Recovering Interval Probabilities from Mixed Representations

Why Endpoints Matter

For a continuous law, changing whether an interval includes one endpoint does not change its probability. For a law with atoms, it can. An interval probability may combine a jump at an endpoint, atoms inside the interval, and probability accumulated continuously across it. Keeping these contributions distinct makes calculations reliable.

The cumulative distribution function (CDF) is often the quickest route: subtract its values using the correct endpoint convention. When a law is given instead by its atoms and density, add the relevant atom masses and integrate the density over the interval. For a mixture, first calculate each component's interval probability, then weight and add those probabilities.

Interval Probabilities From a CDF

The left limit distinguishes strictly less than \(x\) from less than or equal to \(x\). For \(a\le b\), the four usual interval conventions therefore give:

These identities follow by subtracting probabilities of nested events. For example, \(\{X\le b\}\) is the disjoint union of \(\{X\le a\}\) and \(\{a<X\le b\}\), so subtracting \(P(X\le a)=F(a)\) from \(F(b)\) gives the first identity. To include \(a\), subtract \(P(X<a)=F(a-)\) instead. Replacing \(\{X\le b\}\) by \(\{X<b\}\), whose probability is \(F(b-)\), gives the two formulas with an open right endpoint. Thus each endpoint's inclusion is handled by choosing the appropriate CDF value or left limit.

The argument uses only event inclusion and the definition of the CDF. In particular, it does not assume that \(F\) is continuous. If \(a=b\), the formula for the closed interval gives \(P(X=a)=F(a)-F(a-)\); each of the other three intervals is empty and has probability zero. The formula \(F(b-)-F(a)\) for \(P(a<X<b)\) applies only when \(a<b\).

Intervals From Atoms and a Density

A law may be given directly by its atom locations and masses, together with a density for its absolutely continuous part. Let \(A\) be the set of atom locations, and write \(p_x=P(X=x)\) for \(x\in A\). Suppose the rest of the law has density \(f\) with respect to length, with no additional singular component. For a measurable set \(B\), the interval formula extends to

For an interval, include an atom in the sum exactly when its location belongs to that interval, and integrate the density over the interval. A countable sum is required if there are countably many atoms. Including or excluding an endpoint does not affect the density integral, but it can change the atom sum. The assumption of no additional singular component matters: if there is singular probability not represented by atoms or by \(f\), it must be added separately. A density formula alone does not account for it.

To justify the result, the event that \(X\) lies in the interval splits into disjoint events: \(X=x\) for each atom location \(x\) in that interval, and \(X\) falls in the interval through the absolutely continuous part of the law. Countable additivity gives the sum of the atom probabilities. By the definition of a density, the absolutely continuous part assigns probability \(\int_a^b f(t)\,dt\) to the interval. Adding these disjoint contributions yields the formula.

Intervals Under a Mixture

In a finite mixture, a component is selected with a specified probability, and the observation is drawn from that component's distribution. Let the component distributions be \(P_1,\ldots,P_k\), and let their weights satisfy \(w_j\ge0\) and \(\sum_{j=1}^k w_j=1\). For any measurable set \(B\), the mixture assigns probability

This applies to intervals as well: find the interval probability under each component, multiply by its weight, and add. The identity can be justified using a component label \(J\). For each \(j\), the joint probability of selecting component \(j\) and then observing \(X\in B\) is \(w_jP_j(X\in B)\), including when \(w_j=0\): in that case the joint probability is zero. The events \(\{X\in B,J=j\}\) are disjoint and their union is \(\{X\in B\}\), so additivity gives the displayed identity. This argument does not condition on any event of probability zero.

Common Mistakes and Why the Calculations Matter

• Using \(F(a)\) when the interval includes \(a\). \(F(a)=P(X\le a)\) already includes the atom at \(a\). To include that endpoint in an interval calculation, subtract \(F(a-)=P(X<a)\) instead.
• Adding an endpoint atom twice. If using CDF differences, the jump is already included or excluded by the chosen values. Do not add it again unless working from an atom-and-density formula.
• Integrating density as if it represented every part of the law. The density integral misses both atoms and any singular component. Add atoms explicitly, and account for singular probability separately when present.
• Conditioning on a zero-weight mixture component. A component with weight zero contributes zero to the mixture. Use its weighted probability term \(0\cdot P_j(X\in B)=0\), not a conditional probability given that the zero-probability component was selected.

These calculations matter whenever a probability statement has strict or inclusive thresholds, such as a probability of being below a cutoff or at most a specified value. The CDF handles arbitrary laws, while an atom-and-density representation can make the separate contributions transparent. Mixture calculations then preserve those endpoint choices component by component.