Tunjukkan bahwa memiliki distribusi miring normal

Biarkan dan independen. Tunjukkan bahwa memiliki distribusi condong-normal dan temukan parameter distribusi ini. $Y_1\sim SN(\mu_1,\sigma_1^2,\lambda)$ $Y_2\sim N(\mu_2,\sigma_2^2)$ $Y_1+Y_2$

Karena variabel acak independen saya mencoba menggunakan konvolusi. Biarkan $Z=Y_1+Y_2$

f_{Z} (z) = \int_{- \infty}^{\infty} 2 ϕ (y_{1} | μ_{1}, σ_{1}) Φ (λ (\frac{y_{1} - μ_{1}}{σ_{1}})) ϕ (z - y_{1} | μ_{2}, σ_{2}^{2}) d y_{1}

$f_Z(z)=\int_{-\infty}^{\infty}2\phi(y_1|\mu_1,\sigma_1)\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\phi(z-y_1|\mu_2,\sigma_2^2)\,\text{d}y_1$

Di sini dan masing-masing adalah pdf dan cdf normal standar. $\phi()$ $\Phi()$

f_{Z} (z) = \int_{- \infty}^{\infty} 2 \frac{1}{\sqrt{2 π σ_{1}}} \frac{1}{\sqrt{2 π σ_{2}}} e x hal (- \frac{1}{2 σ_{1}^{2}} (y_{1} - μ)^{2} - \frac{1}{2 σ_{2}^{2}} ((z - y_{1})^{2} - μ)^{2}) Φ (λ (\frac{y_{1} - μ_{1}}{σ_{1}})) d y_{1}

$f_Z(z)=\int_{-\infty}^{\infty}2\frac{1}{\sqrt{2\pi\sigma_1}}\frac{1}{\sqrt{2\pi\sigma_2}}exp\Big(-\frac{1}{2\sigma_1^2}(y_1-\mu)^2-\frac{1}{2\sigma_2^2}((z-y_1)^2-\mu)^2\Big)\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1$

Untuk notasi yang disederhanakan, misalkan $k=2\frac{1}{\sqrt{2\pi\sigma_1}}\frac{1}{\sqrt{2\pi\sigma_2}}$

\begin{aligned} f_{Z} (z) & = k \int_{- \infty}^{\infty} \exp (\frac{- 1}{2 σ_{1}^{2} σ_{2}^{2}} (σ_{1}^{2} (y_{1} - μ_{1})^{2} + σ_{2}^{2} ((z - y_{1}) - μ_{2})^{2})) Φ (λ (\frac{y_{1} - μ_{1}}{σ_{1}})) d y_{1} \\ = k \int_{- \infty}^{\infty} \exp (\frac{- 1}{2 σ_{1}^{2} σ_{2}^{2}} (σ_{2}^{2} (y_{1}^{2} - 2 y_{1} μ_{1} + μ_{1}) + σ_{1}^{2} ((z - y_{1})^{2} - 2 (z - y_{1}) μ_{2} + μ_{2}^{2}))) \\ \times Φ (λ (\frac{y_{1} - μ_{1}}{σ_{1}})) d y_{1} = k \int_{- \infty}^{\infty} \exp \\ (\frac{- 1}{2 σ_{1}^{2} σ_{2}^{2}} (σ_{2}^{2} (y_{1}^{2} - 2 y_{1} μ_{1} + μ_{1}) + σ_{1}^{2} (z^{2} - 2 z y_{1} + y_{1}^{2} - 2 z μ_{2} + 2 y_{1} μ_{2} + μ_{2}^{2}))) \\ \times Φ (λ (\frac{y_{1} - μ_{1}}{σ_{1}})) d y_{1} \end{aligned}

$\begin{align*}f_Z(z)&=k\int_{-\infty}^{\infty}\exp\Big(\frac{-1}{2\sigma_1^2\sigma_2^2}\Big(\sigma_1^2(y_1-\mu_1)^2+\sigma_2^2((z-y_1)-\mu_2)^2\Big)\Big)\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1\\ &=k\int_{-\infty}^{\infty}\exp\Big(\frac{-1}{2\sigma_1^2\sigma_2^2}\Big(\sigma_2^2(y_1^2-2y_1\mu_1+\mu_1)+\sigma_1^2((z-y_1)^2-2(z-y_1)\mu_2+\mu_2^2)\Big)\Big)\\&\quad\times\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1=k\int_{-\infty}^{\infty} \exp\\&\Big(\frac{-1}{2\sigma_1^2\sigma_2^2}\Big(\sigma_2^2(y_1^2-2y_1\mu_1+\mu_1)+\sigma_1^2(z^2-2zy_1+y_1^2-2z\mu_2+2y_1\mu_2+\mu_2^2)\Big)\Big)\\&\quad\times\Phi\Big(\lambda(\frac{y_1-\mu_1}{\sigma_1})\Big)\,\text{d}y_1 \end{align*}$

Tapi saya terjebak pada titik ini.

