Radon–Nikodym derivative of $\alpha$-measure

Question

I'm struggling a bit to understand how the Radon–Nikodym derivative is computed in pag.3 of this paper written by Mark Joshi titled "THE USE OF POWER NUMERAIRES IN OPTION PRICING". Given a process defined as $dS_t=\mu S_tdt + \sigma S_t dW_t$, the change of measure $\frac{\mathrm{d}\mathbb Q^1}{\mathrm d\mathbb Q^0}$ from bank account numeraire $(\alpha=0)$ to the stock measure $(\alpha=1)$ is clear:

\begin{align*} \frac{\mathrm{d}\mathbb Q^1}{\mathrm d\mathbb Q^0}=\frac{S_T}{S_0}\frac{B_0}{B_T}=\frac{S_T}{S_0}e^{-rT}=\frac{S_0e^{(r-\frac{1}{2}\sigma^2)T+\sigma W_T -rT}}{S_0}=e^{-\frac{1}{2}\sigma^2T+\sigma W_T} \end{align*}

I'm confused by the definition of the change of measure $\frac{\mathrm{d}\mathbb Q^\alpha}{\mathrm d\mathbb Q^0}$ from the risk neutral measure $(\alpha=0)$ to the $\alpha$-measure, given that $S^\alpha_T=S^\alpha_0e^{\alpha(r-\frac{1}{2}\sigma^2)T+\alpha \sigma W_T}$:

\begin{align*} \frac{\mathrm{d}\mathbb Q^\alpha}{\mathrm d\mathbb Q^0} = \frac{N_{T,T}^\alpha B_0}{N_{0,T}^\alpha B_T}=\frac{N_{T,T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\frac{S_{T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\color{red}{e^{-\frac{1}{2}\alpha^2\sigma^2T+\alpha\sigma W_T}}. \end{align*}

Answer 1

\begin{align*} \frac{\mathrm{d}\mathbb Q^\alpha}{\mathrm d\mathbb Q^0} = \frac{N_{T,T}^\alpha B_0}{N_{0,T}^\alpha B_T}=\frac{N_{T,T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\frac{S_{T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\color{red}{e^{-\frac{1}{2}\alpha^2\sigma^2T+\alpha\sigma W_T}}. \end{align*}

I was confused by the definition of $N^\alpha_0=S_0^\alpha \exp\left(rT(a-1)+0.5\sigma^2(a^2-\alpha)T\right)$ given that $S_T^\alpha=S_0^\alpha \exp\left(\alpha(r-0.5\sigma^2)T+\alpha\sigma W_T\right)$. To be specific I did not understood the red step in the definition of $N^\alpha_0$:

\begin{align*} N_0^\alpha &= e^{-rT}\mathbb{E}^{\mathbb Q}[S_T^\alpha|\mathcal{F}_0] \\ &= e^{-rT}\mathbb{E}^{\mathbb Q}\left[S_0^\alpha\exp\left(\alpha\left(r-\frac{1}{2}\sigma^2\right)T+\alpha\sigma W_T \right)\bigg|\mathcal{F}_t\right] \\ &= e^{-rT}S_0^\alpha\exp\left(\alpha\left(r-\frac{1}{2}\sigma^2\right)T+\frac{1}{2}\alpha^2\sigma^2T\right) \\ &= S_0^\alpha \exp\left(rT(\alpha-1)+0.5\sigma^2(\alpha^2-\alpha)T\right) \end{align*}

Turns out to to be the expectation of a lognormal variable: $$\mathbb EY^{\alpha}=\mathbb Ee^{{\alpha}\mu+{\alpha}\sigma U}=e^{{\alpha}\mu}\mathbb Ee^{{\alpha}\sigma U}=e^{{\alpha}\mu}M_U({\alpha}\sigma)=e^{{\alpha}\mu+\frac12{\alpha}^2\sigma^2}$$

Afterwards it is pretty straightforward:

\begin{align*} \frac{N_{T,T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\frac{S_0^\alpha \exp\left(\alpha(r-0.5\sigma^2)T+\alpha\sigma W_T\right) e^{-rT}}{S_0^\alpha \exp\left(rT(\alpha-1)+0.5\sigma^2(\alpha^2-\alpha)T\right)}= e^{-\frac{1}{2}\alpha^2\sigma^2T+\alpha\sigma W_T} \end{align*}