The Normal distribution of generating data g(y) and specified model f(y) is-
g(y)~N(m , t 2 m,t^2 m , t 2 )
f(y)~N( μ , o 2 ) (\mu,o^2) ( μ , o 2 )
The probability density function is-
Y = 1 2 π e − x 2 2 Y=\dfrac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}} Y = 2 π 1 e 2 − x 2
l o g Y = l o g 1 2 π + l o g e − x 2 2 = l o g 1 2 π + − x 2 2 − ( 1 ) logY=log\dfrac{1}{\sqrt{2\pi}}+loge^{\frac{-x^2}{2}}=log\dfrac{1}{\sqrt{2\pi}}+\dfrac{-x^2}{2}~~~~~-(1) l o g Y = l o g 2 π 1 + l o g e 2 − x 2 = l o g 2 π 1 + 2 − x 2 − ( 1 )
So E ( l o g Y ) = ∫ 0 r x l o g Y d x E(logY)=\int_{0}^{r}xlogYdx E ( l o g Y ) = ∫ 0 r x l o g Y d x
= ∫ 0 r x l o g 1 2 π + − x 3 2 d x = x 2 2 l o g 1 2 π ∣ 0 r − x 4 8 ∣ 0 r = r 2 2 l o g 1 2 π − r 4 8 = − l o g ( 2 r t 2 ) − 1 =\int_{0}^{r} xlog\dfrac{1}{\sqrt{2\pi}}+\dfrac{-x^3}{2}dx\\[9pt] =\dfrac{x^2}{2}log\dfrac{1}{\sqrt{2\pi}}|_0^r-\dfrac{x^4}{8}|_0^r\\[9pt]=\dfrac{r^2}{2}log\dfrac{1}{\sqrt{2\pi}}-\dfrac{r^4}{8}\\[9pt]=-log(2rt^2)-1 = ∫ 0 r x l o g 2 π 1 + 2 − x 3 d x = 2 x 2 l o g 2 π 1 ∣ 0 r − 8 x 4 ∣ 0 r = 2 r 2 l o g 2 π 1 − 8 r 4 = − l o g ( 2 r t 2 ) − 1
Hence E ( l o g Y ) = − l o g ( 2 r t 2 ) − 1 E(logY)= -log(2rt^2)-1 E ( l o g Y ) = − l o g ( 2 r t 2 ) − 1
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