EDIT: Mengikuti saran dalam komentar, mengambil dan $\mu_1=\mu_2=0$ $\sigma_1^2=\sigma_2^2=1$

\begin{aligned} \int_{- \infty}^{\infty} 2 \frac{1}{\sqrt{2 π}} \frac{1}{\sqrt{2 π}} \exp (- \frac{1}{2} [y_{1}^{2} + z^{2} - 2 z y_{1} + y_{1}^{2}]) Φ (λ y_{1}) d y_{1} \\ \int_{- \infty}^{\infty} 2 \frac{1}{\sqrt{2 π}} \frac{1}{\sqrt{2 π}} \exp (- \frac{1}{2} y_{1}^{2}) Φ (λ y_{1}) \exp (- \frac{1}{2} (z - y_{1})^{2}) d y_{1} \end{aligned}

$\begin{align*} &\int_{-\infty}^\infty 2\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}}\exp\Big(-\frac{1}{2}[y_1^2+z^2-2zy_1+y_1^2]\Big)\Phi(\lambda y_1)dy_1 \\&\int_{-\infty}^\infty 2\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}}\exp\Big(-\frac{1}{2}y_1^2\Big)\Phi(\lambda y_1) \exp\Big(-\frac{1}{2}(z-y_1)^2\Big)dy_1\end{align*}$

condong-normal.

— kjetil b halvorsen
sumber

Mencoba kasus yang lebih sederhana dari , akan mengurangi sedikit kekacauan dan membuat Anda melihat hutan alih-alih pohon?

μ_{1} = μ_{2} = 0

$\mu_1 = \mu_2 = 0$

σ_{1} = σ_{2} = 1

$\sigma_1 = \sigma_2 = 1$

— Dilip Sarwate

Saya pikir saran Dilip adalah saran yang bagus, tetapi Anda mungkin ingin memeriksa ekspansi Anda dari istilah kuadratik dengan hati-hati. (Ini tidak akan memperbaiki masalah langsung Anda tetapi pada akhirnya akan menjadi masalah)

— Glen_b -Reinstate Monica

Ulangi kemiringan dalam hal dan gunakan mgf dari condong normal (lihat di bawah), karena dan independen, memiliki mgf yaitu , mgf dari condong normal dengan parameter , dan mana $\delta=\lambda/\sqrt{1+\lambda^2}$ $Y_1$ $Y_2$ $Z=Y_1 + Y_2$

\begin{aligned} M_{Z} (t) & = M_{Y_{1}} (t) M_{Y_{2}} (t) \\ = 2 e^{μ_{1} t + σ_{1}^{2} t^{2} / 2} Φ (σ_{1} δ t) e^{μ_{2} t + σ_{2}^{2} t^{2} / 2} \\ = 2 e^{(μ_{1} + μ_{2}) t + (σ_{1}^{2} + σ_{2}^{2}) t^{2} / 2} Φ (σ_{1} δ t) \\ = 2 e^{μ t + σ^{2} t^{2} / 2} Φ (σ δ^{'} t), \end{aligned}

$\begin{align} M_Z(t) &= M_{Y_1}(t)M_{Y_2}(t) \\ &= 2e^{\mu_1 t +\sigma_1^2 t^2/2}\Phi(\sigma_1\delta t)e^{\mu_2t +\sigma_2^2 t^2/2} \\ &= 2e^{(\mu_1+\mu_2)t + (\sigma_1^2+\sigma_2^2)t^2/2}\Phi(\sigma_1 \delta t) \\ &= 2e^{\mu t + \sigma^2 t^2/2}\Phi(\sigma \delta' t), \end{align}$

μ = μ_{1} + μ_{2}

$\mu=\mu_1+\mu_2$

σ^{2} = σ_{1}^{2} + σ_{2}^{2}

$\sigma^2=\sigma_1^2+\sigma_2^2$

σ δ^{'} = σ_{1} δ

$\sigma\delta'=\sigma_1\delta$

δ^{'}

$\delta'$ adalah parameter kemiringan baru. Karenanya, Dalam parameterisasi lain, parameter miring baru dapat ditulis, setelah beberapa aljabar, misalnya sebagai

δ^{'} = δ \frac{σ_{1}}{σ} = δ \frac{σ_{1}}{\sqrt{σ_{1}^{2} + σ_{2}^{2}}} .

