Find $[E(Y^2)]$. When $Y = 3 * X - 5$ and $X$ is distributed in range $[0, 5]$

Question

Need help with exercise.

Random variable $X$ is evenly distributed in range $[0, 5]$. Need to find $E[Y^2]$ when $Y = 3X - 5$

Every hint/tip will be appreciated.

Thank you

Alternative solution

$$E[Y^2] = E[(3X-5)^2] = E[9X^2-30X+25] = 9E[X^2]-30E[X]+25 =9(E[X]^2 + D[X]) - 30 *2.5 +25 =9*(2.5^2 + 25/12) - 75+25 = 25$$

Answer 1

I assume that $X$ is uniformly distributed. The CDF of $X\sim\mathcal{U}(0,5)$ is $$ F_X(x)=\Pr[X\le x]=\frac{x-0}{5-0}=\frac x5. $$ Therefore $$ \begin{align} \Pr[Y\le y]&=\Pr[3X-5\le y]\\ F_Y(y)&=\Pr[3X\le y+5]\\ &=\Pr\left[X\le \frac{y+5}{3}\right]\\ &=F_X\left(\frac{y+5}{3}\right)\\ &=\frac{y+5}{15}. \end{align} $$ The region $0\le x<5$ is corresponding to $-5\le y<10$. Hence $Y\sim\mathcal{U}(-5,10)$. The pdf of $Y$ is $$ f_Y(y)=\frac1{10-(-5)}=\frac1{15}, $$ then $$ \begin{align} \text{E}\left[Y^2\right]&=\int_{-5}^{10} y^2f_Y(y)\ dy\\ &=\frac1{15}\int_{-5}^{10} y^2\ dy\\ &=\frac1{15}\left[\frac13y^3\right]_{-5}^{10}\\ &=\frac1{45}(10^3-(-5)^3)\\ &=\Large\color{blue}{25}. \end{align} $$

Answer 2

It should be evident that $Y$ is uniformly distributed on $[-5,10]$ since $X$ is uniform on $[0,5]$, hence $3X$ is uniform on $[0,15]$ and thus $3X-5$ is uniform on $[-5,10]$.

Therefore $$EY^{2}=\frac{1}{15}\int_{-5}^{10}y^{2}\;dy=\Bigg[\frac{1}{15}\frac{y^{3}}{3}\Bigg]_{y=-5}^{y=10}=\frac{1}{45}(1000+125)=25.$$

Other methods:

(1) Compute directly the distribution of $Y$ or $Y^{2}$ as in the other answer.

(2) Apply the measure-theoretic change of variable formula as in the comment.