Finding the probability density function of $Y=e^X$, where $X$ is standard normal Let the random variable $X$ have the $N(0,1)$ distribution for which the probability function is:
$$
f(x)= \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right), -\infty< x <\infty
$$
Let $Y=e^X$.
A. Find the probability density function for $Y$,
B. Find $E(Y)$,
C. Find $E(Y^2)$ and deduce $\mathrm{Var}(Y)$.
B and C I can do if I find A but can anybody explain to me how this is done. The logic behind it.
 A: The cumulative distribution function of $Y$ is
$$F_Y(y)=P(Y\leq y)=P(e^X\leq y)$$
which is $0$ when $y\leq 0$, and when $y>0$ then
$$
P(e^X\leq y)=P(X\leq\log(y))=\int_{-\infty}^{\log(y)}f_X(x)\,\mathrm dx=F_X(\log(y)).
$$
Differentiating $F_Y(y)$ we obtain the density of $Y$:
$$
f_Y(y)=F'_X(\log(y))\frac{1}{y}=f(\log(y))\frac{1}{y}=\frac{1}{y\sqrt{2\pi}}\exp\left(-\frac{\log(y)^2}{2}\right),\quad y>0,
$$
and $f_Y(y)=0$ for $y\leq 0$.
A: To find the density function $f_Y(y)$ of $Y$, one strategy is to find the cumulative distribution function $F_Y(y)$, and then differentiate. Note that $Y$ is always positive, so $F_Y(y)=0$ if $y\le 0$. Now suppose that $y\gt 0$. Then 
$$F_Y(y)=\Pr(Y\le y)=\Pr(e^X\le y)=\Pr(X\le \ln y).$$
Thus
$$F_Y(y)=\int_{-\infty}^{\ln y}f(x)\,dx.$$
To find the density, differentiate. We do this by differentiating under the integral sign, that is, by using the Fundamental Theorem of Calculus. We get
$$f_Y(y)=\frac{1}{y}f(\ln y).$$ 
Remark: To find $E(Y)$, I would suggest not using the density. Easier, I think, is to use
$$\int_{-\infty}^\infty e^x f(x)\,dx.$$
Apart from a constant, we are integrating $e^{-(x^2-2x)/2}$. Complete the square, and make the substitution $u=x-1$.
A: The simplest way is to use the change of variable formula for continous random variables 
$$f_Y(y) = f_x(x) \; \left| \frac{dx}{dy}\right|$$ where $\left| \frac{dx}{dy}\right|$ is the abolute of the Jacobian determinant, and $f_X(x)$ and $f_Y(y)$ are the PDFs of $X$ and $Y$, respectively. Note: This formula is valid, if the transformation is invertible.
Now calc:


*

*$\left| \frac{dy}{dx}\right| = y \Rightarrow \left| \frac{dx}{dy}\right| =\left| \frac{1}{\frac{dy}{dx}} \right| = \frac{1}{y}$

*$f_Y(y) = f_x(x) \; \left| \frac{dx}{dy}\right| = \varphi(x) \; \frac{1}{y} = \varphi(\ln{(y)}) \; \frac{1}{y} $, where $f_x(x) = \varphi(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$. As it should, this yields the same solution as Stefans wrote.

A: As other answers already covered the probability density function, I want to answer parts 2 and 3 of the question: Namely the mean and variance of $E(e^X)$:
For the meant we see that the following calculation holds:
\begin{equation*}
E(e^X)=\int_{-\infty}^\infty e^{x}\frac{1}{\sqrt{2\pi}}\cdot e^{x^2/2}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty exp\left(-\frac{x(x-2)}{2}\right) dx
\end{equation*}
Now make the substitution $t=x+1$ to obtain:
\begin{equation*}
E(e^X)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(-\frac{t^2-1}{2}\right)dt=e^{1/2}
\end{equation*}
Now we shall calculate $E(Y^2)=E(e^{2X})$:
\begin{align*}
E(e^{2X})=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(-\frac{x(x-4)}{2}\right)dx\overset{t=x+2}{=}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(-\frac{t^2-4}{2}\right)dt=e^2
\end{align*}
In the end we hence obtain:
\begin{align*}
Var(e^X)=E(e^{2x})-(E(e^X))^2=e^2-e=e(e-1).
\end{align*}
