Statistics Graduate Level Quiz: Transformations And Jacobians

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Transformations And JacobiansQuestion 1 of 10

Let $X$ and $Y$ be independent standard exponential random variables. Define $U=X+Y$ and $V=X/(X+Y)$ . Which expression gives the joint density of $(U,V)$ ?

f_{U,V}(u,v)=e^{-u},\quad u>0,\ 0<v<1

f_{U,V}(u,v)=u e^{-u},\quad u>0,\ 0<v<1

f_{U,V}(u,v)=v(1-v)e^{-u},\quad u>0,\ 0<v<1

f_{U,V}(u,v)=u^{-1}e^{-u},\quad u>0,\ 0<v<1

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Transformations And Jacobians

Practice Transformations And Jacobians in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Transformations And Jacobians, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let $X$ and $Y$ be independent standard exponential random variables. Define $U=X+Y$ and $V=X/(X+Y)$ . Which expression gives the joint density of $(U,V)$ ?

$f_{U,V}(u,v)=e^{-u},\quad u>0,\ 0<v<1$
$f_{U,V}(u,v)=u e^{-u},\quad u>0,\ 0<v<1$ (correct answer)
$f_{U,V}(u,v)=v(1-v)e^{-u},\quad u>0,\ 0<v<1$
$f_{U,V}(u,v)=u^{-1}e^{-u},\quad u>0,\ 0<v<1$

Explanation: When you encounter a transformation problem like this, your first instinct should be the change-of-variables technique: express the original densities in terms of the new variables, then multiply by the absolute Jacobian determinant. Here,

X

and

Y

are independent standard exponentials, so

f_{X,Y}(x,y) = e^{-x}e^{-y} = e^{-(x+y)}

for

x,y > 0

. The transformation is

U = X+Y

and

V = X/(X+Y)

, which inverts to

X = UV

and

Y = U(1-V)

. The Jacobian is:

J = \begin{vmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{vmatrix} = \begin{vmatrix} v & u \\ 1-v & -u \end{vmatrix} = -uv - u(1-v) = -u

|J| = u

. Substituting into the joint density:

f_{U,V}(u,v) = e^{-uv} \cdot e^{-u(1-v)} \cdot u = u e^{-u}

for

u > 0,\ 0 < v < 1

. This confirms answer B is correct, and notice that

U

and

V

are independent —

U \sim \text{Gamma}(2,1)

and

V \sim \text{Uniform}(0,1)

. Choice A forgets the Jacobian entirely, writing just

e^{-u}

— a classic omission error. Choice C introduces

v(1-v)

, which has no basis in the calculation; this might reflect confusion with a Beta density. Choice D uses

u^{-1}e^{-u}

, which corresponds to a Gamma(0,1) — a non-normalizable, nonsensical density. As a strategy, always write out the inverse transformation first, compute the Jacobian explicitly, and verify your result integrates to 1. The Jacobian step is the most commonly dropped piece under exam pressure.

Question 2

Let $X$ and $Y$ be independent standard normal random variables, and define $U=X^2+Y^2$ and $V=X^2/(X^2+Y^2)$ . What is the joint density of $(U,V)$ ?

$\frac{e^{-u/2}}{4\pi\sqrt{v(1-v)}},\quad u>0,\ 0<v<1$
$\frac{u\, e^{-u/2}}{2\pi\sqrt{v(1-v)}},\quad u>0,\ 0<v<1$
$\frac{e^{-u/2}}{2\pi\sqrt{v(1-v)}},\quad u>0,\ 0<v<1$ (correct answer)
$\frac{e^{-u/2}}{2\pi\, v(1-v)},\quad u>0,\ 0<v<1$

Explanation: When you see a transformation problem like this, your first instinct should be the change-of-variables technique: express the original variables in terms of the new ones, compute the Jacobian, and multiply by the original joint density. Here,

X

and

Y

are independent standard normals, so their joint density is

f_{X,Y}(x,y) = \frac{1}{2\pi}e^{-(x^2+y^2)/2}

. Now invert the transformation. From

U = X^2+Y^2

and

V = X^2/(X^2+Y^2)

, you get

X^2 = UV

and

Y^2 = U(1-V)

. Each combination of signs for

(x,y)

gives one branch, and there are four such branches, each contributing equally. For one branch, the Jacobian of

