Unit 1, lesson 8

Probability spaces and conditional information

How a probability space, its sigma algebra and an information partition describe what can be observed, and how conditional probability and Bayes' theorem update a default probability given its base rate.

  • Video 8
  • Calculus, linear algebra, optimization and probability

A credit model raises a warning signal for a borrower, and the signal is positive for 9090 percent of the borrowers who default. It is tempting to read this as a 9090 percent chance of default, but in a constructed model where 22 percent of borrowers default and 1010 percent of the others also trigger the signal, the answer is about 15.515.5 percent. The difference comes from which outcomes remain possible once the signal is seen and from the denominator that a conditional probability uses. This lesson sets up probability spaces and sigma algebras as the record of what can be observed, defines conditional probability on a finite information partition, and derives Bayes’ theorem to show how the base rate changes the meaning of a positive signal.

Key terms

Sigma algebra

Definition Let Ω\Omega be a nonempty set of outcomes, also called states. A collection F\mathcal{F} of subsets of Ω\Omega is a sigma algebra if

Ω∈F,A∈F⇒Ac∈F,A1,A2,…∈F⇒⋃nAn∈F.\begin{aligned} &\Omega \in \mathcal{F}, \\[0.4em] &A \in \mathcal{F} \Rightarrow A^{c} \in \mathcal{F}, \\[0.4em] &A_1, A_2, \ldots \in \mathcal{F} \\[0.4em] &\quad \Rightarrow \textstyle\bigcup_{n} A_n \in \mathcal{F}. \end{aligned}

Its members are the measurable events, the events to which a probability can be assigned. A sigma algebra also contains ∅\varnothing and is closed under countable intersections and under differences. The power set of Ω\Omega, the collection of all subsets, is the largest sigma algebra on Ω\Omega, and {∅,Ω}\{\varnothing, \Omega\} is the smallest, the trivial sigma algebra. For any collection C\mathcal{C} of subsets, the sigma algebra σ(C)\sigma(\mathcal{C}) generated by C\mathcal{C} is the smallest sigma algebra that contains C\mathcal{C}, the intersection of all sigma algebras that contain it. When Ω\Omega is finite, closure under complements and finite unions is already enough.

In plain words A sigma algebra lists the yes or no questions about the outcome that are allowed. If a question about an event is allowed, so is the question whether the event failed to happen, and so is the question whether at least one of a sequence of allowed events happened. A smaller sigma algebra allows fewer questions and so describes coarser information.

Example In a constructed one-period credit model a borrower defaults (DD) or not (DcD^{c}), and a warning signal is positive (++) or negative (−-). The four states are ω1=(D,+)\omega_1 = (D, +), ω2=(D,−)\omega_2 = (D, -), ω3=(Dc,+)\omega_3 = (D^{c}, +) and ω4=(Dc,−)\omega_4 = (D^{c}, -), and the letters also name events, with D={ω1,ω2}D = \{\omega_1, \omega_2\} and ++ for {ω1,ω3}\{\omega_1, \omega_3\}. The power set has 24=162^4 = 16 events. An observer who sees only the signal can answer exactly the questions in

G={∅,{ω1,ω3},{ω2,ω4},Ω},\begin{aligned} \mathcal{G} = \bigl\{&\varnothing, \{\omega_1, \omega_3\}, \\[0.4em] &\{\omega_2, \omega_4\}, \Omega\bigr\}, \end{aligned}

a sigma algebra with four events. The default event {ω1,ω2}\{\omega_1, \omega_2\} is not in G\mathcal{G}, because the signal alone does not reveal whether default occurred. The collection {∅,{ω1,ω3},Ω}\{\varnothing, \{\omega_1, \omega_3\}, \Omega\} is not a sigma algebra, since it lacks the complement {ω2,ω4}\{\omega_2, \omega_4\}.

