"Let $X$ be a random variable with density function $f_X$ , and let $Y=X+b$, $Z=aX$, and $W=aX+b$, where $a≠0$. Find the density functions $f_Y , f_Z$, and $f_W$"
In general if $U=g(X)$ where $g$ is a bijection mapping the support of $X$ to the support of $U$ we apply a change of variables transformation:
$$ f_U(u) = f_X(g^{-1}(u))\cdot\left\lvert\frac{\mathrm d g^{-1}(u)}{\mathrm d u}\right\rvert$$
[Note: the modulus is because density is an unsigned measure of the gradient of the CDF.]
But for these three, it is just a linear scale and shift transformation:
$Y$ is $X$ shifted by $b$. So the density of $Y$ at $y$ is that of $X$ at $x-b$.
$$f_Y(y) = f_X(y-b)$$
$Z$ is $X$ scaled by $a$. So the density of $Z$ at $z$ is $\lvert 1/a\rvert$ times the density of $X$ at $z/a$. $$f_Z(z) = \tfrac 1 {\lvert a\rvert}~f_X(\tfrac z a)$$
$W$ is $X$ scaled by $a$ and shifted by $b$. So...
$$f_W(w) = \tfrac 1 {\lvert a\rvert}\cdot f_X\left(\tfrac{w-b}{a}\right)$$
NB: as stated, the density function must be non-negative.
"Let $X$ be a random variable with cumulative distribution function $F_X$, and blah blah blagh"
$$F_Y(y) = \Pr(X+b\leq y) = F_X(y-b)$$
$$F_Z(z) = \Pr(aX\leq z) = \begin{cases}F_X(z/a) & : a>0 \\ 1-F_X(z/a) & : a< 0 \\ \mathbf 1_{0\leq z} & : a=0\end{cases}$$
$$F_W(w) = ....$$
You are given "$X$ is uniformly distributed on $[0,2]^2$", so joint PDF is
$$f_{X_1,X_2}(x_1,x_2)=\frac{1}{4}$$
inside $[0,2]^2$, zero outside.
Now,
$$f_{X_1}(x_1)=\int f_{X_1,X_2}(x_1,x_2) dx_2=\int_0^2 \frac{1}{4} dx_2=\frac{1}{2}$$
inside $[0,2]$, zero outside. The same is true for $f_{X_2}(x_2)$.
Now we see that
$$f_{X_1,X_2}(x_1,x_2)=f_{X_1}(x_1)f_{X_2}(x_2)$$
so $X_1$ and $X_2$ are independent, and the PDF of $Y=X_1+X_2$ is convolution of $f_{X_1}(x_1)$ and $f_{X_2}(x_2)$.
Best Answer
$$P(Y \leq y)=P(X^{2} \geq \frac 1y -1)$$ $$=2P(X \geq \sqrt {\frac 1 y -1})$$ $$=2\int_{\sqrt {\frac 1 y -1}}^{\infty} \frac 1 {\pi} \frac 1 {1+t^{2}} dt.$$ Now just differentiate w.r.t. $y$.