(x,y)\mapsto(u,v)

satisfies

|J|^{-1} = \frac{1}{4\sqrt{uv}\cdot\sqrt{u(1-v)}} \cdot 2u

, which simplifies so that the combined contribution from all four branches yields:

f_{U,V}(u,v) = \frac{e^{-u/2}}{2\pi\sqrt{v(1-v)}}, \quad u>0,\ 0<v<1

This is choice C. Notice this factors into a

\text{Gamma}(1, 1/2)

(i.e., Exponential) density in

u

and an

\text{arcsine}(0,1)

density in

v

, confirming

U

and

V

are independent. Choice A is missing the factor of 2 in the denominator — a Jacobian arithmetic error. Choice B introduces an extra

u

in the numerator, which would correspond to a

\chi^2_4

rather than

\chi^2_2

marginal. Choice D replaces

\sqrt{v(1-v)}

with

v(1-v)

, reflecting a failure to take the square root when computing the Jacobian. Strategy tip: Always track all branches of the inverse transformation — missing sign combinations is the most common Jacobian mistake on graduate-level transformation problems.

Question 3

Let $X$ have a standard exponential distribution and let $\Theta$ be independent and uniform on $[0,2\pi)$ . Define $U=e^X\cos\Theta$ and $V=e^X\sin\Theta$ . Writing $r=\sqrt{u^2+v^2}$ , which is the joint density of $(U,V)$ ?

$f_{U,V}(u,v)=\frac{1}{2\pi r^2},\quad r>1$
$f_{U,V}(u,v)=\frac{1}{2\pi r},\quad r>1$
$f_{U,V}(u,v)=\frac{1}{2\pi r^3},\quad 0<r<1$
$f_{U,V}(u,v)=\frac{1}{2\pi r^3},\quad r>1$ (correct answer)

Explanation: When you see a transformation from polar-like coordinates to Cartesian coordinates involving exponentials, your first instinct should be to apply the change-of-variables formula carefully, tracking both the Jacobian and the support. Here, set

R = e^X

, so since

X \sim \text{Exp}(1)

, we have

X = \ln R

, and

dX/dR = 1/R

. Since

X > 0

, we get

R > 1

. The joint density of

(R, \Theta)

f_{R,\Theta}(r,\theta) = e^{-\ln r} \cdot \frac{1}{2\pi} \cdot \frac{1}{r} = \frac{1}{2\pi r^2}

for

r > 1

\theta \in [0, 2\pi)

. Now transform to

(U, V) = (R\cos\Theta, R\sin\Theta)

, a standard polar-to-Cartesian map. The Jacobian of the inverse transformation gives

|J| = 1/r

, so:

f_{U,V}(u,v) = \frac{1}{2\pi r^2} \cdot \frac{1}{r} = \frac{1}{2\pi r^3}, \quad r > 1.

This confirms D is correct. A uses

1/(2\pi r^2)

, which is the density of

(R,\Theta)

before the polar-to-Cartesian Jacobian is applied — a classic step-skipping error. B gives

1/(2\pi r)

, suggesting only one Jacobian factor was used instead of both (forgetting the exponential change-of-variables). C gets the density formula right but places the support at

0 < r < 1

, ignoring that

X > 0

forces

R = e^X > 1

— a critical support error. As a study tip: always decompose multi-step transformations sequentially, tracking the support and applying each Jacobian factor explicitly. Support errors and missing Jacobian terms are the two most common traps in transformation problems.

Question 4

Let $(X,Y)$ have a centered bivariate normal distribution with unit marginal variances and correlation $\rho$ , where $|\rho|<1$ . Define the polar angle $\Theta=\operatorname{atan2}(Y,X)$ with $0\leq\Theta<2\pi$ . Which density does $\Theta$ have?

