Finding the pdf of Y from that of X, linear transformation

Question

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.

Answer 1

There are several standard approaches for deriving the density of a transform $g(X)$ of a random variable, including:

1. 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)
2. 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 $$\dfrac{\text d}{\text dy}F_Y(y)=f_Y(y)$$
3. 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)
4. the "Jacobian formula", which is a consequence of 2. and only applies when $g$ is invertible and differentiable $$f_Y(y)=f_X(g^{-1}(y))\left| \dfrac{\text d g^{-1}}{\text d x}(y)\right|\tag{2}$$
5. 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.

Answer 2

let me propose an intuitive approach for linear transformation

let's treat first scale and then shift.

1. 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)

2. 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.