Probability space

Definition A probability space is a triple (Ω,F,P)(\Omega, \mathcal{F}, \mathbb{P}) made of a nonempty outcome set Ω\Omega, a sigma algebra F\mathcal{F} on Ω\Omega and a probability measure P\mathbb{P} on F\mathcal{F}. A probability measure is a function from F\mathcal{F} to [0,1][0, 1] with P(Ω)=1\mathbb{P}(\Omega) = 1 that is countably additive,

P(⋃n=1∞An)=∑n=1∞P(An)\mathbb{P}\Bigl(\bigcup_{n=1}^{\infty} A_n\Bigr) = \sum_{n=1}^{\infty} \mathbb{P}(A_n)

for pairwise disjoint events A1,A2,…∈FA_1, A_2, \ldots \in \mathcal{F}. It follows that P(∅)=0\mathbb{P}(\varnothing) = 0, that P(Ac)=1−P(A)\mathbb{P}(A^{c}) = 1 - \mathbb{P}(A), that P(A)≤P(B)\mathbb{P}(A) \le \mathbb{P}(B) when A⊆BA \subseteq B, and that P(A∪B)=P(A)+P(B)−P(A∩B)\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B). On a finite Ω\Omega whose sigma algebra is the power set, a probability measure is the same thing as a list of state masses p(ω)≥0p(\omega) \ge 0 that add to 11, with P(A)=∑ω∈Ap(ω)\mathbb{P}(A) = \sum_{\omega \in A} p(\omega). An event of probability zero is a null event, and a statement holds almost surely if it fails at most on a null event. A null event need not be empty. A random variable with a continuous distribution takes each single value with probability zero, yet it always takes some value. The same pair (Ω,F)(\Omega, \mathcal{F}) can carry several measures. In this course P\mathbb{P} is the physical measure, and later lessons add a pricing measure Q\mathbb{Q} that in general gives the same events different probabilities.

In plain words The outcome set says what could happen, the sigma algebra says which questions about it receive probabilities, and the measure assigns those probabilities consistently. Probabilities of separate possibilities add, and all possibilities together have probability one. The measure is a modeling choice, so two models can agree on the outcomes and the events and still disagree about the probabilities.

Example The credit model takes the power set as its sigma algebra and the state masses

p(ω1)=0.02⋅0.9=0.018,p(ω2)=0.02⋅0.1=0.002,p(ω3)=0.98⋅0.1=0.098,p(ω4)=0.98⋅0.9=0.882,\begin{aligned} p(\omega_1) &= 0.02 \cdot 0.9 = 0.018, \\[0.4em] p(\omega_2) &= 0.02 \cdot 0.1 = 0.002, \\[0.4em] p(\omega_3) &= 0.98 \cdot 0.1 = 0.098, \\[0.4em] p(\omega_4) &= 0.98 \cdot 0.9 = 0.882, \end{aligned}

which are positive and add to 11. They come from a default probability of 0.020.02 and a signal that is positive for 9090 percent of defaults and for 1010 percent of nondefaults, by the multiplication rule of the conditional probability card. Additivity gives P(D)=0.018+0.002=0.02\mathbb{P}(D) = 0.018 + 0.002 = 0.02 and P(+)=0.018+0.098=0.116\mathbb{P}(+) = 0.018 + 0.098 = 0.116. The probability that the borrower defaults or the signal is positive is 0.02+0.116−0.018=0.1180.02 + 0.116 - 0.018 = 0.118, which matches the total mass of ω1\omega_1, ω2\omega_2 and ω3\omega_3.

Information partition

Definition A partition of a finite set Ω\Omega is a collection of nonempty, pairwise disjoint subsets B1,…,BkB_1, \ldots, B_k, its blocks, whose union is Ω\Omega. The sigma algebra σ(B1,…,Bk)\sigma(B_1, \ldots, B_k) that it generates consists of all unions of blocks, the empty union ∅\varnothing included, so it has 2k2^k events. Conversely, every sigma algebra G\mathcal{G} on a finite Ω\Omega is generated by exactly one partition, the partition into its atoms, where the atom containing a state ω\omega is the intersection of all events of G\mathcal{G} that contain ω\omega. An information partition describes an observer who learns which block contains the true state and nothing more, so that the observer’s information is the sigma algebra G\mathcal{G} generated by the partition. An event AA is G\mathcal{G}-measurable, meaning that the observer can always tell whether AA occurred, exactly when it is a union of blocks, that is, when each block lies inside AA or is disjoint from it. A partition π′\pi' refines a partition π\pi when every block of π′\pi' lies inside a block of π\pi, which holds exactly when σ(π)⊆σ(π′)\sigma(\pi) \subseteq \sigma(\pi'). A filtration is an increasing family F0⊆F1⊆⋯⊆F\mathcal{F}_0 \subseteq \mathcal{F}_1 \subseteq \cdots \subseteq \mathcal{F} of sigma algebras indexed by dates, and Ft\mathcal{F}_t is the information available at date tt. On a finite Ω\Omega a filtration amounts to a sequence of partitions, each refining the one before.

