Finding the pdf of Y from that of X, linear transformation The question is

Let $X$ be a continuous random variable with pdf $f_X(x) = 2(1 − x)$, $0 ≤ x ≤ 1$. If $Y = 2X − 1$, find the pdf of $Y$.

I understand these steps$$F_Y(Y ≤ y) = P(2X-1 ≤ y) = P(X ≤ (y+1)/2) = F_X((y+1)/2)$$
I do not understand how to get the pdf of $Y$ from this. I know that we are supposed to differentiate both sides with respect to $y$, but I do not understand what that means.
 A: There are several standard approaches for deriving the density of a transform $g(X)$ of a random variable, including:

*

*the "push-forward" technique, when looking at
$$\int_A f_Y(y)\text dy=\mathbb P(g(X)\in A)=\mathbb P(X\in g^{-1}(A))=\int_{g^{-1}(A)} f_X(x)\text dx$$
for a generic (measurable) set $A$ and identifying $f_Y$ (this technique applies even when $g$ is not invertible)

*the cdf technique, which is a special case of the above (when $g$ is invertible and increasing):
$$F_Y(y)=\mathbb P(g(X)\le y)=\mathbb P(x\le g^{-1}(y))\tag{1}$$
and taking the derivative of $F_Y$ to find the density
$$\dfrac{\text d}{\text dy}F_Y(y)=f_Y(y)$$

*the "mute function" technique, of which 1. is a special case, where $f_Y$ is identified by
$$\mathbb E^Y[h(Y)]=\mathbb E^X[h(g(X))]=\int_\mathcal X (h\circ g)(x)f_X(x)\text dx=\int_\mathcal Y h(y)f_Y(y)\text dy$$
(this technique applies even when $g$ is not invertible)

*the "Jacobian formula", which is a consequence of 2. and only applies when $g$ is invertible and differentiable
$$f_Y(y)=f_X(g^{-1}(y))\left| \dfrac{\text d g^{-1}}{\text d x}(y)\right|\tag{2}$$

*the moment generating approach, assuming $\mathbb E^X[\exp\{t g(X)\}]$ exists for an open interval of $t$'s, which is a particular case of 3., with
$$\mathbb E^X[\exp\{t g(X)\}]=\int_\mathcal X e^{tg(x)}f_X(x)\text dx\varphi(t)$$
returning a function of $t$ that identifies uniquely the distribution of $Y$ and hence its density.

A: let me propose an intuitive approach for linear transformation
let's treat first scale and then shift.

*

*scale:



*

*what values Y = scaled random variable X,  can get? 
in this case, Y = 2X, X goes from 0 to 1 so Y will get values from 0 to 2.


*how the distribution of Y will look? 
Y proportional to X, its probability function will be of the same form, just stretched 
(straight line, stretched to the boundaries of the support)


*what are the probability values? 
straight line, going from point (y=0,P=M) to (y=2,P=0) 
remember that the area under it should be 1 
M then will be 1 and the line equation is P(y) = 0.5*(2-y)


*shift:



*

*what are new possible values for Y? 
now, instead of lets say 0, Y will get -1, instead of 2, will be 1. 
Support: y = [-1,1]

*how this affects the PDF? 
same as the support, the pdf should be shifted to the left by 1: 
P(y) <- P(y+1) = 0.5*(2-(y+1)) 
P(y) = 0.5*(1-y) for y between -1 and 1.

