\] dan \[Q=\frac{1}{1-\rho^2}\left[\left(\frac{x-\mu_x}{\sigma_x}\right)^2-2\rho\left(\frac{x-\mu_x}{\sigma_x}\right)\left(\frac{y-\mu_y}{\sigma_y}\right)+\left(\frac{y-\mu_y}{\sigma_y}\right)^2\right]\] untuk semua nilai \(x\) dan \(y.\) Parameter \(\mu_x\) dan \(\mu_y\) adalah bilangan real, \(\sigma_x>0,\) \(\sigma_y>0,\) dan \(\rho\) adalah koefisien korelasi antara \(X\) dan \(Y,\) dimana \(-1\leq\rho\leq1.\)

Sifat-sifat distribusi bivariate normal:
  • Sebaran marginal dari \(X\) dan \(Y\) adalah distribusi normal univariate,
  • Sebaran \(X\) dengan syarat \(Y=y\) atau \(Y\) dengan syarat \(X=x\) juga merupakan distribusi normal univariate.
  • Jika \(X\) dan \(Y\) adalah independen maka koefisien korelasi \(\rho=0.\)

Fungsi Distribusi Marjinal Distribusi Bivariate Normal


Pada fungsi kepatan peluang distribusi bivariate normal, dimisalkan \[c=\frac{1}{2\pi\sqrt{(1-\rho^2)\sigma_x^2\sigma_y^2}},\,\, u=\frac{x-\mu_x}{\sigma_x},\,\, v=\frac{y-\mu_y}{\sigma_y}\] maka \[du=\frac{1}{\sigma_x}dx,\,\, dv=\frac{1}{\sigma_y}dy\] dengan demikian fungsi distribusi marjinal \(f(x)\) adalah \[\begin{aligned} f(x)&=\int f(x,y)\,dy\\ &=c\int\exp{\left\{-\frac{1}{2(1-\rho^2)}(u^2-2\rho uv+v^2)\right\}}\sigma_y\,dv \end{aligned}\] Susunan pada komponen \(u^2-2\rho uv+v^2\) dapat ditulis menjadi \(v^2-2\rho uv+u^2,\) selanjutnya dapat ditulis dalam persamaan kuadrat \[\left(v^2-2(\rho u)v+{(\rho u)}^2\right)-{(\rho u)}^2+u^2\] yang persamaan sederhananya adalah \[{(v-\rho u)}^2+u^2(1-\rho^2)\] Dengan demikian penurunan fungsi distribusi marjinalnya menjadi \[\begin{aligned} f(x)&=c\,\sigma_y\int\exp{\left\{-\frac{1}{2(1-\rho^2)}\left[{(v-\rho u)}^2+u^2(1-\rho^2)\right]\right\}}\,dv\\ &=c\,\sigma_y\,\exp{\left\{-\frac{u^2}{2}\right\}} \int\exp{\left\{-\frac{1}{2(1-\rho^2)}\left[{(v-\rho u)}^2\right]\right\}}\,dv \end{aligned}\] Komponen \[\int\exp{\left\{-\frac{1}{2(1-\rho^2)}\left[{(v-\rho u)}^2\right]\right\}}\] sama dengan komponen distribusi normal dengan rata-rata (mean) sama dengan \(\rho u\) dan varian \(1-\rho^2,\) dimana dengan persamaan distribusi normal diperoleh \[\int\exp{\left\{-\frac{1}{2(1-\rho^2)}\left[{(v-\rho u)}^2\right]\right\}}=\sqrt{2\pi(1-\rho^2)}\] Penurunan fungsi distribusi marjinal menjadi \[\begin{aligned} f(x) &=c\,\sigma_y\,\exp{\left\{-\frac{u^2}{2}\right\}}\sqrt{2\pi(1-\rho^2)}\\ &=\frac{1}{2\pi\sqrt{(1-\rho^2)\sigma_x^2\sigma_y^2}}\,\sigma_y\,\exp{\left\{-\frac{u^2}{2}\right\}}\sqrt{2\pi(1-\rho^2)}\\ f(x)&=\frac{1}{\sqrt{2\pi\sigma_x^2}}\exp{\left\{-\frac{1}{2}\left(\frac{x-\mu_x}{\sigma_x}\right)^2\right\}} \end{aligned}\] Fungsi distribusi tersebut merupakan fungsi distribusi normal \(N(\mu_x,\sigma_x^2).\) Cara yang sama juga dapat dilakukan terhadap \(y\) sehingga diperoleh fungsi distribusi normal \(N(\mu_y,\sigma_y^2).\)