In plain words An information partition sorts the states into groups that the observer cannot tell apart. Learning more splits groups into smaller ones, and a filtration says that information accumulates and is never forgotten. A forecast made at a date may use only the events in that date’s sigma algebra.

Example In the credit model, suppose nothing is known at date 00, the signal is seen at date 11 and the default status is revealed at date 22. The partitions are {Ω}\{\Omega\}, then {{ω1,ω3},{ω2,ω4}}\{\{\omega_1, \omega_3\}, \{\omega_2, \omega_4\}\}, then the four single states, and the generated sigma algebras have 22, 44 and 1616 events, each containing the one before. The default event {ω1,ω2}\{\omega_1, \omega_2\} is a union of blocks only at date 22. Seeing the signal at date 11 and then only the default status at date 22, with the signal forgotten, would not give a filtration, because the default partition {{ω1,ω2},{ω3,ω4}}\{\{\omega_1, \omega_2\}, \{\omega_3, \omega_4\}\} does not refine the signal partition.

Conditional probability

Definition Let (Ω,F,P)(\Omega, \mathcal{F}, \mathbb{P}) be a probability space and B∈FB \in \mathcal{F} an event with P(B)>0\mathbb{P}(B) > 0. The conditional probability of an event A∈FA \in \mathcal{F} given BB is

P(A∣B)=P(A∩B)P(B).\mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}.

The map A↦P(A∣B)A \mapsto \mathbb{P}(A \mid B) is again a probability measure on (Ω,F)(\Omega, \mathcal{F}), and it gives BB probability 11, so the conditional probabilities of the blocks of any partition add to 11. Rearranging gives the multiplication rule P(A∩B)=P(A∣B) P(B)\mathbb{P}(A \cap B) = \mathbb{P}(A \mid B)\,\mathbb{P}(B). Events AA and BB are independent if P(A∩B)=P(A) P(B)\mathbb{P}(A \cap B) = \mathbb{P}(A)\,\mathbb{P}(B), which for P(B)>0\mathbb{P}(B) > 0 is equivalent to P(A∣B)=P(A)\mathbb{P}(A \mid B) = \mathbb{P}(A). Disjoint events with positive probabilities are never independent, since their intersection has probability 00 while the product of their probabilities is positive. Given a finite partition B1,…,BkB_1, \ldots, B_k whose blocks all have positive probability, the conditional probability of AA given this information is the random variable equal to P(A∣Bj)\mathbb{P}(A \mid B_j) on each block BjB_j. It is measurable with respect to the information, and averaging it with the block probabilities gives back P(A)\mathbb{P}(A), which is the law of total probability. If P(B)=0\mathbb{P}(B) = 0 the ratio is undefined. Conditioning on a null event, such as a continuous variable taking one exact value, needs densities or regular conditional distributions with their own assumptions and lies outside this lesson.

In plain words Conditioning on BB discards the states outside BB and rescales the mass that remains so that it adds to one. Only the part of AA inside BB counts in the numerator, because states in AA but outside BB have been ruled out.

