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VEE Mathematical Statistics · Functions of a Random Variable · Tutorial 261 of 880

Choosing Between CDF and Jacobian Methods

Choose a reliable transformation method by examining the inverse map, the event of interest, and the shape of the support.

Intermediate 12 min read

What You'll Learn

  • Use event probabilities to derive a transformed CDF, especially when a scalar map is not one-to-one.
  • Apply the scalar change-of-variable formula and account for every inverse branch.
  • Decide when a bivariate Jacobian is appropriate by checking invertibility and the transformed support.
  • Set integration limits from the transformed support rather than assuming it is rectangular.
  • Check transformed densities by integrating over their full support.
  • Compare transformed-coordinate calculations with direct event probabilities when useful.

Choosing a Method for a Transformation

A transformed random variable can be described by its cumulative distribution function (CDF), by a density obtained from a change of variables, or sometimes by either method. The main decision is not a matter of preference: it depends on how difficult the event or inverse map is. A CDF calculation starts from the event defining the transformed variable. A Jacobian calculation starts from an inverse map and must account for the transformed support.

The CDF route is especially useful when a scalar transformation has several inverse branches or when the event \(g(X)\leq t\) is easy to describe. For a bivariate transformation, a Jacobian is often effective when the map is one-to-one and the transformed support is manageable. If that support is awkward, working directly with the event in the original variables may be simpler.

Definition: If \(T=g(X)\), then \(F_T(t)=P(T\leq t)=P(g(X)\leq t)\). If a differentiable scalar transformation is one-to-one on the support of \(X\), then, wherever the inverse is differentiable, \(f_T(t)=f_X(g^{-1}(t))\left|\frac{d}{dt}g^{-1}(t)\right|\). If several inverse branches \(x_j(t)\) map to the same \(t\), sum their contributions: \(f_T(t)=\sum_j f_X(x_j(t))/|g'(x_j(t))|\), at values where the branches are differentiable and \(g'(x_j(t))\ne 0\).

These formulas require attention to support. The density is zero outside the values that the transformation can actually attain. At a turning point, a branch derivative may be zero, so the branch formula is used away from that point; a density may become unbounded near it without failing to be integrable.

Start With the Event: A Scalar Transformation With Two Branches

For a non-monotone scalar function, finding the inverse of the transformation alone does not identify which original values contribute. The CDF makes the set of contributing values explicit and includes all branches at once.

Worked Example: Squaring a Uniform Random Variable

Let \(X\sim\operatorname{Uniform}(-2,2)\), with density \(f_X(x)=1/4\) for \(-2<x<2\). Set \(T=X^2\). Find the support, CDF, and density of \(T\), and verify that the density integrates to one.

Method: calculate the CDF from the event. Since \(-2<X<2\), the square satisfies \(0\leq T\leq4\). For \(0\leq t\leq4\),

$$ F_T(t)=P(X^2\leq t) =P(-\sqrt{t}\leq X\leq\sqrt{t}) =\frac{2\sqrt{t}}{4} =\frac{\sqrt{t}}{2}. $$

For \(t<0\), \(F_T(t)=0\); for \(t\geq4\), \(F_T(t)=1\). Differentiating in the interior of the support gives

$$ f_T(t)=\frac{1}{4\sqrt{t}},\qquad 0<t<4. $$

Check. The density integrates to \(\int_0^4 \frac{1}{4\sqrt{t}}\,dt =\frac{1}{4}[2\sqrt{t}]_0^4=1\).

The branch formula gives the same result: for \(0<t<4\), the two inverse values are \(x=\sqrt{t}\) and \(x=-\sqrt{t}\), and \(|g'(x)|=|2x|=2\sqrt{t}\) on either branch. Thus each contributes \((1/4)/(2\sqrt{t})\), and their sum is \(1/(4\sqrt{t})\). Common mistake avoided: using only the positive square root would omit half the probability and produce a density integrating to \(1/2\).

