Why do we need to use the jacobian when finding distribution function of variables?

Question

Say we have $X\sim \mathcal{N}(\mu ,\sigma)$ and we are told that $Y=2X$.

So every $X$ we get, we double and that is $Y$. So the density at $y=f_Y(y)$ is equal to the density at $f_X(y/2)$. So we plug in $y/2$ into $f_X()$ and we get $f_Y()$.

I think this makes perfect sense. why do we need to multiply it by the Jacobian?

Answer 1

In a formal argument, $$ F_Y(y) = \mathbb{P}[Y \le y] = \mathbb{P}[2X \le y] = \mathbb{P}[X\le y/2] = F_X(y/2), $$ and differentiation yields $$ f_Y(y) = F'_Y(y) = \frac{d}{dy} \left[ F_X(y/2) \right] = f_X(y/2) \frac{d}{dy} \left[y/2\right] = \frac{f_X(y/2)}{2}. $$

Answer 2

You need to have $f_Y$ such that $\int_\Bbb R f_Y(y)\mathsf d y=1$

If you just substitute $f_X(y/2)$ for $f_Y(y)$ you'd get $\int_\Bbb R f_Y(y)\mathsf d y~{=\int_\Bbb R f_X(y/2)\mathsf d y \\= \int_\Bbb R f_X(x)\mathsf d(2x) \\= 2}$

Answer 3

Think back to integral calculus and recall working out the integral $$ \int f(x)\,dx\tag1 $$ using the substitution $x=g(y)$. It's wrong to plug $x=g(y)$ into $f(x)$ alone, obtaining $$ \int f(g(y))\,dy; $$ you need to also replace $dx$ by $g'(y)\,dy$, so that the integral becomes $$ \int f(g(y))\,g'(y)\,dy.\tag2 $$ The $g'(y)$ bit is essentially the Jacobian. It needs to be present in order for the integration to yield the same result before and after the change of variables. The change of variables represents a distortion of the original coordinates and the Jacobian is a measure of how much a unit volume is distorted by the transformation. The $g'(y)$ piece plays the same role in probability when obtaining the density of a random variable $Y$ such that $X=g(Y)$.