$f_\Theta(\theta)=\frac{1-\rho^2}{2\pi\left(1-\rho\sin 2\theta\right)},\quad 0\leq\theta<2\pi$
$f_\Theta(\theta)=\frac{\sqrt{1-\rho^2}}{2\pi\left(1-\rho\cos 2\theta\right)},\quad 0\leq\theta<2\pi$
$f_\Theta(\theta)=\frac{1}{2\pi\sqrt{1-\rho^2}}\left(1-\rho\sin 2\theta\right),\quad 0\leq\theta<2\pi$
$f_\Theta(\theta)=\frac{\sqrt{1-\rho^2}}{2\pi\left(1-\rho\sin 2\theta\right)},\quad 0\leq\theta<2\pi$ (correct answer)

Explanation: When you encounter a transformation of a bivariate normal vector, your instinct should be to marginalize out the radial component using the joint density in polar coordinates. Start with the bivariate normal density:

f(x,y) = \frac{1}{2\pi\sqrt{1-\rho^2}}\exp\!\left(-\frac{x^2 - 2\rho xy + y^2}{2(1-\rho^2)}\right).

Substitute

x = r\cos\theta,\, y = r\sin\theta

and include the Jacobian

r

, then integrate out

r \in [0,\infty)

. The quadratic form in the exponent becomes

r^2(1 - \rho\sin 2\theta)/(1-\rho^2)

, so the integral over

r

is a Gaussian integral:

\int_0^\infty r\exp\!\left(-\frac{r^2(1-\rho\sin 2\theta)}{2(1-\rho^2)}\right)dr = \frac{1-\rho^2}{1-\rho\sin 2\theta}.

Multiplying by the prefactor

\frac{1}{2\pi\sqrt{1-\rho^2}}

yields

f_\Theta(\theta) = \frac{\sqrt{1-\rho^2}}{2\pi(1-\rho\sin 2\theta)},

which is answer D. Choice A omits the square root, writing

1-\rho^2

in the numerator instead of

\sqrt{1-\rho^2}

—this fails the normalization check. Choice B replaces

\sin 2\theta

with

\cos 2\theta

; the cross-term

2\rho xy = 2\rho r^2\cos\theta\sin\theta = \rho r^2\sin 2\theta

, so the angular dependence must involve

\sin 2\theta

, not

\cos 2\theta

. Choice C inverts the structure entirely, placing

(1-\rho\sin 2\theta)

in the numerator, which cannot be a valid density since it grows unboundedly with

\rho

near certain angles. The key study tip: always verify a candidate density by checking that it integrates to 1 and that the trigonometric term matches the cross-term

2\rho xy = \rho r^2\sin 2\theta

from the exponent—this immediately rules out two of the four choices.

Question 5

A random positive-definite symmetric matrix is represented by $M=\begin{pmatrix}A&C\\C&B\end{pmatrix}$ and has a density $g(a,b,c)$ with respect to $da\,db\,dc$ . Define $S=A$ , $T=B$ , and $R=C/\sqrt{AB}$ . What is the joint density of $(S,T,R)$ ?

$g\!\left(s,t,\,r\sqrt{st}\right)(st)^{-1/2},\quad s>0,\ t>0,\ |r|<1$
$g\!\left(s,t,\,r\sqrt{st}\right)(st)^{1/2},\quad s>0,\ t>0,\ |r|<1$ (correct answer)
$g\!\left(s,t,\,rst\right)\cdot st,\quad s>0,\ t>0,\ |r|<1$
$g\!\left(s,t,\,r\sqrt{st}\right)(st)^{1/2},\quad s>0,\ t>0,\ r\in\mathbb{R}$

Explanation: When you see a transformation of random variables problem like this, your instinct should be: find the Jacobian. The joint density of the new variables equals the original density evaluated at the inverse transformation, multiplied by the absolute value of the Jacobian determinant. Here, the transformation is

S = A,\ T = B,\ R = C/\sqrt{AB}

. To find the joint density of

(S, T, R)

, first invert:

A = S,\ B = T,\ C = R\sqrt{ST}

. The original density becomes

g(s, t, r\sqrt{st})

. Now compute the Jacobian

\left|\partial(a,b,c)/\partial(s,t,r)\right|

. Since

a = s

and

b = t

transform trivially, the Jacobian matrix is essentially triangular, and the only nontrivial entry is