This example also explains why a CDF is often the clearest first step for a many-to-one transformation: the event automatically includes every input that maps into the required range. The Jacobian branch formula is an efficient alternative once all inverse branches and their derivatives have been identified.

Use a Jacobian When the Inverse Map Is Manageable

For a differentiable, one-to-one transformation of a continuous random vector, the change-of-variables formula multiplies the original density by the absolute determinant of the inverse Jacobian. The transformed density must be restricted to the image of the original support. A transformation can be one-to-one and still produce a nonrectangular support.

Theorem: Suppose \((X,Y)\) has joint density \(f_{X,Y}\) on a support \(S\), and a differentiable transformation \((U,V)=h(X,Y)\) has a differentiable inverse \((x(u,v),y(u,v))\) on the relevant image. Then \(f_{U,V}(u,v)=f_{X,Y}(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\) on the transformed support, and the density is zero outside it. If the map has multiple inverse points, the contributions from all valid inverse points must be summed.

The determinant appears because a small region in the transformed coordinates corresponds approximately to a region in the original coordinates whose area is scaled by the absolute determinant. Equating the probability assigned to corresponding regions gives the density formula: original density times original area must equal transformed density times transformed area.

Worked Example: A Ratio and Its Denominator

Suppose \((X,Y)\) has density \(f_{X,Y}(x,y)=1/2\) on \(0<x<y<2\), and density zero elsewhere. Define \(U=X/Y\) and \(V=Y\). Find the joint density of \((U,V)\), then calculate \(P(V>1)\).

Method: invert the transformation and track its support. From \(U=X/Y\) and \(V=Y\), the inverse is \(x=uv,\ y=v\). The original conditions \(0<x<y<2\) become \(0<uv<v<2\). Since \(v>0\), this is equivalent to \(0<u<1,\ 0<v<2\).

The inverse Jacobian is

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

Therefore,

$$ f_{U,V}(u,v)=\frac{v}{2},\qquad 0<u<1,\quad 0<v<2. $$

As a support check, its integral is \(\int_0^2\int_0^1(v/2)\,du\,dv =\int_0^2(v/2)\,dv=1\). For the requested probability, the condition \(V>1\) gives

$$ P(V>1)=\int_1^2\int_0^1\frac{v}{2}\,du\,dv =\int_1^2\frac{v}{2}\,dv =\frac{1}{4}(4-1) =\frac{3}{4}=0.7500. $$

Check. The same event is \(Y>1\) in the original support. Its probability is \(\int_1^2\int_0^y(1/2)\,dx\,dy =\int_1^2 y/2\,dy=3/4\). The two calculations agree. Common mistake avoided: the transformed density is \(v/2\), not the original constant \(1/2\); omitting the Jacobian factor \(v\) would give the wrong probability and a density that does not integrate to one.

Use Support Geometry to Set the Limits

For a bivariate transformation, the integration limits are part of the solution, not a detail to be filled in later. A convenient inverse does not make the support rectangular automatically. One reliable approach is to write the original support conditions in terms of the inverse variables, simplify them, and only then choose integration limits.

Worked Example: Sum and Difference Coordinates

Let \(X\) and \(Y\) be independent \(\operatorname{Uniform}(0,1)\) random variables, so \(f_{X,Y}(x,y)=1\) on \(0<x<1,\ 0<y<1\). Define \(U=X+Y\) and \(V=X-Y\). Find the transformed density and calculate \(P(U>1,\ V>0)\).

Method: apply the Jacobian, then integrate over the image of the square. Solving for the original variables gives \(x=(u+v)/2\) and \(y=(u-v)/2\). The absolute inverse Jacobian determinant is \(\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=1/2\), so the transformed density is \(1/2\) wherever the inverse lies inside the unit square.

The conditions \(0<x<1\) and \(0<y<1\) become \(0<u+v<2\) and \(0<u-v<2\). Equivalently, the support is \(0<u<2\) and \(-\min(u,2-u)<v<\min(u,2-u)\). This support narrows toward both \(u=0\) and \(u=2\); it is not the whole rectangle \(0<u<2,\ -1<v<1\).

