A random variable $X$ has the p.d.f
$$f_X(x)=\begin{cases}
2x & 0<x<1 \\
0 & \text{ else}
\end{cases}
$$
Then find the p.d.f of $Y=4X+3.$
My attempt
$y=4x+3\implies x=(y-3)/4.$ Apply in $f_X(x)$. We get $f_Y(y)=3[(y-3)/4], $ for $0<(y-3)/4<1\implies 3<y<7.$ and $f_Y(y)=0$ else. But answer was given was $f_Y(y)=(3/16)(y-3).$ I am learning the basics of statistics. I am facing this type of problem for the first time. I am not sure about the logic. Please help me.
Best Answer
Hint $:$ Let $F_X$ and $F_Y$ respectively denote the probability distribution functions for $X$ and $Y.$ Then we have $$F_Y(y) = P(Y \leq y) = P \left (X \leq \frac {y - 3} {4} \right ) = F_X \left (\frac {y - 3} {4} \right ).$$ Now take the derivative in both sides with respect to $y$ to find the probability density function of $Y.$ It will be the following $:$ $$f_Y (y) = \begin{cases} \frac {1} {8} (y - 3) & 3 \lt y \lt 7 \\ 0 & \text{otherwise} \end{cases}$$