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.
Best Answer
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 $u=x-1$.