$\delta'=\delta\frac{\sigma_1}\sigma=\delta\frac{\sigma_1}{\sqrt{\sigma_1^2+\sigma_2^2}}.$

λ^{'}

$\lambda'$

λ^{'} = \frac{δ^{'}}{\sqrt{1 - δ^{' 2}}} = \frac{λ}{\sqrt{1 + \frac{σ_{2}^{2}}{σ_{1}^{2}} (1 + λ^{2})}} .

$\lambda' = \frac{\delta'}{\sqrt{1-\delta'^2}}=\frac{\lambda}{\sqrt{1 + \frac{\sigma_2^2}{\sigma_1^2}(1+\lambda^2)}}.$

Mgf dari standar kemiringan normal dapat diturunkan sebagai berikut: \ end {align} mgf dari condong normal dengan parameter lokasi dan skala

\begin{aligned} M_{X} (t) & = E e^{t X} \\ = \int_{- \infty}^{\infty} e^{x t} 2 \frac{1}{\sqrt{2 π}} e^{- x^{2} / 2} Φ (λ x) d x \\ = 2 \int_{- \infty}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} (x^{2} - 2 t x)} Φ (λ x) d x \\ = 2 \int_{- \infty}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} ((x - t)^{2} - t^{2})} Φ (λ x) d x \\ = 2 e^{t^{2} / 2} \int_{- \infty}^{\infty} \frac{1}{\sqrt{2 π}} e^{- \frac{1}{2} (x - t)^{2}} P (Z \leq λ x) d x, & where Z \sim N (0, 1) \\ = 2 e^{t^{2} / 2} P (Z \leq λ U), & where U \sim N (t, 1) \\ = 2 e^{t^{2} / 2} P (Z - λ U \leq 0) \\ = 2 e^{t^{2} / 2} P (\frac{Z - λ U + λ t}{\sqrt{1 + λ^{2}}} \leq \frac{λ t}{\sqrt{1 + λ^{2}}}) \\ = 2 e^{t^{2} / 2} Φ (\frac{λ}{\sqrt{1 + λ^{2}}} t) . \end{aligned}

$\begin{align} M_X(t)&=Ee^{tX} \\ &=\int_{-\infty}^\infty e^{xt}2\frac1{\sqrt{2\pi}}e^{-x^2/2}\Phi(\lambda x)dx \\ &=2\int_{-\infty}^\infty \frac1{\sqrt{2\pi}}e^{-\frac12(x^2-2tx)}\Phi(\lambda x)dx\\ &=2\int_{-\infty}^\infty \frac1{\sqrt{2\pi}}e^{-\frac12((x-t)^2-t^2)}\Phi(\lambda x)dx \\ &=2e^{t^2/2} \int_{-\infty}^\infty \frac1{\sqrt{2\pi}}e^{-\frac12(x-t)^2}P(Z\le \lambda x)dx, & \text{where }Z \sim N(0,1) \\ &=2e^{t^2/2} P(Z\le \lambda U), & \text{where }U \sim N(t,1)\\ &=2e^{t^2/2} P(Z - \lambda U \le 0) \\ &=2e^{t^2/2} P(\frac{Z - \lambda U +\lambda t}{\sqrt{1+\lambda^2}} \le \frac{\lambda t}{\sqrt{1+\lambda^2}}) \\ &=2e^{t^2/2}\Phi(\frac\lambda{\sqrt{1+\lambda^2}}t). \end{align}$

μ

$\mu$ dan adalah

σ

$\sigma$

M_{μ + σ X} (t) = E e^{(μ + σ X) t} = e^{μ t} M_{X} (σ t) = 2 e^{μ t + σ^{2} t^{2} / 2} Φ (\frac{λ}{\sqrt{1 + λ^{2}}} σ t) .

$M_{\mu + \sigma X}(t)=Ee^{(\mu+\sigma X)t} = e^{\mu t}M_X(\sigma t) = 2e^{\mu t+\sigma^2 t^2/2}\Phi(\frac\lambda{\sqrt{1+\lambda^2}}\sigma t).$

— Jarle Tufto
sumber

Saya tidak mengerti bagaimana Anda mendapatkan ini dapatkah Anda memberi saya detail lebih lanjut?

δ^{'} = δ \frac{σ_{1}}{σ}

$\delta'=\delta\frac{\sigma_1}\sigma$

Anda hanya menyamakan jumlah yang muncul sebelum dan dalam eksponensial dan dalam argumen fungsi untuk menemukan parameter baru.

t

$t$

t^{2}

$t^2$

Φ

$\Phi$

— Jarle Tufto