Question bank — Transformations of random variables — change-of-variable technique

False — the Jacobian is a function of $y$, so it reshapes the density; e.g. for $Y=X^2$ the $\frac{1}{2\sqrt{y}}$ factor is exactly what creates the blow-up near $y=0$, which is shape, not just a constant.

True — $\left|dx/dy\right|=1/|a|$ is constant in $y$, so $f_Y$ is just $f_X$ shifted by $b$, stretched by $|a|$, and scaled in height by $1/|a|$; the family (e.g. normal) is preserved.

False — you need the derivative of the inverse, $\left|dx/dy\right|=1/|g^{\prime }(x)|$, so it should be divided by $|g^{\prime }|$, not multiplied; multiplying by $|g^{\prime }|$ inverts the stretch factor.

False — discrete masses just relabel: $P(Y=g(x))=P(X=x)$ with no stretch factor, because there is no "length" to dilute; the Jacobian is purely a density phenomenon.

True — a monotonic $g$ has exactly one root $x_1=g^{-1}(y)$ for each $y$, so the sum has one term and collapses to $f_X(g^{-1}(y))|dx/dy|$.

True — equal MGFs mean identical distributions, and applying the same deterministic $g$ to identically distributed inputs gives identically distributed outputs.

False — dropping the branch halves the density everywhere, so it integrates to $1/2$, not 1; both symmetric branches pour mass into the same $y$.

True — conservation of probability mass means $\int f_{Y}dy=\int f_{X}dx=1$; a failed normalization check is a reliable signal of a missing Jacobian or branch.

The error is the missing absolute value; it must be $\frac{1}{|a|}$, otherwise for $a<0$ the density comes out negative, which is impossible.

The error is calling $x^3$ non-injective — cubing is strictly increasing over all reals, so there is exactly one real root $x=y^{1/3}$ and only one branch.

Two errors: absolute value is non-monotonic with roots $x=\pm y$, and the Jacobian $|dx/dy|=1$ per branch must multiply each; the correct answer is $f_Y(y)=2f_X(y)$ for $y>0$.

The exponential maps $(-\infty ,\infty )$ onto $(0,\infty )$, so $f_Y(y)=0$ for $y\le 0$; forgetting to restrict the range assigns density where no mass can land.

This one is actually correct — both $\pm \sqrt{y}$ lie in $(-1,1)$, so summing both branches is right; the trap is that it looks like the earlier "one-sided" uniform case where only $+\sqrt{y}$ was valid.

The formula uses the Jacobian of the inverse map $h=g^{-1}$; using the forward one gives the reciprocal area-factor, off by $1/|\det J|$ unless you take the reciprocal.

On the restricted domain $(-\pi /2,\pi /2)$, tan is strictly increasing and one-to-one, so there is a single branch; the periodicity only matters if the support spans multiple periods.

The CDF is always well-defined and monotone even where a density is awkward, and differentiating $F_Y(y)=F_X(g^{-1}(y))$ delivers the Jacobian automatically via the chain rule — no hand-waving about "$dx$" needed.

An integrable singularity: ${\int }_0^1\frac{1}{2\sqrt{y}}dy=[\sqrt{y}{]}_0^1=1$, so infinite height over a vanishingly thin sliver still holds finite mass — squaring compresses many small $x$ into a tiny $y$-neighborhood.

Densities are nonnegative mass-per-length; the sign of $dx/dy$ only records direction of the map, and the inequality flip in the CDF derivation already handles orientation, leaving the magnitude as the true stretch factor.

The mechanism is forced: two symmetric normal branches feed each $y>0$, and the $e^{-x^2/2}$ shape combined with the $\frac{1}{2\sqrt{y}}$ Jacobian produces exactly $\frac{1}{\sqrt{2\pi y}}{e}^{-y/2}$, which is the ${\chi }_1^2$ density by definition.

Because $F(Y)$ is uniformly distributed for any continuous $Y$; running that backwards, feeding a uniform into $F^{-1}$ reconstructs the target distribution exactly — a direct application of change-of-variable with $g^{-1}=F$.

The formula needs a computable inverse $g^{-1}$ and its derivative; if $g$ has no elementary inverse (e.g. $g(x)=x+e^x$), the Jacobian and density exist but cannot be written in closed form.

LOTUS computes $\int g(x)f_X(x)dx$ directly with no inversion or Jacobian; the change-of-variable machinery is only worth invoking when you genuinely need $f_Y$ itself, not a single number.

It breaks down — $g$ has no inverse and $Y$ is degenerate (a point mass at 5), so no density exists; change-of-variable requires $g$ to be non-constant and invertible on the support.

The inverse $x=y^{1/3}$ has infinite $dx/dy$ at $y=0$, so $f_Y$ may spike there; the formula still holds pointwise for $y\ne 0$, and the single measure-zero point does not affect probabilities.

Only $x=-\sqrt{y}$ lies in the support, so despite the square being non-monotonic in general, on this domain it is one-to-one and you use a single branch — always intersect roots with the actual support.

Exactly zero — no $x$ in the support maps there, so no mass can arrive; forgetting this is the most common way a "valid-looking" $f_Y$ fails to integrate to 1.

You sum the point masses of all $x$ mapping to a given $y$: $P(Y=y)={\sum }_{x:g(x)=y}P(X=x)$ — the discrete echo of "sum over branches," but with masses, not densities.

$Y$ becomes a mixed variable: a point mass at $0$ (all $X\le 0$ pile up) plus a continuous density for $y>0$; pure change-of-variable handles only the continuous part, and you treat the atom separately.

Yes — a measure-zero exceptional set carries no probability, so the Jacobian determinant formula holds almost everywhere and the density is unaffected by the excluded curve.