We have a random variable $Y$ that has a uniform distribution on the interval $[1,5]$. The cost of delay is given by $U = 2Y^2 + 3$. Use the method of transformations to derive the density function of $U$.
[Math] density function by the method of transformations
probabilitystatistics
Best Answer
First, if $Y$ takes values in $[1,5]$, then $U$ takes values in the range...? So, for $x$ in that range, you should consider the distribution function $F_U {(x)} := {\rm P}(U \le x)$. Substituting $U=2Y^2+3$ leads straightforwardly to $F_U {(x)} = F_Y {(\varphi(x))}$ for some function $\varphi(x)$, where $F_Y$ is the distribution function of $Y$. What is $\varphi(x)$? Note that $\varphi(x)$ takes values in $[1,5]$. Now, the density $f_U$ of $U$ is given by $$ f_U (x) = \frac{{\rm d}}{{{\rm d}x}}F_U (x) = \frac{{\rm d}}{{{\rm d}x}}F_Y (\varphi (x)) =...? $$ Final hint: the density $f_Y$ of $Y$ is constant on $[1,5]$; what is that constant?
Suggestion: verify that the $f_U {(x)}$ you find integrates to $1$.