Minimize $V(s)=e^{-s\alpha}\frac{1}{2}(1+e^s)$ for fixed $\alpha$, where $0 < \alpha < 1$.

$$ \begin{align} V(s) && = && e^{-s\alpha}\frac{1}{2}(1+e^s) \\ && = && \frac{1}{2}\frac{1 + e^s}{e^{s\alpha}} \\ && = && \frac{1}{2}(\frac{1}{e^{s\alpha}}+\frac{e^s}{e^{s\alpha}}) \end{align} $$

Therefore

$$ \begin{align} V'(s) && = && \frac{1}{2}( \frac{-ln(e^\alpha)e^{s\alpha}}{e^{2s\alpha}} + \frac{e^{s}e^{s\alpha} - e^{s}ln(e^{\alpha})e^{s\alpha}}{e^{2s\alpha}} ) \\ && = && \frac{1}{2}( \frac{-\alpha e^{s\alpha}}{e^{2s\alpha}} + \frac{e^{s+s\alpha}-\alpha e^{s+s\alpha}}{e^{2s\alpha}} ) \\ && = && \frac{1}{2}( \frac{-\alpha}{e^{s\alpha}} + \frac{(1-\alpha)e^{s+s\alpha}}{e^{2s\alpha}} ) \\ && = && \frac{1}{2}( \frac{-\alpha}{e^{s\alpha}} + \frac{(1-\alpha)e^s}{e^{s\alpha}} ) \end{align} $$

We want $V'(s) = 0$, so $$ \begin{align} 0 && = && \frac{1}{2}( \frac{-\alpha}{e^{s\alpha}} + \frac{(1-\alpha)e^s}{e^{s\alpha}} ) \\ 0 && = && \frac{-\alpha}{e^{s\alpha}} + \frac{(1-\alpha)e^s}{e^{s\alpha}} \\ 0 && = && -\alpha + (1-\alpha)e^s \\ \alpha && = && (1-\alpha)e^s \\ \frac{\alpha}{1-\alpha} && = && e^s \\ ln(\frac{\alpha}{1-\alpha}) && = && s \end{align} $$

We also need to prove that $V''(s)$ is positive for all $s \in \mathbb{R}$:

$$ \begin{align} V''(s) && = && \frac{1}{2}( \frac{-\alpha \cdot -ln(e^\alpha)e^{s\alpha}}{e^{2s\alpha}} + (1-\alpha)\frac{e^{s}e^{s\alpha}-ln(e^\alpha)e^{s\alpha}e^s}{e^{2s\alpha}} ) \\ && = && \frac{1}{2}( \frac{\alpha^{2} e^{s\alpha}}{e^{2s\alpha}} + (1-\alpha)\frac{(1-\alpha)e^{s+s\alpha}}{e^{2s\alpha}} ) \\ && = && \frac{1}{2}( \frac{\alpha^2}{e^{s\alpha}} + (1-\alpha)^2\frac{e^s}{e^{s\alpha}} ) \end{align} $$

Because $0 < \alpha < 1$:

• $\alpha^2$ is always positive
• $(1-\alpha)^2$ is always positive

$e^x$ is also always positive for any $x \in \mathbb{R}$

Therefore $V''(s)$ is always positive for any $s \in \mathbb{R}$.

Therefore $V(s)$ is minimum at $s=ln(\frac{\alpha}{1-\alpha})$.