If X has uniform distribution on (0,1), how do I calculate probability density functions of $Y=\exp(X)$ and $Y=\sqrt{X}$?
I can assume the shape of these pdfs when I make histogram of mentioned variables in MATLAB, but i don't know how to calculate exact functions. Can i get some help, please?
Thanks in advance!
The method for the first part of this question seems to be similar to the question addressed in the first part of the question here.
There, many of the answers suggest finding the CDF (cumulative distribution function) and taking the derivative, which is the PDF (probability distribution function). The CDF of $Y=e^X$ is defined as follows: $$ F_Y(y)=P(Y\leq y)=P(e^X \leq y)$$ which is 0 when $y\leq0$, and for y>0 is as follows: $$ P(e^X \leq y)=P\left(X\leq \ln(y)\right) = F_X(\ln(y))$$ Differentiating $F_X(\ln(y))$ with respect to $y$ to get $f_y(y)$, we obtain the following: $$ f_y(y) = F'_X(\ln(y))\frac{1}{y}=f_x(\ln(y))\frac{1}{y}$$ Since $X$ is uniformly distributed over the interval $(0,1)$, it has a PDF of $f_x(x)=1$ for $ 0<x<1$. $$ f_x(\ln(y))=1*\frac{1}{y}=\frac{1}{y}, 0<\frac{1}{y}<1$$ Thus, $f_y(y)=\frac{1}{y}$ for $0<\ln(y)<1$ and $f_y(y)=0$ otherwise.
Similarly for $Y=\sqrt{X}$, $$F_Y(y)=P(Y\leq y)=P(\sqrt{X}\leq y)=P(X\leq y^2)=F_X(y^2)$$ Then, differentiating, $$ f_y(y)=F'_X(y^2)2y=f_x(y^2)2y $$ Again, since $X$ is uniformly distributed over the interval $(0,1)$, it has a PDF of $f_x(x)=1$ for $ 0<x<1$. Then, $$ f_x(y^2)=1*2y =2y, 0<y^2<1 $$ Thus, $f_y(y)=2y$ for $0<y^2<1$ and $f_y(y)=0$ otherwise.
