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X is a random variable with the density

$$F_X(x) = \begin{cases}2 x^{-2}, & x \in (1,2)\\ 0 & \operatorname{otherwise}\end{cases}$$

How can I find the density of $ Y= 0.5 \, X - 1$?

I thought about transforming it into

$X =2Y+2$

but I do not know how to evaluate it from there. Should I just put it into integral bounds? But from the formula $\int_{-\infty}^Xf(x)$ i get distribution not the density.

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$\mathbb{P}(Y\leq y)=\mathbb{P}(X\leq 2y+2)$

If $2y+2\geq2$, i.e. $y\geq 0$ then $f_Y(y)=0$ by differentiation

If $2y+2\leq1$, i.e. $y\leq -\frac{1}{2}$, then again $f_Y(y)=0$

For $1<2y+2<2$ we have that $\mathbb{P}(X\leq 2y+2)=\int_1^{2y+2}\frac{2}{x^2}dx=\frac{2y+1}{y+1}$

Differentiate this (with respect to $y$) to geth that pdf for $y\in (-\frac{1}{2}, 0)$

In general, differentiating the cdf gives the pdf

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By the fundamental theorem of calulus: $\tfrac{\mathsf d~~}{\mathsf d y}\int_c^{g(y)}f(x)\mathsf d x=g'(y)\cdot f(g(y))$.

So you just need to use the chain rule to change the variable - or simply notice that it is merely a linear change of scale and a shift.   $Y$ will be twice as dense as $X$, and its support bounds will be transformed by the given formulae.

$$\begin{align}f_X(x) &=2 x^{-2}~\mathbf 1_{x\in(1;2)}\\ f_Y(y)&= 2(2y+2)^{-2}\lvert\tfrac{\mathsf d ~}{\mathsf dy}(2y+2)\rvert~\mathbf 1_{2y+2\in{(1;2)}}\\ &= (y+1)^{-2}~\mathbf 1_{y\in(-1/2;0)}\end{align}$$

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