标准正态密度 f ( x ) = 1 2 π e − x 2 / 2 f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2} f ( x ) = 2 π 1 e − x 2 /2 P ( X > 0 ) = 1 / 2 P(X>0)=1/2 P ( X > 0 ) = 1/2
E [ X ∣ X > 0 ] = ∫ 0 ∞ x f ( x ) d x 1 / 2 = 2 ∫ 0 ∞ 1 2 π x e − x 2 / 2 d x . E[X\mid X>0]=\frac{\int_0^{\infty} x f(x)\,dx}{1/2}=2\int_0^{\infty} \frac{1}{\sqrt{2\pi}}x e^{-x^2/2}dx. E [ X ∣ X > 0 ] = 1/2 ∫ 0 ∞ x f ( x ) d x = 2 ∫ 0 ∞ 2 π 1 x e − x 2 /2 d x . 令 u = − x 2 / 2 u=-x^2/2 u = − x 2 /2 1 2 π \frac{1}{\sqrt{2\pi}} 2 π 1
E [ X ∣ X > 0 ] = 2 π . E[X\mid X>0]=\sqrt{\frac{2}{\pi}}. E [ X ∣ X > 0 ] = π 2 . 英文解析
For the standard normal pdf f ( x ) = 1 2 π e − x 2 / 2 , f(x)=\tfrac{1}{\sqrt{2\pi}}\,e^{-\,x^2/2}, f ( x ) = 2 π 1 e − x 2 /2 , P ( X > 0 ) = 1 2 . P(X>0)=\tfrac12. P ( X > 0 ) = 2 1 . E [ X ∣ X > 0 ] = ∫ 0 ∞ x 1 2 π e − x 2 / 2 d x 1 2 = 2 ∫ 0 ∞ 1 2 π x e − x 2 / 2 d x . E[X\mid X>0]
= \frac{\int_{0}^{\infty} x\,\tfrac{1}{\sqrt{2\pi}}e^{-\,x^2/2}\,dx}{\,\tfrac12\,}
= 2\int_{0}^{\infty}\frac{1}{\sqrt{2\pi}}\,x\,e^{-\,x^2/2}\,dx. E [ X ∣ X > 0 ] = 2 1 ∫ 0 ∞ x 2 π 1 e − x 2 /2 d x = 2 ∫ 0 ∞ 2 π 1 x e − x 2 /2 d x . u = − 1 2 x 2 u=-\tfrac12\,x^2 u = − 2 1 x 2 1 2 π , \tfrac{1}{\sqrt{2\pi}}, 2 π 1 , E [ X ∣ X > 0 ] = 2 × 1 2 π = 2 π . E[X\mid X>0]
= 2 \times \frac{1}{\sqrt{2\pi}}
= \sqrt{\frac{2}{\pi}}. E [ X ∣ X > 0 ] = 2 × 2 π 1 = π 2 .