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Let $(X,Y)$ be a pair of random variables with joint pdf $f_{XY}$. Let $(U,V)$ be two random variables obtained from $(X,Y)$ by $U = u(X,Y)$ and $V = v(X,Y)$ where $u$ and $v$ are, say, nice invertible functions. Then there is a method to compute the pdf of $(U,V)$ using Jacobians, as stated, for instance, as Theorem 2.9 here: https://www.utstat.toronto.edu/mikevans/jeffrosenthal/book.pdf

I have a confusion when applying this method and haven't been able to figure out what I am missing. Obviously something very basic.

Suppose in $(X,Y)$, the second random variable $Y=X$ and is thus "spurious"; so the pdf of $f_{XY}$ is simply the pdf $f_X$. More precisely, $f_{XY}(x,y) = 0$ when $x\neq y$ and equal to $f_X(x)$ when $x=y$.

Now suppose $(U,V)$ is defined as $(aX, aY)$ for some constant $a > 0$. Once again, the "$V$" is spurious, and so the pdf of $(U,V)$ should simply be the pdf of $aX$; that is $f_{UV}(x,y) = 0$ when $x\neq y$ and $f_{UV}(x) = \frac{1}{a}\cdot f_X(\frac{x}{a})$ when $x=y$.

However, if we look at the Jacobian approach, then we get $f_{UV}(x,y) = \frac{1}{a^2}f_{XY}(x/a, y/a)$ where we got the $\frac{1}{a^2}$ as the determinant of the $2\times 2$ diagonal matrix with entries $1/a$ on the diagonal.

What am I missing in the application of the "Jacobian method"?

A clarification would be greatly appreciated. Thanks!

PS: This question arose when I was trying to figure out the pdf of a "scaled" Dirichlet distrubution. More precisely, say $(X_1, \ldots, X_k) \sim \mathrm{Dir}(\alpha_1, \ldots, \alpha_k)$ be a vector on the $k$-dimensional simplex, what is the pdf of $(aX_1, aX_2, \ldots, aX_k)$ for some scalar $a>0$. My answer and what I believe is true is off by a factor $1/a$...and I think this is because the sum of the $X_i$'s is $1$ (which means $X_k$ is determined by the others). This led to the question above. Thanks!

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    $\begingroup$ (1) In your first "suppose," $(X,Y)$ does not have a pdf. (2) In the second "suppose," the $V$ is not "spurious" at all. When you multiply all the random values by the same positive constant, you are merely changing their units of measurement. You are not otherwise changing the shape or the general characteristics of the distribution. $\endgroup$
    – whuber
    Commented Jul 26, 2023 at 21:18
  • $\begingroup$ Thank you! @dimitriy: the pair $(X,Y)$ in my mind is formed by sampling $X$ and then duplicating it. That is what I meant by $Y=X$. So, seeing $(3,6)$ is indeed a $0$ probability event. For simplicity let's say $X \sim \mathrm{Unif}(0,1)$. And as far as I understand, the joint distribution of this pair is given by $f(x,y) = 0$ if $x\neq y$ and $f(x,y) = 1$ for $x = y$ and $0\leq x\leq 1$. Isn't that correct, @whuber? Why doesn't $(X,Y)$ have a pdf? $\endgroup$
    – DeepC
    Commented Jul 27, 2023 at 1:45
  • $\begingroup$ I think I understand now, so deleted my previous comment. Per @whuber’s comment, the case of measuring shares that sum to 1 and shares’ that sum to 100 is illuminating. $\endgroup$
    – dimitriy
    Commented Jul 27, 2023 at 2:30
  • $\begingroup$ $(X,Y)$ has no pdf because there exists no function which, when integrated, gives the correct probabilities. This is immediate from the fact that the support of $(X,Y)$ has measure zero. Of course $(X,Y)$ has a cumulative distribution function and it can be represented by a measure, but that measure is singular with respect to Lebesgue measure. $\endgroup$
    – whuber
    Commented Jul 27, 2023 at 12:37
  • $\begingroup$ Thank you @whuber! FWIW, this link math.stackexchange.com/questions/2910315/… considered the exact same thing. This indeed resolves my "silly" question, and also the question in my "PS". I am new-ish to stackexchange; should I be moving @whuber's comment as an "answer". In any case, many thanks! $\endgroup$
    – DeepC
    Commented Jul 27, 2023 at 15:55

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