Fungsi Distribusi Bersyarat


Fungsi distribusi \(Y\) dengan syarat \(X=x\) adalah \[f(y|x)=\frac{f(x,y)}{f(x)}\] Pada distribusi bivariate normal, jika dimisalkan \[u=\frac{x-\mu_x}{\sigma_x},\,\, v=\frac{y-\mu_y}{\sigma_y}\] maka \[\begin{aligned} f(y|x)&= \frac{\frac{1}{2\pi\sqrt{(1-\rho^2)\sigma_x^2\sigma_y^2}}\int\exp{\left\{-\frac{1}{2(1-\rho^2)}(u^2-2\rho uv+v^2)\right\}}}{\frac{1}{\sqrt{2\pi\sigma_x^2}}\exp{\left\{-\frac{1}{2}u^2\right\}}}\\ &=\frac{1}{\sqrt{2\pi(1-\rho^2)\sigma_y^2}}\exp{\left\{-\frac{1}{2(1-\rho^2)}(u^2-2\rho uv+v^2)+\frac{1}{2}u^2\right\}}\\ &=\frac{1}{\sqrt{2\pi(1-\rho^2)\sigma_y^2}}\exp{\left\{-\frac{1}{2(1-\rho^2)}(u^2-2\rho uv+v^2-(1-\rho^2)u^2)\right\}}\\ &=\frac{1}{\sqrt{2\pi(1-\rho^2)\sigma_y^2}}\exp{\left\{-\frac{1}{2(1-\rho^2)}\left(v^2-2\rho uv+(\rho u)^2\right)\right\}}\\ &=\frac{1}{\sqrt{2\pi(1-\rho^2)\sigma_y^2}}\exp{\left\{-\frac{1}{2(1-\rho^2)}{(v-\rho u)}^2\right\}} \end{aligned}\] Selanjutnya dapat dikembalikan kebentuk \(x\) dan \(y\) sehingga bentuknya menjadi \[\begin{aligned} f(y|x)&=\frac{1}{\sqrt{2\pi(1-\rho^2)\sigma_y^2}}\exp{\left\{-\frac{1}{2(1-\rho^2)}{\left(\frac{y-\mu_y}{\sigma_y}-\rho \frac{x-\mu_x}{\sigma_x}\right)}^2\right\}}\\ &=\frac{1}{\sqrt{2\pi(1-\rho^2)\sigma_y^2}}\exp{\left\{-\frac{1}{2(1-\rho^2)\sigma_y^2}{\left((y-\mu_y)-\rho \frac{\sigma_y}{\sigma_x}(x-\mu_x)\right)}^2\right\}}\\ &=\frac{1}{\sqrt{2\pi(1-\rho^2)\sigma_y^2}}\exp{\left\{-\frac{1}{2(1-\rho^2)\sigma_y^2}{\left(y-\left[\mu_y+\rho \frac{\sigma_y}{\sigma_x}(x-\mu_x)\right]\right)}^2\right\}} \end{aligned}\] Bentuk fungsi di atas adalah bentuk fungsi distribusi normal \[Y|X \sim N\left(\mu_y+\rho \frac{\sigma_y}{\sigma_x}(x-\mu_x),(1-\rho^2)\sigma_y^2\right)\]

Kovarian


Kovarian \(X\) dan \(Y\) didefinisikan oleh \[\begin{aligned} \text{Cov}(XY)&=E[(X-\mu_x)(Y-\mu_y)]\\ &=\iint(x-\mu_x)(y-\mu_y)f(x,y)\,dx\,dy \end{aligned}\] Jika pada fungsi distribusi bivariate normal dimisalkan \[u=\frac{x-\mu_x}{\sigma_x},\,\, v=\frac{y-\mu_y}{\sigma_y}\] maka \[du=\frac{1}{\sigma_x}dx,\,\, dvu=\frac{1}{\sigma_y}dy.\] Dengan demikian \[\begin{aligned} \text{Cov}(XY)&=c\iint u\sigma_x v\sigma_y \exp{\left\{-\frac{1}{2(1-\rho^2)}(u^2-2\rho uv+v^2)\right\}}\sigma_xdu\, \sigma_ydv\\ &=c\,\sigma_x^2\sigma_y^2\iint uv \exp{\left\{-\frac{1}{2(1-\rho^2)}(u^2-2\rho uv+v^2)\right\}}\, du\,dv\\ &=c\,\sigma_x^2\sigma_y^2\iint uv \exp{\left\{-\frac{1}{2(1-\rho^2)}(u^2-2\rho uv+(\rho v)^2-(\rho v)^2+v^2)\right\}}\, du\,dv\\ &=c\,\sigma_x^2\sigma_y^2\iint uv \exp{\left\{-\frac{1}{2(1-\rho^2)}\left[(u-\rho v)^2+v^2(1-\rho^2)\right]\right\}}\, du\,dv\\ &=c\,\sigma_x^2\sigma_y^2\iint uv \exp{\left\{-\frac{1}{2(1-\rho^2)}(u-\rho v)^2-\frac{1}{2}v^2\right\}}\, du\,dv\\ &=c\,\sigma_x^2\sigma_y^2\iint uv \exp{\left\{-\frac{1}{2(1-\rho^2)}(u-\rho v)^2\right\}}\exp{\left\{-\frac{1}{2}v^2\right\}}\, du\,dv\\ &=c\,\sigma_x^2\sigma_y^2\int v \exp{\left\{-\frac{1}{2}v^2\right\}} \left[ \int u\exp{\left\{-\frac{1}{2(1-\rho^2)}(u-\rho v)^2\right\}}\, du\right]\,dv \end{aligned}\] Komponen \[\int u\exp{\left\{-\frac{1}{2(1-\rho^2)}(u-\rho v)^2\right\}}\, du\] sama dengan komponen distribusi normal dimana \(\rho v=\mu\) dan \((1-\rho^2)=\sigma^2.\) \[\begin{aligned} \text{E}(X)&=\mu\\ \frac{1}{\sqrt{2\pi\sigma^2}} \int x \exp \left\{-\frac{1}{2\sigma^2}{(x-\mu)}^2\right\}\,dx&=\mu\\ \int x \exp \left\{-\frac{1}{2\sigma^2}{(x-\mu)}^2\right\}\,dx&=\mu\sigma\sqrt{2\pi}\\ \int u\exp{\left\{-\frac{1}{2(1-\rho^2)}(u-\rho v)^2\right\}}\, du &=\rho v\sqrt{1-\rho^2}\sqrt{2\pi} \end{aligned}\] Sehingga \[\text{Cov}(XY)=c\,\sigma_x^2\sigma_y^2 \rho \sqrt{1-\rho^2}\sqrt{2\pi} \int v^2 \exp{\left\{-\frac{1}{2}v^2\right\}}\,dv.\] Dalam kalkulus \[\int x^2 \exp{\left\{-\frac{1}{2}x^2\right\}}\,dx=\sqrt{2\pi},\] sehingga \[\begin{aligned} \text{Cov}(XY)&=c\,\sigma_x^2\sigma_y^2 \rho\sqrt{1-\rho^2}\sqrt{2\pi}\sqrt{2\pi}\\ &=\frac{1}{2\pi\sigma_x\sigma_y\sqrt{1-\rho^2}}\sigma_x^2\sigma_y^2 \rho\,2\pi\sqrt{1-\rho^2}\\ &=\sigma_x\sigma_y \rho \end{aligned}\]