# density function by the method of transformations

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 $U$.

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You remember the formula no? For the densities $f_U(u)=f_Y(y)\left|\frac{dy}{du}\right|$. – Raskolnikov Dec 9 '10 at 16:37

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 $1$.
If you find $f_U$, I can tell you if it is right. – Shai Covo Dec 9 '10 at 17:08