Find the distribution of a random variable

Question

Let $\Omega=[0,1]$, $\mathcal{F}=\mathcal{B} \cap [0,1]$, and $P$ be the Lebesgue measure restricted to $[0,1]$. Let $\Phi_{\mu,\sigma^2}(x)=\mathcal{N}_{\mu,\sigma^2}((-\infty,x]) $. Then it is clear that $\xi=\Phi^{-1}_{\mu,1}$ is a random variable on $(\Omega,\mathcal{F},P)$. Can anyone show me how to find the distribution of $\xi$ under $P$ ? and also $\Bbb{E}(\eta)=1$ under $P$, where $\eta=e^{-\mu\xi+\mu^2/2}$.

Answer 1

The distribution of $\xi$ under $P$ is $\Phi_{\mu, 1}$. \begin{eqnarray*} P \left\{ \omega : \xi \left( \omega \right) \leqslant x \right\} & = & P \left\{ \omega : \Phi^{- 1}_{\mu, 1} \left( \omega \right) \leqslant x \right\}\\ & = & P \left\{ \omega : \omega \leqslant \Phi_{\mu, 1} \left( x \right) \right\}\\ & = & \Phi_{\mu, 1} \left( x \right) \end{eqnarray*} The second line follows by strict monotonicity and the third by the properties of Lebesgue measure on $\left[ 0, 1 \right]$ (uniform distribution). Then that distribution has probability density function $\frac{1}{\sqrt{2\pi}}\mathrm{e}^{- \frac{x^2}{2}}$, resulting in \begin{eqnarray*} \frac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty} \mathrm{e}^{- \mu x + \frac{\mu^2}{2}} \mathrm{e}^{- \frac{x^2}{2}} \mathrm{d} x & = & \frac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty} \mathrm{e}^{- \frac{\left( x - \mu \right)^2}{2}} \mathrm{d} x\\ & = & 1 \end{eqnarray*}