Soal:
Misalkan \(U \sim N(0, 1)\) dan \(V \sim \chi^2(r)\) saling independen. Gabungkan kedua distribusi tersebut.
Misalkan terdapat variabel acak \(Q\) dengan
\[ Q = \frac{U}{\sqrt{V/r}} \]
dan \(u = q_1 \sqrt{z}\), dengan \(z = v\). Tentukanlah \(g(q_1, z) = h \left( \frac{q_1 \sqrt{z}}{\sqrt{r}}, z \right) \cdot |J|\).
Penyelesaian:
Langkah 1: Distribusi gabungan \(U\) dan \(V\)
Karena \(U\) dan \(V\) independen:
\[ f_{U,V}(u,v) = f_U(u) \cdot f_V(v) \]
dengan
\[ f_U(u) = \frac{1}{\sqrt{2\pi}} e^{-u^2/2}, \quad f_V(v) = \frac{1}{2^{r/2} \Gamma(r/2)} v^{r/2 - 1} e^{-v/2}, \quad v > 0. \]
Maka:
\[ f_{U,V}(u,v) = \frac{1}{\sqrt{2\pi}} e^{-u^2/2} \cdot \frac{1}{2^{r/2} \Gamma(r/2)} v^{r/2 - 1} e^{-v/2}. \]
Langkah 2: Transformasi variabel
Didefinisikan:
\[ Q = \frac{U}{\sqrt{V/r}}, \quad Z = V. \]
Invers transformasi:
\[ U = Q \sqrt{Z / r}, \quad V = Z. \]
Langkah 3: Menghitung Jacobian
\[ J = \frac{\partial(u,v)}{\partial(q_1,z)} = \begin{vmatrix} \frac{\partial u}{\partial q_1} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial q_1} & \frac{\partial v}{\partial z} \end{vmatrix} \]
\[ \frac{\partial u}{\partial q_1} = \sqrt{\frac{z}{r}}, \quad \frac{\partial u}{\partial z} = \frac{q_1}{2\sqrt{z r}}, \quad \frac{\partial v}{\partial q_1} = 0, \quad \frac{\partial v}{\partial z} = 1. \]
\[ J = \left( \sqrt{\frac{z}{r}} \right)(1) - \left( \frac{q_1}{2\sqrt{z r}} \right)(0) = \sqrt{\frac{z}{r}}. \]
Jadi:
\[ |J| = \sqrt{\frac{z}{r}}. \]
Langkah 4: Distribusi gabungan \(Q\) dan \(Z\)
Substitusi \(u = q_1 \sqrt{z/r}\) dan \(v = z\) ke \(f_{U,V}(u,v)\):
\[ f_{U,V}(q_1 \sqrt{z/r}, z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{(q_1 \sqrt{z/r})^2}{2}} \cdot \frac{1}{2^{r/2} \Gamma(r/2)} z^{r/2 - 1} e^{-z/2}. \]
\[ e^{-\frac{q_1^2 z / r}{2}} = e^{-q_1^2 z / (2r)}. \]
Maka:
\[ g(q_1,z) = f_{U,V}(q_1 \sqrt{z/r}, z) \cdot |J| \]
\[ = \frac{1}{\sqrt{2\pi}} e^{-q_1^2 z / (2r)} \cdot \frac{1}{2^{r/2} \Gamma(r/2)} z^{r/2 - 1} e^{-z/2} \cdot \sqrt{\frac{z}{r}}. \]
Gabungkan suku-suku:
\[ g(q_1,z) = \frac{1}{\sqrt{2\pi r} \cdot 2^{r/2} \Gamma(r/2)} \cdot z^{\frac{r+1}{2} - 1} \cdot \exp\left[ -\frac{z}{2} \left( 1 + \frac{q_1^2}{r} \right) \right]. \]
Langkah 5: Marginalkan terhadap \(Z\) untuk mendapatkan distribusi \(Q\)
\[ f_Q(q_1) = \int_0^\infty g(q_1,z) \, dz \]
\[ = \frac{1}{\sqrt{2\pi r} \, 2^{r/2} \, \Gamma(r/2)} \int_0^\infty z^{\frac{r+1}{2} - 1} \exp\left[ -\frac{z}{2} \left( 1 + \frac{q_1^2}{r} \right) \right] dz. \]
Gunakan rumus:
\[ \int_0^\infty x^{\alpha-1} e^{-\beta x} dx = \Gamma(\alpha) \beta^{-\alpha}. \]
Di sini:
\[ \alpha = \frac{r+1}{2}, \quad \beta = \frac{1}{2} \left( 1 + \frac{q_1^2}{r} \right). \]
Maka:
\[ \int_0^\infty z^{\frac{r+1}{2} - 1} \exp\left[ -\frac{z}{2} \left( 1 + \frac{q_1^2}{r} \right) \right] dz = \Gamma\left( \frac{r+1}{2} \right) \left[ \frac{1}{2} \left( 1 + \frac{q_1^2}{r} \right) \right]^{-\frac{r+1}{2}}. \]
Substitusi:
\[ f_Q(q_1) = \frac{1}{\sqrt{2\pi r} \, 2^{r/2} \, \Gamma(r/2)} \cdot \Gamma\left( \frac{r+1}{2} \right) \cdot \left[ \frac{1}{2} \left( 1 + \frac{q_1^2}{r} \right) \right]^{-\frac{r+1}{2}}. \]
Langkah 6: Sederhanakan konstanta
\[ 2^{-r/2} \cdot \left( \frac12 \right)^{-(r+1)/2} = 2^{-r/2} \cdot 2^{(r+1)/2} = 2^{1/2} = \sqrt{2}. \]
Jadi:
\[ f_Q(q_1) = \frac{1}{\sqrt{2\pi r} \, \Gamma(r/2)} \cdot \Gamma\left( \frac{r+1}{2} \right) \cdot \sqrt{2} \cdot \left( 1 + \frac{q_1^2}{r} \right)^{-\frac{r+1}{2}}. \]
\[ \frac{\sqrt{2}}{\sqrt{2\pi r}} = \frac{1}{\sqrt{\pi r}}. \]
Langkah 7: Hasil akhir fungsi densitas distribusi \(t\)-Student
Ganti \(q_1\) dengan \(t\) dan \(r\) dengan \(\nu\):
\[ \boxed{f_T(t) = \frac{\Gamma\left( \frac{\nu+1}{2} \right)}{\sqrt{\nu\pi} \, \Gamma\left( \frac{\nu}{2} \right)} \left( 1 + \frac{t^2}{\nu} \right)^{-\frac{\nu+1}{2}}} \]
untuk \(t \in \mathbb{R}\), \(\nu > 0\).
Kesimpulan:
Dari variabel normal baku \(U\) dan chi-square \(V\) yang independen, kita peroleh distribusi \(t\)-Student dengan \(\nu\) derajat kebebasan melalui transformasi dan integrasi.