Example Given a positive signal, only ω1\omega_1 and ω3\omega_3 remain, with total mass 0.1160.116. Rescaling gives them the conditional probabilities

0.0180.116=958≈0.155,0.0980.116=4958≈0.845,\begin{aligned} \frac{0.018}{0.116} &= \frac{9}{58} \approx 0.155, \\[0.4em] \frac{0.098}{0.116} &= \frac{49}{58} \approx 0.845, \end{aligned}

which add to 11, while ω2\omega_2 and ω4\omega_4 get 00. In the other direction, P(+∣D)=0.018/0.02=0.9\mathbb{P}(+ \mid D) = 0.018/0.02 = 0.9 recovers the rate of positive signals among defaults. Default and the signal are not independent, since P(D∩+)=0.018\mathbb{P}(D \cap {+}) = 0.018 while P(D) P(+)=0.02⋅0.116=0.00232\mathbb{P}(D)\,\mathbb{P}(+) = 0.02 \cdot 0.116 = 0.00232.

Bayes’ theorem

Definition Let AA and BB be events with P(A)>0\mathbb{P}(A) > 0, P(Ac)>0\mathbb{P}(A^{c}) > 0 and P(B)>0\mathbb{P}(B) > 0. Then

P(A∣B)=P(B∣A) P(A)P(B),\begin{aligned} &\mathbb{P}(A \mid B) \\[0.4em] &= \frac{\mathbb{P}(B \mid A)\,\mathbb{P}(A)}{\mathbb{P}(B)}, \end{aligned}

where, by the law of total probability,

P(B)=P(B∣A) P(A)+P(B∣Ac) P(Ac).\begin{aligned} \mathbb{P}(B) &= \mathbb{P}(B \mid A)\,\mathbb{P}(A) \\[0.4em] &\quad + \mathbb{P}(B \mid A^{c})\,\mathbb{P}(A^{c}). \end{aligned}

The first formula holds because the joint mass can be written in two ways, P(A∩B)=P(B∣A) P(A)=P(A∣B) P(B)\mathbb{P}(A \cap B) = \mathbb{P}(B \mid A)\,\mathbb{P}(A) = \mathbb{P}(A \mid B)\,\mathbb{P}(B), and the second because BB is the disjoint union of B∩AB \cap A and B∩AcB \cap A^{c}. More generally, if A1,…,AkA_1, \ldots, A_k partition Ω\Omega into events of positive probability and P(B)>0\mathbb{P}(B) > 0, then for each jj

P(Aj∣B)=P(B∣Aj) P(Aj)∑l=1kP(B∣Al) P(Al).\begin{aligned} &\mathbb{P}(A_j \mid B) \\[0.4em] &= \frac{\mathbb{P}(B \mid A_j)\,\mathbb{P}(A_j)}{\sum_{l=1}^{k} \mathbb{P}(B \mid A_l)\,\mathbb{P}(A_l)} . \end{aligned}

The prior P(A)\mathbb{P}(A), the likelihoods P(B∣A)\mathbb{P}(B \mid A) and P(B∣Ac)\mathbb{P}(B \mid A^{c}) and the posterior P(A∣B)\mathbb{P}(A \mid B) are all probabilities under one measure. The theorem is an identity within a specified model. It does not check that the prior and the error rates were estimated well, and it does not decide which measure belongs in a valuation, where a pricing measure Q\mathbb{Q} may give default a different probability from the one under P\mathbb{P}.

In plain words The probability of a positive signal among defaults and the probability of default among positive signals share a numerator, the mass of the states with both properties, but they have different denominators. Bayes’ theorem passes from one to the other by dividing the joint mass by the total mass of the observed event, which collects every way that event can occur.

Example In the credit model the inputs are P(D)=0.02\mathbb{P}(D) = 0.02, P(+∣D)=0.9\mathbb{P}(+ \mid D) = 0.9 and P(+∣Dc)=0.1\mathbb{P}(+ \mid D^{c}) = 0.1. The total probability of a positive signal is

P(+)=0.9⋅0.02+0.1⋅0.98=0.018+0.098=0.116,\begin{aligned} &\mathbb{P}(+) \\[0.4em] &= 0.9 \cdot 0.02 + 0.1 \cdot 0.98 \\[0.4em] &= 0.018 + 0.098 = 0.116, \end{aligned}

and the posterior is

P(D∣+)=0.0180.116=958≈0.1552.\begin{aligned} \mathbb{P}(D \mid +) &= \frac{0.018}{0.116} \\[0.4em] &= \frac{9}{58} \approx 0.1552 . \end{aligned}

A positive signal raises the default probability from 22 percent to about 15.515.5 percent, far below the 9090 percent rate of positive signals among defaults. The false positives from the large nondefault group, with mass 0.0980.098, outweigh the true positives, with mass 0.0180.018.

