Answer to Question #189972 in Statistics and Probability for Dheerajparadkar

Question

Suppose that the true distribution g(y) generating data and the specified model f (y) have normal distribution N(m, t2) and N(u, o²), respectively.

(a) Show that

Ec[logg (Y)] = -log(2rt²) - 1

1

where EG[] is an expectation with respect to the true distribution N(m, t²).

Answer 1

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

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

f(y)~N$(\mu,o^2)$

The probability density function is-

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

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

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

     $=\int_{0}^{r} xlog\dfrac{1}{\sqrt{2\pi}}+\dfrac{-x^3}{2}dx\\[9pt] =\dfrac{x^2}{2}log\dfrac{1}{\sqrt{2\pi}}|_0^r-\dfrac{x^4}{8}|_0^r\\[9pt]=\dfrac{r^2}{2}log\dfrac{1}{\sqrt{2\pi}}-\dfrac{r^4}{8}\\[9pt]=-log(2rt^2)-1$

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