Find $E(Y)$ and $Var(Y)$ of $\log Y \sim N (\mu,\sigma^2)$
I tried solving this in 2 different ways. The second way is what I am stuck on:
1st Way:
Let $Y=e^X$ where $X \sim N (\mu,\sigma^2)$. Then $X=\mu+\sigma Z$
$$E(e^{tZ})=\int e^{tz}\frac{1}{\sqrt{2\pi \sigma^2}}\exp{\frac{-z^2}{2}}dz$$ $$=\int \frac{1}{\sqrt{2\pi \sigma^2}}\exp({\frac{-z^2+2tz-t^2}{2}+t^2/2)}dz$$ $$=e^{t^2/2}$$
Hence $$E(e^X)=E(e^{\mu+\sigma Z})=e^\mu E(e^{\sigma Z})=e^\mu e^{\sigma^2/2}$$ and $$E(e^{2X})=E(e^{2\mu+2\sigma Z})=e^{2\mu} E(e^{2\sigma Z})=e^{2\mu} e^{2\sigma^2}$$
However. The second way:
I begin by writing the pdf of $\log Y$:
Let $f(x)$ be the pdf of a normally distributed variable with $(\mu,\sigma ^2)$
$$P(Y<y)=P(\log Y<\log y)=P(X< \log y)=\int_{-\infty}^{\log y}f(x)dx$$
Hence the pdf of $Y$ is $$d/dy \int_{-\infty}^{\log y}f(x)dx=\frac{f(\log y)}{y}$$
Now to find the $E(Y)$ I simply need to integrate:
$$E(Y)=\int \frac{f(\log y)}{y} \times y dy$$
Can anyone help me solve this integral? And also $$E(Y^2)=\int \frac{f(\log y)}{y} \times y^2 dy$$
$$\mathbb{P} (Y<y)=\mathbb{P} ({{e}^{X}}<y)=\mathbb{P} (X<\ln y)=\frac{1}{\sigma \sqrt{2\pi }}\int_{-\infty }^{\ln y}{\exp \left( -\frac{{{(x-\mu )}^{2}}}{2{{\sigma }^{2}}} \right)}\,dx$$ set $u=e^x$, or $x=\ln u\,$, we have $$\mathbb{P} (Y<y)=\frac{1}{\sigma \sqrt{2\pi }}\int_{0 }^{ y}\frac{1}{u}{\exp \left( -\frac{{{(\ln u-\mu )}^{2}}}{2{{\sigma }^{2}}} \right)}\,du$$ Therefore $$f_Y(y)=\frac{1}{\sigma \sqrt{2\pi }\,y} \exp \left( -\frac{{{(\ln y-\mu )}^{2}}}{2{{\sigma }^{2}}}\right)\,\,,\,\,y>0$$ thus $$\mathbb{E}[Y]=\frac{1}{\sigma \sqrt{2\pi }}\int_{0 }^{ +\infty}y\cdot\frac{1}{y}{\exp \left( -\frac{{{(\ln y-\mu )}^{2}}}{2{{\sigma }^{2}}} \right)}\,dy=\frac{1}{\sigma \sqrt{2\pi }}\int_{0 }^{ +\infty}\exp \left( -\frac{{{(\ln y-\mu )}^{2}}}{2{{\sigma }^{2}}} \right)dy$$ set $\frac{-\mu+\ln y}{\sqrt{2}\sigma}=x\,$,thus $dy=\sqrt{2}\sigma\exp\left(\sqrt{2}\sigma x+\mu\right)dx$, and $$\mathbb{E}[Y]=\frac{ e^{\mu}}{ \sqrt{\pi }}\int_{-\infty }^{ +\infty}\exp\left(\sqrt{2}\sigma x-x^2\right)dx$$ $$\mathbb{E}[Y]=\frac{ e^{\mu+\frac{1}{2}\sigma^2}}{ \sqrt{\pi }}\int_{-\infty }^{ +\infty}\exp\left(-\left(x-\frac{\sigma}{\sqrt{2}}\right)^2\right)dx$$ set $u=\sqrt{2}x-\sigma$, we have
$$\mathbb{E}[Y]=e^{\mu+\frac{1}{2}\sigma^2}\frac{ 1}{ \sqrt{2\pi }}\int_{-\infty }^{ +\infty}e^{-\frac{u^2}{2}}du=e^{\mu+\frac{1}{2}\sigma^2}$$