<s>[INST] Let $X\sim\mathrm{Normal}(1,4)$. If $Y=0.5^X$, find $\Bbb E[Y^2]$. 
Let $X\sim\mathrm{Normal}(1,4)$. If $Y=0.5^X$, find $\Bbb E[Y^2]$.

So I have $\Bbb E[Y^2]=M^{\prime\prime}_Y(0)$
So I need to compute $M_Y(t)=M_{0.5^X}(t)=\Bbb E\left[e^{0.5^X}\right]$
$$\Bbb E\left[e^{0.5^X}\right]=\int_{-\infty}^\infty e^{0.5^x t}\cdot f_X(x)\,dx=\int_{-\infty}^\infty e^{0.5^x t}\cdot \frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2\cdot 4}}\,dx$$
So from here do I have 
$$M_Y^{\prime\prime}(t)=\int_{-\infty}^\infty (0.5^x)^2 e^{0.5^x t}\cdot \frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2\cdot 4}}\,dx$$ or $$M_Y^{\prime\prime}(0)\int_{-\infty}^\infty (0.5^x)^2\cdot \frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2\cdot 4}}dx\,?$$
 [/INST] Let $f(x)$ and $F(x)$ denote the p.d.f. and c.d.f. of $X\sim\operatorname N(1,2^2)$ respectively. 
Let $g(z)$ and $G(z)$ denote the p.d.f. and c.d.f of $Z=0.5^{2X}$ respectively.
Then $$G(z)=\operatorname P(Z\le z)=\operatorname P\left(2^{2X}\ge\frac1z\right)=\operatorname P\left(2X\ge-\frac{\ln z}{\ln2}\right)=1-F\left(-\frac{\ln z}{2\ln2}\right)$$ so its derivative becomes $$g(z)=\frac1{2z\ln2}f\left(-\frac{\ln z}{2\ln 2}\right)=\frac1{2z\ln2}\cdot\frac1{2\sqrt{2\pi}}\exp\left(-\frac1{2\cdot2^2}\left(-\frac{\ln z}{2\ln2}-1\right)^2\right).$$ Hence \begin{align}\Bbb E[Z]&=\int_{\Omega_Z} zg(z)\,dz=\frac1{(4\ln2)\sqrt{2\pi}}\int_0^\infty\exp\left(-\frac18\left(\frac1{4\ln^22}\ln^2z+\frac1{\ln2}\ln z+1\right)\right)\,dy\\&=\frac{e^{-1/8}}{(4\ln2)\sqrt{2\pi}}\int_0^\infty z^{-(\ln z+4\ln2)/(32\ln^22)}\,dz\end{align} and the substitution $u=\ln z$ yields \begin{align}\Bbb E[Z]&=\frac{e^{-1/8}}{(4\ln2)\sqrt{2\pi}}\int_{-\infty}^\infty(e^u)^{1-(u+4\ln2)/(32\ln^22)}\,du\\&=\frac{e^{-1/8}e^{(8\ln2-1)^2/8}}{(4\ln2)\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(-\frac12\left(\frac{u-2(8\ln2-1)\ln2}{4\ln2}\right)^2\right)\,du\\&=e^{-1/8}e^{(8\ln2-1)^2/8}=e^{(8\ln2-2)\ln2}=2^{8\ln2-2}.\end{align} Thus $$\Bbb E[Y^2]=\Bbb E[Z]=4^{4\ln2-1}.$$
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