Question #189978

Show that


Ec[logƒ(Y)] = —— log(26²) —


7²+(m_µ)²


202


(c) Show that Kullback-Leibler divergence of f (y) with respect to g(y) is given by


D(glf) EG[logg (Y)]-EG[log f(Y)]


0² t² + (m-μ)² - 1}


1
Expert's answer
2021-05-07T14:39:47-0400

The Normal distribution of generating data g(y) and specified model f(y) is-

g(y)~N(m,t2m,t^2)

f(y)~N(μ,o2)(\mu,o^2)



The probability density function is-


Y=12πex22Y=\dfrac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}


logY=log12π+logex22=log12π+x22     (1)logY=log\dfrac{1}{\sqrt{2\pi}}+loge^{\frac{-x^2}{2}}=log\dfrac{1}{\sqrt{2\pi}}+\dfrac{-x^2}{2}~~~~~-(1)



So E(logY)=0rxlogYdxE(logY)=\int_{0}^{r}xlogYdx


     =0rxlog12π+x32dx=x22log12π0rx480r=r22log12πr48=log(2rt2)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


Hence E(logY)=log(2rt2)1E(logY)= -log(2rt^2)-1


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Comments

Assignment Expert
10.05.21, 12:52

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Dheerajparadkar
08.05.21, 17:34

Thank U

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