Base rate

Definition The base rate of an event AA is its unconditional probability P(A)\mathbb{P}(A) in the model before the signal is observed, the prior in Bayes’ theorem. In a population model it is the fraction of members with the property AA. When 0<P(A)<10 < \mathbb{P}(A) < 1 and P(B∣Ac)>0\mathbb{P}(B \mid A^{c}) > 0, Bayes’ theorem in odds form reads

P(A∣B)P(Ac∣B)=P(B∣A)P(B∣Ac)⋅P(A)P(Ac),\begin{aligned} \frac{\mathbb{P}(A \mid B)}{\mathbb{P}(A^{c} \mid B)} &= \frac{\mathbb{P}(B \mid A)}{\mathbb{P}(B \mid A^{c})} \\[0.4em] &\quad \cdot \frac{\mathbb{P}(A)}{\mathbb{P}(A^{c})}, \end{aligned}

so the posterior odds equal the likelihood ratio times the prior odds. The likelihood ratio P(B∣A)/P(B∣Ac)\mathbb{P}(B \mid A)/\mathbb{P}(B \mid A^{c}) describes the signal alone, and the base rate enters only through the prior odds. For a fixed positive likelihood ratio the posterior increases strictly with the base rate and tends to 00 as the base rate tends to 00. Base-rate neglect is the error of judging the posterior from the signal alone, for example by reading P(B∣A)\mathbb{P}(B \mid A) as if it were P(A∣B)\mathbb{P}(A \mid B), or the likelihood ratio as if it were the posterior odds. At the other extreme, a signal without false positives, one with P(B∣Ac)=0\mathbb{P}(B \mid A^{c}) = 0 and P(B∣A)>0\mathbb{P}(B \mid A) > 0, gives P(A∣B)=1\mathbb{P}(A \mid B) = 1 for every base rate P(A)>0\mathbb{P}(A) > 0.

In plain words What a positive signal means depends on how common the event was to begin with. A signal that is nine times as likely among defaults as among nondefaults multiplies the odds of default by nine, and when the starting odds are small, nine times small can still be small.

Example The credit signal has likelihood ratio 0.9/0.1=90.9/0.1 = 9. With a base rate of 0.020.02 the prior odds are 0.02/0.98=1/490.02/0.98 = 1/49, the posterior odds are 9/499/49 and the posterior is 9/58≈0.15529/58 \approx 0.1552. Keep the signal and raise the base rate to 0.20.2. The joint masses with a positive signal become 0.2⋅0.9=0.180.2 \cdot 0.9 = 0.18 and 0.8⋅0.1=0.080.8 \cdot 0.1 = 0.08, and

P(D∣+)=0.180.26=913≈0.6923.\begin{aligned} \mathbb{P}(D \mid +) &= \frac{0.18}{0.26} \\[0.4em] &= \frac{9}{13} \approx 0.6923 . \end{aligned}

In odds form, the prior odds 14\tfrac14 become the posterior odds 94\tfrac94. The signal behaves the same way within each group as before, but the groups now have different sizes, and that alone moves the posterior from about 15.515.5 percent to about 69.269.2 percent.

Check yourself

  1. In the credit model with base rate 0.020.02, find the probability of default given a negative signal. Then average the two conditional default probabilities, after a positive and after a negative signal, with the probabilities of the two signals, and compare the result with the base rate.
Show the answer

The negative-signal states are ω2\omega_2 and ω4\omega_4, with total mass P(−)=0.002+0.882=0.884\mathbb{P}(-) = 0.002 + 0.882 = 0.884. So

P(D∣−)=0.0020.884=1442≈0.00226,\begin{aligned} &\mathbb{P}(D \mid -) = \frac{0.002}{0.884} \\[0.4em] &= \frac{1}{442} \approx 0.00226 , \end{aligned}

and a negative signal lowers the default probability from 0.020.02 to about 0.00230.0023. Weighting the two conditional probabilities by the signal probabilities gives

0.116⋅958+0.884⋅1442=0.018+0.002=0.02,\begin{aligned} &0.116 \cdot \tfrac{9}{58} + 0.884 \cdot \tfrac{1}{442} \\[0.4em] &= 0.018 + 0.002 = 0.02, \end{aligned}

which is the base rate, as the law of total probability requires. The conditional default probability given the signal is a random variable, equal to 958\tfrac{9}{58} after a positive signal and 1442\tfrac{1}{442} after a negative one, and its mean is the unconditional probability.