\partial c/\partial r = \sqrt{st}

. The full Jacobian determinant is therefore

\sqrt{st} = (st)^{1/2}

. Multiplying gives the joint density

g(s, t, r\sqrt{st})\cdot(st)^{1/2}

, confirming answer B. A is wrong because it uses

(st)^{-1/2}

— this would be the Jacobian of the inverse transformation, i.e., inverting the roles of old and new variables. A common trap is flipping the Jacobian direction. C is wrong on two counts: the argument

rst

treats the transformation as

C = RST

rather than

C = R\sqrt{ST}

, and the factor

st

doesn't match the correct Jacobian. D uses the correct density form and Jacobian, but allows

r \in \mathbb{R}

. Since

M

is positive definite, its determinant

AB - C^2 > 0

, which forces

|R| = |C|/\sqrt{AB} < 1

. Study tip: Always track the support of transformed variables — positive definiteness constraints often restrict correlation-like parameters to

|r| < 1

, and forgetting this is a classic exam mistake.

Question 6

Let $X\sim\operatorname{Gamma}(a,\lambda)$ and $Y\sim\operatorname{Gamma}(b,\mu)$ be independent, where the second argument is a rate. Define $U=X+Y$ and $V=X/(X+Y)$ . Under which condition are $U$ and $V$ independent?

They are independent exactly when $a=b$ , regardless of the two rate parameters
They are independent exactly when $\lambda=\mu$ , regardless of the two shape parameters (correct answer)
They are independent exactly when $a+b=1$ and the rates are both positive
They are independent for all positive shapes and rates because the support is rectangular

Explanation: Whenever you see a question pairing a sum and a ratio of independent Gamma random variables, you should immediately think of the Beta-Gamma algebra and the classical result about independence. The key theorem: if

X \sim \text{Gamma}(a, \lambda)

and

Y \sim \text{Gamma}(b, \lambda)

share the same rate, then

U = X + Y

and

V = X/(X+Y)

are independent, with

U \sim \text{Gamma}(a+b, \lambda)

and

V \sim \text{Beta}(a, b)

. The proof uses the Jacobian of the transformation and shows the joint density factors into a product of marginals — but only when the rates are equal, allowing the exponential terms to combine cleanly into a function of

u

alone. When

\lambda \neq \mu

, the joint density of

(U, V)

does not factor, so independence fails. This confirms B is correct: independence holds exactly when

\lambda = \mu

, for any positive shapes

a

and

b

. A is wrong because matching shape parameters is irrelevant — equal shapes with unequal rates still breaks independence. The shape parameters determine the Beta marginal of

V

, not the factorization condition. C is wrong because

a + b = 1

has no special role here; this confuses the Gamma parameterization with a probability-simplex constraint. D is wrong on two counts: a rectangular support does not imply independence (that's a common misconception), and independence genuinely fails when rates differ. Study tip: Memorize the equal-rate condition as the price of admission for the Gamma-Beta independence result — it appears frequently on graduate qualifying exams and is easy to misremember as a shape condition.

Question 7

Let $X$ and $Y$ be independent standard normal random variables. Define $U=XY$ and $V=X/Y$ , ignoring the probability-zero event $Y=0$ . Which expression gives the joint density of $(U,V)$ ?

$\frac{1}{4\pi|v|}\exp\!\left[-\frac{1}{2}\left(uv+\frac{u}{v}\right)\right],\quad uv>0$
$\frac{1}{2\pi|u|}\exp\!\left[-\frac{1}{2}\left(uv+\frac{u}{v}\right)\right],\quad uv>0$
$\frac{1}{2\pi|v|}\exp\!\left[-\frac{1}{2}\left(uv+\frac{u}{v}\right)\right],\quad uv>0$ (correct answer)
$\frac{1}{2\pi|v|}\exp\!\left[-\frac{1}{2}\left(uv-\frac{u}{v}\right)\right],\quad uv<0$

Explanation: When you need the joint density of a transformation like

(U,V) = (XY, X/Y)

, your go-to tool is the change-of-variables formula:

f_{U,V}(u,v) = f_{X,Y}(x,y)\cdot|J|^{-1}

, where you invert the Jacobian determinant from

(x,y)

(u,v)

. Start by inverting the transformation. From

u = xy

and

v = x/y

, you get

x^2 = uv

and

y^2 = u/v

, so

x = \pm\sqrt{uv}

and

y = \pm\sqrt{u/v}

. For real solutions, you need

uv > 0

(both same sign). There are two solution branches, each contributing equally. Computing the Jacobian

\partial(u,v)/\partial(x,y)

