Choose X uniformly from (−3,3) and set Y=X². A small positive y can arise from both a positive and a negative x. Predict why using only x=√y would lose half the probability. A transformation changes both support and how source intervals map into destination intervals.
Predict
Predict the support of Y and whether Y has a point mass at zero. Only X=0 maps to zero under squaring, and that singleton has probability zero. A density can become unbounded near zero without creating an atom there. Compare this with a flat map that sends an entire interval to one point.
Experiment
For 0<y<9, F_Y(y)=P(−√y≤X≤√y)=2√y/6=√y/3. Differentiating gives f_Y(y)=1/(6√y) on (0,9). Its integral is [√y/3] from zero to nine, equal to one.
Feature
Result
Source support
(−3,3)
Interior destination support
(0,9)
Inverse branches
√y and −√y
Atom at zero
None
Sketch the two source intervals mapping to one narrow destination interval. Their probabilities add; the sign of an inverse derivative does not subtract probability.
Understand
For a differentiable one-to-one transformation Y=g(X) with a differentiable inverse on the relevant interior, the density is f_Y(y)=f_X(g⁻¹(y)) times the absolute derivative of g⁻¹(y). The inverse derivative measures how much source length corresponds to a unit of destination length. Support restrictions are part of the formula.
For a transformation with several regular inverse branches, sum the contributions over all preimages. Squaring has two branches whenever both signs belong to the source support. A CDF calculation often handles such support issues more reliably than memorizing a formula. Start from the event g(X)≤y and solve it in the original variable.
The absolute value is essential for a decreasing transformation. Probability mass remains nonnegative when orientation reverses. For Y=a+bX with nonzero b, density scales by 1/|b| and the support endpoints may reverse order. If b=0, the result is a point mass, so the ordinary inverse-density formula does not apply.
A flat segment can create an atom. If X is uniform on (−2,2) and Y=max(X,0), half the source probability collapses to zero. The remaining half has density 1/4 on (0,2). A density-only answer would integrate to one half and omit the atom. Record the full mixed law or CDF.
Derivative singularities need careful interpretation. In the square example the inverse derivative diverges near zero, but the resulting density is integrable. The change-of-variables argument applies away from that endpoint, and the CDF confirms the total law. Infinite density height is not infinite probability at a point.
Do not assume a nonlinear transform preserves a familiar family or its mean. A uniform variable squared is not uniform, and E(g(X)) generally differs from g(E(X)). LOTUS can check a transformed moment independently of the new density calculation.
Apply
Derive the practice square law first by its CDF and then by summing inverse-branch contributions. Compare their normalizations. For the clipped variable, explicitly state both the atom and continuous part. A complete answer should not hide missing mass behind a graph that shows only the smooth portion.
Continue
Next move to joint laws. A support region can couple coordinates even when the density has a simple constant height.
Apply · explain · revise
Practice and reasoning
Drafts save locally as you type. Reveal worked solutions when useful; the site remembers that exposure. Automatic matches check the requested answer, while written arguments require review against the rubric.
X is uniform on (0,1), Y=3X+2. Find the density of Y at 3.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
X is uniform on (−1,1), Y=X². Find P(Y≤1/4).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
The monotone density transformation uses the inverse derivative with:
For Y=X² when X can be positive or negative, the inverse calculation generally needs:
For X uniform on (−2,2), derive the CDF and density of Y=X², and check normalization.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
X is uniform on (−1,1), Y=max(X,0). Describe the distribution, including any atom.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