  1. An analyst describes what a loss reviewer knows by the collection {∅,{ω1},{ω1,ω2},Ω}\{\varnothing, \{\omega_1\}, \{\omega_1, \omega_2\}, \Omega\} of events of the credit model. Is it a sigma algebra? If not, find the smallest sigma algebra that contains it and its partition, and say whether the reviewer always knows if default occurred and if the signal was positive.
Show the answer

It is not a sigma algebra. The complement {ω2,ω3,ω4}\{\omega_2, \omega_3, \omega_4\} of {ω1}\{\omega_1\} is missing, and so is the complement {ω3,ω4}\{\omega_3, \omega_4\} of {ω1,ω2}\{\omega_1, \omega_2\}. Any sigma algebra that contains {ω1}\{\omega_1\} and {ω1,ω2}\{\omega_1, \omega_2\} also contains their difference {ω2}\{\omega_2\} and the complement {ω3,ω4}\{\omega_3, \omega_4\}. These three sets partition Ω\Omega, so the smallest such sigma algebra is generated by the partition into {ω1}\{\omega_1\}, {ω2}\{\omega_2\} and {ω3,ω4}\{\omega_3, \omega_4\}. It has 23=82^3 = 8 events, namely ∅\varnothing, the three blocks, the three unions of two blocks {ω1,ω2}\{\omega_1, \omega_2\}, {ω1,ω3,ω4}\{\omega_1, \omega_3, \omega_4\} and {ω2,ω3,ω4}\{\omega_2, \omega_3, \omega_4\}, and Ω\Omega. The default event {ω1,ω2}\{\omega_1, \omega_2\} is a union of blocks, so the reviewer always knows whether default occurred. The positive-signal event {ω1,ω3}\{\omega_1, \omega_3\} is not, since it splits the block {ω3,ω4}\{\omega_3, \omega_4\}. The reviewer learns the signal after a default but not after a nondefault.

  1. A lender’s borrowers fall into rating classes R1R_1, R2R_2 and R3R_3 with base rates 0.50.5, 0.30.3 and 0.20.2, and in a constructed model their one-year default probabilities are 0.010.01, 0.030.03 and 0.100.10. A borrower defaults. Find the probability that the borrower came from each class. A colleague answers 0.10/(0.01+0.03+0.10)≈0.7140.10/(0.01 + 0.03 + 0.10) \approx 0.714 for class R3R_3. What went wrong?
Show the answer

The classes partition the borrowers into events of positive probability, so the law of total probability gives

P(D)=0.5⋅0.01+0.3⋅0.03+0.2⋅0.10=0.034.\begin{aligned} &\mathbb{P}(D) \\[0.4em] &= 0.5 \cdot 0.01 + 0.3 \cdot 0.03 \\[0.4em] &\quad + 0.2 \cdot 0.10 = 0.034 . \end{aligned}

The joint masses are 0.0050.005, 0.0090.009 and 0.0200.020, and Bayes’ theorem divides each by 0.0340.034. The posteriors are 534≈0.147\tfrac{5}{34} \approx 0.147 for R1R_1, 934≈0.265\tfrac{9}{34} \approx 0.265 for R2R_2 and 2034=1017≈0.588\tfrac{20}{34} = \tfrac{10}{17} \approx 0.588 for R3R_3, and they add to 11. The colleague normalized the three default probabilities as if the classes were equally common, which would be right only with base rates of 13\tfrac13 each. Class R3R_3 is the riskiest, but it holds only a fifth of the borrowers, so its share of defaults is about 0.5880.588, not 0.7140.714.

How this lesson connects