: the determinant is

2x/y + 2y \cdot x/y^2 = 2x(1/y + 1/y) \cdots

— more cleanly,

|J_{x,y\to u,v}| = 2|x/y| = 2|v|^{1/2}/|v|^{1/2}\cdot\ldots

. The standard result gives

|\partial(x,y)/\partial(u,v)| = \frac{1}{2|v|}

. Since

X,Y

are independent standard normals,

f_{X,Y} = \frac{1}{2\pi}e^{-(x^2+y^2)/2}

. Note

x^2 + y^2 = uv + u/v

. Multiplying the two branches and the Jacobian factor, the joint density is:

f_{U,V}(u,v) = \frac{1}{2\pi|v|}\exp\!\left[-\frac{1}{2}\!\left(uv + \frac{u}{v}\right)\right], \quad uv > 0

confirming C is correct. A uses

4\pi|v|

in the denominator — this mishandles the factor of 2 from summing both branches. B replaces

|v|

with

|u|

in the denominator, misidentifying which variable appears in the Jacobian. D has the correct structure but uses

uv - u/v

in the exponent with

uv < 0

, which doesn't match the exponent

x^2 + y^2 = uv + u/v > 0

required for a valid density. Strategy tip: Always track the Jacobian factor carefully — most wrong answers in transformation problems differ only in the normalizing constant or the Jacobian variable, so computing

|\partial(x,y)/\partial(u,v)|

explicitly (rather than guessing) is essential.

Question 8

A random pair $(X,Y)$ has joint density $f_{X,Y}(x,y)=2$ on the region $0<x<y<1$ and zero elsewhere. Define $U=X/Y$ and $V=Y$ . Which statement about $(U,V)$ is correct?

$f_{U,V}(u,v)=2v$ on $0<u<1,\ 0<v<1$ , so $U$ and $V$ are independent (correct answer)
$f_{U,V}(u,v)=2u$ on $0<u<v<1$ , so $U$ and $V$ are dependent
$f_{U,V}(u,v)=2/v$ on $0<u<1,\ 0<v<1$ , so the proposed density is improper
$f_{U,V}(u,v)=2$ on $0<u<v<1$ , so $U$ and $V$ are conditionally uniform

Explanation: When you see a change-of-variables problem, your first instinct should be to find the Jacobian and determine the new support region — both steps are equally important. Here, set

U = X/Y

and

V = Y

, so the inverse transformation is

X = UV

and

Y = V

. The Jacobian is:

\left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \begin{vmatrix} v & u \\ 0 & 1 \end{vmatrix} = v

So the joint density transforms as

f_{U,V}(u,v) = f_{X,Y}(uv, v) \cdot |J| = 2 \cdot v = 2v

. Now check the support. The original region

0 < x < y < 1

becomes

0 < uv < v < 1

. Since

v > 0

, dividing

uv < v

v

gives

u < 1

, and from

uv > 0

you get

u > 0

. Combined with

0 < v < 1

, the new support is simply

0 < u < 1,\ 0 < v < 1

— a rectangle. Since

f_{U,V}(u,v) = 2v

factors into a function of

u

alone times a function of

v

alone (trivially,

1 \cdot 2v

) over a rectangular support,

U

and

V

are independent. This confirms answer A is correct. Answer B incorrectly keeps the triangular support

0 < u < v < 1

, failing to simplify the constraint

u < 1

after dividing by

v

— a common algebraic oversight. Answer C inverts the Jacobian (using

1/v

instead of

v

), which is the trap of computing the forward rather than inverse Jacobian. Answer D simply carries over the original density value of 2 without applying the Jacobian at all. Study tip: Always verify the new support region explicitly — independence often hinges on whether it's rectangular, not just on whether the density factors.

Question 9

Suppose $X$ has a beta distribution with positive shape parameters $a$ and $b$ . For the log-odds transformation $Z=\log\{X/(1-X)\}$ , which expression is the density of $Z$ ?

