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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.

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  • $\begingroup$ Thank you for your response! I see the formula for how to do this in the "Scalar to Scalar" section. I guess that I really struggle with using the formulas in practice without seeing an example first. (such as, why are there absolute value around the derivative of the inverse function?) I will continue to work on it, thanks again. $\endgroup$
    – katie
    Jan 18, 2021 at 7:26
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    $\begingroup$ The absolute value is necessary for transforms that have a negative derivative, like $Y=-X+2$. Otherwise the resulting density would be negative. $\endgroup$
    – Xi'an
    Jan 18, 2021 at 7:28
  • $\begingroup$ $$f_Y(y)=\frac{\text d}{\text dy}F_Y(y)=\frac{\text d}{\text dy}F_X((y+1)/2)=\frac12 f_X((y+1)/2)$$ $\endgroup$
    – Xi'an
    Jan 18, 2021 at 7:38
  • $\begingroup$ Thank you!! I am going to ask one more really obvious question (apologies), but for the part that involves plugging g(y)'s inverse into x's pdf, does that mean simply plugging it in to get 1-y? Or do I need to differentiate or integrate in any way? This way gets me a final answer of (1-y)/2, but this method gives me an answer that has the term y^2 in it. I'm really sorry I know that this is supposed to be very obvious and easy... $\endgroup$
    – katie
    Jan 18, 2021 at 7:41
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    $\begingroup$ The link refers to another transform $g$ where $g(x)=x^2$ and $g^{-1}(x)=\sqrt x$ so it is logical that you find other functions. The formula (2) below applies to all $g$'s that are invertible and differentiable but the values of the terms in (2) depend on the context. $\endgroup$
    – Xi'an
    Jan 18, 2021 at 9:37

2 Answers 2

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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.
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  • $\begingroup$ Thank you so much for your help. Infinite gratitude to you. $\endgroup$
    – katie
    Jan 18, 2021 at 9:55
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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)

  1. 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.
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  • $\begingroup$ Thank you for the response! I am still a bit confused about how the math works, but you have given me a nice way of visualizing the problem. $\endgroup$
    – katie
    Jan 18, 2021 at 3:33

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