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 $$M_Y^{\prime\prime}(0)\int_{-\infty}^\infty (0.5^x)^2\cdot \frac{1}{2\sqrt{2\pi}}e^{-\frac{(x-1)^2}{2\cdot 4}}dx\,?$$
Best Answer
You have $Y^2 = 0.5^{2X} = \left( \frac{1}{4} \right)^X,$ so $\mathbb{E}[Y^2] = \mathbb{E}[e^{tX}]$ where $ t= - \log 4.$ The moment generating function of a normal variable with mean $\mu$ and variance $\sigma^2$ is equal to $\exp(\mu t + \sigma^2 t^2/2)$ so $\mathbb{E}[Y^2] = \exp(8 \log^2 2)/4.$