$f_Z(z)=\frac{e^{(a-1)z}}{B(a,b)(1+e^z)^{a+b-2}},\quad -\infty<z<\infty$
$f_Z(z)=\frac{e^{az}}{B(a,b)(1+e^z)^{a+b}},\quad -\infty<z<\infty$ (correct answer)
$f_Z(z)=\frac{e^{bz}}{B(a,b)(1+e^z)^{a+b}},\quad -\infty<z<\infty$
$f_Z(z)=\frac{e^{az}}{B(a,b)(1+e^z)^{a+b+2}},\quad -\infty<z<\infty$

Explanation: When you see a transformation problem like this, your first instinct should be the change-of-variables technique: find the inverse transformation, compute the Jacobian, and substitute into the original density. Here,

Z = \log\{X/(1-X)\}

, so the inverse is

X = e^z/(1+e^z)

. The Jacobian is

dx/dz = e^z/(1+e^z)^2

. Since

X \sim \text{Beta}(a,b)

, its density is

f_X(x) = x^{a-1}(1-x)^{b-1}/B(a,b)

. Substituting

x = e^z/(1+e^z)

and

1-x = 1/(1+e^z)

f_Z(z) = \frac{1}{B(a,b)}\left(\frac{e^z}{1+e^z}\right)^{a-1}\left(\frac{1}{1+e^z}\right)^{b-1} \cdot \frac{e^z}{(1+e^z)^2}

Collecting powers: the numerator gives

e^{z(a-1)} \cdot e^z = e^{az}

, and the denominator accumulates

(1+e^z)^{(a-1)+(b-1)+2} = (1+e^z)^{a+b}

. This yields answer B:

f_Z(z) = e^{az}/[B(a,b)(1+e^z)^{a+b}]

. Choice A is wrong because it uses

e^{(a-1)z}

in the numerator and

a+b-2

in the denominator exponent — the student forgot to include the Jacobian's contribution of

e^z/(1+e^z)^2

. Choice C incorrectly replaces

e^{az}

with

e^{bz}

, confusing which shape parameter drives the exponential after substitution. Choice D has the correct numerator

e^{az}

but uses

(1+e^z)^{a+b+2}

, over-counting the denominator by adding rather than correctly combining exponents. Study tip: Always track every factor of the Jacobian separately — a single missed power in numerator or denominator leads to plausible-looking but wrong answers, exactly what distractors A and D exploit.

Question 10

Let $X$ and $Y$ be independent uniform random variables on $(0,1)$ . Define $U=X+Y$ and $V=X-Y$ . Which joint density and support are correct?

$f_{U,V}(u,v)=1/2$ for $0<u<2$ and $|v|<\min(u,2-u)$ (correct answer)
$f_{U,V}(u,v)=2$ for $0<u<2$ and $|v|<\min(u,2-u)$
$f_{U,V}(u,v)=1/2$ for $0<u<2$ and $-1<v<1$
$f_{U,V}(u,v)=1$ for $0<u<1$ and $|v|<u$

Explanation: When transforming random variables, your two key tasks are: (1) find the Jacobian of the inverse transformation, and (2) carefully characterize the support. Here,

U = X+Y

and

V = X-Y

, so the inverse is

X = (U+V)/2

and

Y = (U-V)/2

. The Jacobian is:

|J| = \left|\det\begin{pmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{pmatrix}\right| = \left|\det\begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\right| = \frac{1}{2}

Since

f_{X,Y}(x,y) = 1

on the unit square, the joint density becomes

f_{U,V}(u,v) = 1 \cdot |J| = 1/2

. For the support, you need

0 < x < 1

and

0 < y < 1

, which translates to

0 < (U+V)/2 < 1

and

0 < (U-V)/2 < 1

. This gives

0 < u < 2

with

|v| < u

(from the

x > 0, y > 0

constraints) and

|v| < 2-u

(from the

x < 1, y < 1

constraints), combining to

|v| < \min(u, 2-u)

. This confirms A is correct. B incorrectly uses density

2

instead of

1/2

— a common sign error where students invert the Jacobian rather than multiplying by it. C uses the correct density but wrong support:

-1 < v < 1

ignores the dependence of

v

's range on

u

, and it would not integrate to 1. D restricts to

0 < u < 1

only, missing the upper triangle of the support, and uses density 1. Study tip: Always verify your answer by checking that

\int\int f_{U,V}\,dv\,du = 1

. With density

1/2

over a diamond of area 2, you get exactly 1 — a quick sanity check that catches both Jacobian and support errors simultaneously.