For the event \(U>1,\ V>0\), the support conditions reduce to \(1<u<2\) and \(0<v<2-u\). Hence

$$ P(U>1,V>0) =\int_1^2\int_0^{2-u}\frac{1}{2}\,dv\,du =\frac{1}{2}\int_1^2(2-u)\,du =\frac{1}{2}\left[\;2u-\frac{u^2}{2}\;\right]_1^2 =\frac{1}{4} =0.2500. $$

Check in the original coordinates. The event is \(x+y>1\) and \(x>y\). Its region has \(1/2<x<1\) and \(1-x<y<x\), so its probability is \(\int_{1/2}^1[x-(1-x)]\,dx =\int_{1/2}^1(2x-1)\,dx=1/4\). Common mistake avoided: integrating over the whole rectangle in \((u,v)\) would include points whose inverse has \(x\) or \(y\) outside \((0,1)\).

A Practical Decision Process

The event-based and Jacobian approaches are complementary. Choose the route that makes both the inverse images and the support easiest to describe, and check that the result has the required probability properties.

1
Write down the target.
For a probability, translate the requested condition into an event involving the original variables or transformed variables. For a full distribution, identify the CDF or density that is needed.
2
Check whether the map is one-to-one.
For a scalar transformation, identify all inverse branches. For a bivariate transformation, solve for the inverse and check whether more than one original point maps to the same transformed point.
3
Choose the simpler geometry.
Use an event-based CDF calculation when the preimage of an event is simple. Use a Jacobian when the inverse and transformed support are manageable. If multiple inverse points exist, include every valid contribution.
4
Check the result.
Confirm that the density is nonnegative, is zero outside the attainable support, and integrates to one. For an event probability, compare with the original-coordinate event when that calculation is convenient.

A common exam trap is to identify a correct inverse formula but then use it on the wrong support, or to use only one branch of a non-one-to-one scalar map. Another is to mistake a simple-looking transformed region for the actual image of the original support. Drawing or describing that image before integrating can prevent both errors.

Key takeaway: Start with the event and the support. A CDF calculation is often clearest when the map folds several inputs onto the same output; a Jacobian is efficient when the inverse is well behaved and its support can be stated exactly. Either method must account for every valid preimage.

Check Your Understanding

For each problem, identify the appropriate event or transformed support before calculating.

  1. Let \(X\sim\operatorname{Uniform}(-2,2)\) and \(T=X^2\). Give the support and CDF of \(T\). (Answer: support \(0\leq t\leq4\); \(F_T(t)=0\) for \(t<0\), \(\sqrt{t}/2\) for \(0\leq t\leq4\), and \(1\) for \(t>4\).)
  2. Let \(X\) have exponential density \(f_X(x)=3e^{-3x}\) for \(x>0\), and set \(W=\ln X\). Find the support and density of \(W\). (Answer: support \(-\infty<w<\infty\); \(f_W(w)=3e^w e^{-3e^w}\).)
  3. Let \(X,Y\) be independent \(\operatorname{Uniform}(0,1)\) variables and set \(T=X+Y\). Calculate \(P(T>1.5)\). (Answer: \(0.1250\).)
  4. Suppose \((X,Y)\) has density \(1/2\) on \(0<x<y<2\), and define \(U=X/Y,\ V=Y\). Give the joint density of \((U,V)\) and calculate \(P(V>1)\). (Answer: \(f_{U,V}(u,v)=v/2\) for \(0<u<1,\ 0<v<2\), and \(P(V>1)=0.7500\).)
  5. Let \(X,Y\) be independent \(\operatorname{Uniform}(0,1)\), with \(U=X+Y,\ V=X-Y\). Give the transformed support and calculate \(P(U>1,\ V>0)\). (Answer: \(0<u<2,\ |v|<\min(u,2-u)\); probability \(0.2500\).)

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