Let's say that Y = aX1 + bX2 + cX3 where all the random variables (X1, X2 and X3) are normally distributed. How can I find the distribution of the function Z = e^Y?
The mean and variance of X1, X2 and X3 are given as N(1,0.3), N(4.1,0.4) and N(6,0.6) respectively.
If $Y\sim N(m,\sigma^2)$ then $\Pr(e^Y<z)=\Pr (Y<\log z)=\int_{-\infty}^{\log z}e^{-(y-m)^2/2\sigma^2}\frac{dy}{\sigma\sqrt{2\pi}}$ and the density of $Z=e^Y$ is $e^{-(\log z-m)^2/2\sigma^2}\frac{1}{\sigma\sqrt{2\pi}}\times \frac{1}{z}$ from the fundamental theorem of calculus and the derivative of composition of functions. If $Y=a_1X_1+a_2X_2+a_3X_3 $ where $X_i\sim N(m_i,\sigma_i^2)$ all independent then $m=a_1m_1+a_2m_2+a_3m_3$ and $\sigma^2=a_1^2\sigma_1^2+a_2^2\sigma_2^2+a_3^2\sigma_3^2.$