Distributions as pushforward measures
Transport probabilities through measurable maps, compute transformed laws, and distinguish marginal distributions from joint information.
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The question for today
What you’ll understand
- Define a distribution without assuming a probability density exists.
- Compute a transformed law from inverse images rather than guessed formulas.
- Distinguish a marginal law from the coupling of multiple variables.
A random variable belongs to an underlying experiment, but many questions depend only on the probabilities of its output values. Its distribution packages exactly this information as a measure on the output space. This viewpoint handles discrete, continuous, singular, and mixed examples without changing the definition. It also identifies what a distribution forgets: the relationship between that variable and other observations made on the same experiment. A shared histogram is not a shared joint model.
Moving the measure rather than the outcomes
Given a measurable map X, define its law by assigning a target event B the probability that X lands in B. The relevant original event is the inverse image of B. Since inverse images preserve complements and countable disjoint unions, the probability axioms transfer to the new measure. No inverse function for X is needed. The map may identify many different outcomes, and the resulting distribution simply combines their probabilities when they produce the same observable value.
For a real random variable, the cumulative distribution function records the probability of each lower half-line. It is nondecreasing, right-continuous, and tends to zero and one at the two infinities. These values determine the Borel probability measure uniquely. A jump represents a point mass. A continuous cumulative distribution function need not possess a density, so differentiating it and assuming the derivative recovers all its probability is not justified without an absolute-continuity hypothesis.
Transformations and their preimages
If Y=g(X) for a measurable function g, its law is the pushforward of the law of X through g. To compute it, describe the set of x values whose transformed value lies in the requested target set. This method works when g is noninjective, where using a single inverse branch loses probability. For a squaring map both positive and negative inputs may contribute to one positive output. Atoms also survive or merge according to the same inverse-image rule.
Integration follows the same transport principle: an expectation of a nonnegative measurable function of X can be integrated against its law. For signed functions the identity applies when the absolute value is integrable, or in the appropriate extended sense when positive and negative parts do not both have infinite integrals. This result is often called the transfer formula. It is more general than a density-based substitution and gives a reliable way to compute expectations for mixtures or transformed discrete variables.
A marginal does not encode dependence
Take a fair sign variable X with values −1 and 1. Setting Y=X and setting Z=−X gives Y and Z exactly the same marginal distribution. Nevertheless X+Y is either −2 or 2, whereas X+Z is always zero. The difference lives in the joint distribution on the product space. A marginal law describes what one coordinate does alone; a coupling specifies how coordinates occur together. Independent coupling is one choice, not an automatic consequence of naming two distributions.
More generally, two variables defined on different probability spaces can have the same law even when comparing their samplewise values makes no sense. Equality in distribution is therefore weaker than almost-sure equality of variables on one common space. When a calculation involves sums, maxima, or simultaneous events, first check whether the joint law is available. Marginal distributions determine expectations of separate integrable terms by linearity, but usually do not determine probabilities of joint events or variances of sums.
Work it through
Squaring a centered uniform observation
Let X be uniform on [−2,2] and Y=X². Determine the cumulative distribution function of Y and its mean.
For y<0 the event X²≤y is empty. For 0≤y≤4 its inverse image is the interval [−√y,√y], whose length is 2√y within an interval of total length four.
Consequently F(y)=√y/2 on [0,4], with F(y)=0 below zero and F(y)=1 above four. Both square-root branches contributed; using only the positive branch would halve the probability.
Use the original uniform law to compute the mean: the integral of x²/4 from −2 to 2 is (1/4)(16/3)=4/3. This also follows by integrating y against the transformed law.
Y has cumulative distribution F(y)=√y/2 for 0≤y≤4 and mean 4/3. It has no atom at zero despite its density becoming unbounded near zero.
The living notebook
Explore the model
Interactive lab · probability
Push a count law through a threshold
Choose n=4 and p=0.5. Define a new observation Y that is one when the success count K is at least three and zero otherwise. Predict the mass at Y=1 by combining the bars at K=3 and K=4. The panel still shows the exact law of K, so perform the two-cell pushforward yourself.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe combined mass is 5/16. Grouping bars constructs a transformed discrete law; the panel does not infer the joint distribution of arbitrary additional variables.
Pause & reason
Check your understanding
Your turn
Try it yourself
A variable X takes −1,0,2 with probabilities 0.2,0.5,0.3. Find the full law and mean of Y=X². Then find the law of Z=1 when X is nonzero and Z=0 otherwise.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
List the output of each map at the three possible input values and add masses only when output values coincide.
Compare with the solution
Squaring gives Y values 1,0,4 with probabilities 0.2,0.5,0.3, so its mean is 0.2+1.2=1.4. The nonzero indicator maps −1 and 2 to the same value one, combining their masses to 0.5; zero retains mass 0.5. Thus Z is a fair Bernoulli variable with mean 0.5. Neither calculation requires a density or an invertible transformation.
Keep the idea
Take it with you
- A law is the pushforward of a probability measure through an observation.
- Inverse images handle many-to-one transformations without losing mass.
- Marginal laws do not specify how different observations are coupled.
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