Represent the following expressions as a power of the number a where a ‡0.

1 answer:

3 0

Properties we are going to use :
1. (a^m)^n = a^mn
2. a^m : a^n = a^(m-n)

((a^-2)^-1)^-1 = (a^(-2 x -1))^-1 = (a^2)^-1 = a^-2

(a : a^-1)^2 = (a^1 - (-1))^2 = (a^2)^2 = a^4

Therefore, the expression is simplified down to a^-2 : a^4

a^-2 : a^4 = a^(-2 - 4) = a^-6.

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The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

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$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :