Toss two fair coins and consider three events: \(A\) , the first coin shows heads; \(B\) , the second coin shows heads; \(C\) , the two coins show different faces. Each has probability \(1/2\) , and each pair satisfies the product rule of a first course: \(\mathbb P(A\cap B)=\mathbb P(A\cap C)=\mathbb P(B\cap C)=1/4\) . By that rule the three events are independent two at a time. Yet whoever knows whether \(A\) and \(B\) occurred knows whether \(C\) occurred. Whatever independence of a family means, it is not a property of its pairs.

The example raises the three questions of this chapter. What is the right definition of independence for a family of random elements, when there may be no densities to multiply and no point masses to enumerate? How can such a definition ever be verified, since it will have to quantify over all events that the random elements can express? And do independent sequences with prescribed laws exist at all, or is the phrase “let \(X_1,X_2,\dots\) be independent with distribution function \(F\) ” a hope and not a hypothesis?

2.1 Independence as a property of information

The definition is about sigma-algebras, that is, about information. Independence of \(X\) and \(Y\) says that every event decidable from \(X\) is independent of every event decidable from \(Y\) : \(\mathbb P(X\in B,\,Y\in C)=\mathbb P(X\in B)\,\mathbb P(Y\in C)\) for all measurable \(B\) and \(C\) . The index set may be infinite, but only finitely many indices enter any one product, so independence of an infinite family is independence of each of its finite subfamilies. Notice also that the product rule is required for every finite subfamily, not only for the full family; for events this amounts to one equation for each subset of the indices (Exercise 1).

One consequence costs nothing. If \(X_i\) are independent random elements and \(f_i\) are measurable maps, then the \(f_i(X_i)\) are independent, because \(\sigma(f_i(X_i))\subseteq\sigma(X_i)\) and subclasses of independent classes are independent.

2.2 Verifying independence on a generating class

Definition 2.1 quantifies over all events of each sigma-algebra. For a random variable, \(\sigma(X)\) contains the preimage of every Borel set, and no one can check a product rule Borel set by Borel set. The theorem that makes the definition usable says that it is enough to check on a generating class, provided the class is closed under intersections. Recall from Chapter 1 that such a class is a \(\pi\) -system, and that two finite measures with the same total mass which agree on a \(\pi\) -system agree on the sigma-algebra it generates.

The hypothesis that the classes be \(\pi\) -systems cannot be dropped; Exercise 18 gives a four-point counterexample. Two corollaries turn the theorem into working tools.

Grouping is used constantly and usually silently. If \(X_1,X_2,X_3,X_4\) are independent, then \(X_1+X_2\) and \(X_3X_4\) are independent, because the first is measurable with respect to \(\sigma(X_1,X_2)\) and the second with respect to \(\sigma(X_3,X_4)\) . If \(X_1,X_2,\dots\) are independent, then for each \(n\) the past \(\sigma(X_1,\dots,X_n)\) is independent of the future \(\sigma(X_{n+1},X_{n+2},\dots)\) ; the zero-one law of Chapter 3 is built on this.

2.3 Independence is product measure

The product of two measurable spaces, the product sigma-algebra \(\mathcal S\otimes\mathcal T\) generated by the measurable rectangles \(B\times C\) , and the product \(\mu\otimes\nu\) of two probability measures are constructions of Measure Theory. With them, independence becomes a statement about a single measure.

The gain is that everything Measure Theory proves about product measures becomes a statement about independent random elements. The most important such result is the pair of theorems of Tonelli and Fubini: an integral against a product measure may be computed as an iterated integral, in either order, when the integrand is nonnegative (Tonelli) or integrable (Fubini).

The inner expression \(\mathbb E\,h(x,Y)\) has a useful reading: freeze the first variable at the value \(x\) , average over the second, and only then average over \(x\) . Chapter 6 will recognise it as a conditional expectation.

The product formula characterizes independence. If \(\mathbb E[f(X)g(Y)]=\mathbb E f(X)\,\mathbb E g(Y)\) for all bounded measurable \(f\) and \(g\) , then taking indicators \(f=\mathbf 1_B\) , \(g=\mathbf 1_C\) recovers the definition. What does not characterize independence is the product formula for one particular pair of functions, and the most common particular pair is \(f(x)=x\) , \(g(y)=y\) .

These are the covariance facts of a first course (math.probability-theory.covariance-independence), now with their measure-theoretic justification. The second statement needs far less than independence, a point that returns in the laws of large numbers.

2.4 Existence of independent sequences

Theorem 1.9 produced, on the unit interval, a random variable with any prescribed distribution function. For independent sequences with prescribed laws the general tool is the infinite product measure, which Measure Theory constructs by extension from finite-dimensional rectangles. For real-valued sequences nothing beyond Lebesgue measure on the unit interval is needed, and the construction is explicit enough to be worth seeing. It starts from the binary digits of a point.

For \(\omega\in(0,1)\) and \(k\ge 1\) let \(D_k(\omega)\) be the \(k\) -th binary digit of \(\omega\) , that is, \(D_k(\omega)=\lfloor 2^k\omega\rfloor-2\lfloor 2^{k-1}\omega\rfloor\in\{0,1\}\) .

So one uniform variable contains infinitely many independent uniform variables, and through the quantile transform infinitely many independent variables with any laws whatever. From now on “let \(X_1,X_2,\dots\) be independent with laws \(\mu_1,\mu_2,\dots\) ” is a hypothesis that can always be met. For random elements of spaces more general than the line the corresponding statement is the existence of infinite product measures, which we take from Measure Theory; nothing in this book depends on the particular space on which a sequence is realized, only on its joint law.

Exercises

1. Prove that events \(A_1,\dots,A_n\) are independent in the sense of Definition 2.1 if and only if \(\mathbb P(\bigcap_{i\in J}A_i)=\prod_{i\in J}\mathbb P(A_i)\) for every nonempty \(J\subseteq\{1,\dots,n\}\) , and count how many of these equations are not automatic.

   Hint

   The sigma-algebra \(\{\emptyset,A_i,A_i^{c},\Omega\}\) is generated by the \(\pi\) -system \(\{A_i,\Omega\}\) , so Theorem 2.2 reduces independence to the product rule for sets chosen from \(\{A_i,\Omega\}\) , which is the stated family of equations. Those with \(\lvert J\rvert=1\) are automatic, leaving \(2^n-n-1\) .

2. On \(\Omega=\{1,\dots,8\}\) with the uniform probability let \(A=\{1,2,3,4\}\) , \(B=\{1,2,3,5\}\) , and \(C=\{1,6,7,8\}\) . Verify that \(\mathbb P(A\cap B\cap C)=\mathbb P(A)\mathbb P(B)\mathbb P(C)\) and decide whether the three events are independent.

   Answer

   \(A\cap B\cap C=\{1\}\) has probability \(1/8=(1/2)^3\) , so the triple product rule holds. The events are not independent: \(A\cap B=\{1,2,3\}\) has probability \(3/8\ne 1/4\) . Also \(\mathbb P(A\cap C)=1/8\ne 1/4\) .

3. Prove that a random variable \(X\) is independent of itself if and only if it is almost surely constant.

   Solution

   Suppose \(X\) is independent of itself, and let \(F\) be its distribution function. For each real \(x\) , the event \(\{X\le x\}\) belongs to \(\sigma(X)\) and is therefore independent of itself: \(F(x)=\mathbb P(\{X\le x\}\cap\{X\le x\})=F(x)^2\) . So \(F(x)\in\{0,1\}\) for every \(x\) . Let \(c=\inf\{x:F(x)=1\}\) . Since \(F\to 0\) at \(-\infty\) and \(F\to 1\) at \(+\infty\) and \(F\) takes only the values \(0\) and \(1\) , the number \(c\) is finite. Right-continuity gives \(F(c)=1\) , and \(F(x)=0\) for \(x<c\) gives \(\mathbb P(X<c)=F(c-)=0\) . Hence \(\mathbb P(X=c)=F(c)-F(c-)=1\) .

   Conversely, if \(X=c\) almost surely, every event \(\{X\in B\}\) has probability \(0\) or \(1\) . An event \(A\) with \(\mathbb P(A)\in\{0,1\}\) satisfies \(\mathbb P(A\cap E)=\mathbb P(A)\,\mathbb P(E)\) for every event \(E\) : both sides are \(0\) if \(\mathbb P(A)=0\) , and both equal \(\mathbb P(E)\) if \(\mathbb P(A)=1\) . So \(\sigma(X)\) is independent of every sigma-algebra, including itself.

4. Let \(X\) and \(Y\) be independent random variables such that \(X+Y\) is almost surely equal to a constant \(c\) . Prove that \(X\) and \(Y\) are both almost surely constant, using the previous exercise.

   Hint

   Almost surely \(Y=c-X\) , so \(\sigma(Y)\) and \(\sigma(X)\) contain the same events up to null sets. Independence of \(X\) and \(Y\) therefore makes \(X\) independent of itself, and Exercise 3 applies.

5. Let \((X,Y)\) be uniformly distributed on the unit disc. Show that \(X\) and \(Y\) are uncorrelated and not independent, and that the polar coordinates \(R=\sqrt{X^2+Y^2}\) and \(\Theta\in[0,2\pi)\) are independent, by computing \(\mathbb P(R\le r,\Theta\le\theta)\) .

   Solution

   The joint law is normalized area on the disc, which is symmetric under \((x,y)\mapsto(-x,y)\) . Hence \(\mathbb E X=0\) and \(\mathbb E[XY]=0\) , so \(\operatorname{Cov}(X,Y)=0\) . The coordinates are not independent: the events \(\{X>0.8\}\) and \(\{Y>0.8\}\) each have positive probability, but they cannot occur together because \(0.8^2+0.8^2>1\) , so the probability of their intersection is \(0\) , not the product.

   For the polar coordinates, fix \(0\le r\le 1\) and \(0\le\theta\le 2\pi\) . The event \(\{R\le r,\Theta\le\theta\}\) is a circular sector of radius \(r\) and opening angle \(\theta\) , with area \(r^2\theta/2\) . Dividing by the area \(\pi\) of the disc,

   \[ \mathbb P(R\le r,\Theta\le\theta)=\frac{r^2\theta}{2\pi}=r^2\cdot\frac{\theta}{2\pi}. \]

   Setting \(\theta=2\pi\) gives \(\mathbb P(R\le r)=r^2\) , and setting \(r=1\) gives \(\mathbb P(\Theta\le\theta)=\theta/(2\pi)\) . The joint distribution function is the product of the marginal ones for all arguments (outside these ranges both sides are \(0\) or reduce to a marginal), so \(R\) and \(\Theta\) are independent by Corollary 2.3. The dependence between \(X\) and \(Y\) was an artefact of the coordinates.

6. Let \(X\) and \(Y\) be independent exponential variables with rate \(1\) . Show that \(X+Y\) and \(X/(X+Y)\) are independent and identify their laws.

   Hint

   Compute \(\mathbb E\,h(X+Y,\,X/(X+Y))\) for bounded measurable \(h\) by Theorem 2.6 as a double integral against \(e^{-x-y}\) , and change variables to \(s=x+y\) , \(v=x/(x+y)\) , with Jacobian \(s\) . The integrand becomes \(h(s,v)\,se^{-s}\) on \((0,\infty)\times(0,1)\) , which is a product: the sum has density \(se^{-s}\) and the ratio is uniform on \((0,1)\) .

7. Let \(X_1,X_2,X_3\) be independent exponential variables with rates \(a_1,a_2,a_3\) . Use Theorem 2.6 to find the probability that \(X_1\) is the smallest of the three and the law of the minimum.

   Answer

   \(\mathbb P(X_1<X_2\wedge X_3)=\int_0^\infty e^{-(a_2+a_3)x}a_1e^{-a_1x}\,dx=a_1/(a_1+a_2+a_3)\) , using that \(X_2\wedge X_3\) is independent of \(X_1\) by grouping. The minimum is exponential with rate \(a_1+a_2+a_3\) .

8. Let \(X\) and \(Y\) be independent random variables and suppose the distribution function of \(X\) is continuous. Prove that \(\mathbb P(X=Y)=0\) , and show by example that independence cannot be dropped.

   Solution

   The diagonal \(\{(x,y):x=y\}\) is a closed subset of the plane, so \(h(x,y)=\mathbf 1_{\{x=y\}}\) is a nonnegative measurable function and Theorem 2.6 applies:

   \[ \mathbb P(X=Y)=\mathbb E\,h(X,Y)=\int_{\mathbb R}\mathbb E\,h(X,y)\,\mathbb P_Y(dy)=\int_{\mathbb R}\mathbb P(X=y)\,\mathbb P_Y(dy). \]

   Here the roles of the two variables in the theorem have been exchanged, which is legitimate since the product measure may be integrated in either order. By Proposition 1.7, \(\mathbb P(X=y)=F_X(y)-F_X(y-)\) , which is \(0\) for every \(y\) because \(F_X\) is continuous. The integrand vanishes identically, so \(\mathbb P(X=Y)=0\) . Nothing was assumed about the law of \(Y\) .

   Without independence the statement is false: take \(Y=X\) with \(X\) uniform on \((0,1)\) . Then \(F_X\) is continuous and \(\mathbb P(X=Y)=1\) .

9. Find the density of the sum of two independent exponential variables with rate \(1\) , and the law of the sum of independent Poisson variables with means \(a\) and \(b\) , in each case from the convolution formula.

   Answer

   For the exponentials, \(\int_0^ze^{-(z-y)}e^{-y}\,dy=ze^{-z}\) for \(z>0\) . For the Poisson variables, \(\sum_{j=0}^{n}e^{-a}\tfrac{a^{j}}{j!}e^{-b}\tfrac{b^{n-j}}{(n-j)!}=e^{-(a+b)}\tfrac{(a+b)^n}{n!}\) by the binomial theorem, so the sum is Poisson with mean \(a+b\) .

10. Let \(X_1,X_2,X_3,X_4\) be independent random variables. Prove carefully that \(X_1+X_2\) and \(X_3X_4\) are independent, naming the \(\pi\) -systems involved and the result that passes from them to the generated sigma-algebras.

    Solution

    Let \(\mathcal G_i=\sigma(X_i)\) for \(i=1,2,3,4\) ; these are independent by hypothesis. Group the indices as \(I_1=\{1,2\}\) and \(I_2=\{3,4\}\) and let \(\mathcal D_1=\{A_1\cap A_2:A_1\in\mathcal G_1,A_2\in\mathcal G_2\}\) and \(\mathcal D_2=\{A_3\cap A_4:A_3\in\mathcal G_3,A_4\in\mathcal G_4\}\) . Each is a \(\pi\) -system, and for \(A_1\cap A_2\in\mathcal D_1\) , \(A_3\cap A_4\in\mathcal D_2\) ,

    \[ \mathbb P(A_1\cap A_2\cap A_3\cap A_4)=\prod_{i=1}^4\mathbb P(A_i)=\mathbb P(A_1\cap A_2)\,\mathbb P(A_3\cap A_4), \]

    using independence of all four sigma-algebras and then of the pairs. So \(\mathcal D_1\) and \(\mathcal D_2\) are independent \(\pi\) -systems, and Theorem 2.2 makes \(\mathcal H_1=\sigma(\mathcal D_1)=\sigma(X_1,X_2)\) and \(\mathcal H_2=\sigma(\mathcal D_2)=\sigma(X_3,X_4)\) independent; this is Corollary 2.4. The map \(\omega\mapsto(X_1(\omega),X_2(\omega))\) is an \(\mathcal H_1\) -measurable random vector, and addition is continuous, so \(X_1+X_2\) is \(\mathcal H_1\) -measurable; likewise \(X_3X_4\) is \(\mathcal H_2\) -measurable. Hence \(\sigma(X_1+X_2)\subseteq\mathcal H_1\) and \(\sigma(X_3X_4)\subseteq\mathcal H_2\) , and subclasses of independent classes are independent.

11. A student claims that for the variables \(U,V,W\) of Example 2.1, pairwise independence gives \(\mathbb E[UVW]=\mathbb E U\,\mathbb E V\,\mathbb E W\) . Compute both sides, and say exactly which hypothesis of which result the student has replaced by a weaker one.

    Solution

    The product \(UVW\) is nonzero only when all three variables equal \(1\) . But \(U=V=1\) forces \(W=0\) , so \(UVW=0\) at every sample point and \(\mathbb E[UVW]=0\) . On the other side, \(\mathbb E U\,\mathbb E V\,\mathbb E W=\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18\) . The two sides differ.

    The student has used the product formula of Theorem 2.6 for three factors. Applied twice, that theorem needs \(U\) to be independent of the pair \((V,W)\) , so that \(\mathbb E[U\cdot(VW)]=\mathbb E U\,\mathbb E[VW]\) , and then \(V\) independent of \(W\) . The second requirement holds. The first does not: \(U\) is independent of \(V\) and of \(W\) separately, but it is a function of the pair \((V,W)\) , since \(U=V\) when \(W=0\) and \(U=1-V\) when \(W=1\) . Pairwise independence has been substituted for independence of the family, which by grouping is what gives independence of \(U\) from \(\sigma(V,W)\) . The correct intermediate value is \(\mathbb E[VW]=\tfrac14\) and \(\mathbb E[U\cdot VW]=0\) .

12. With \(D_k\) the binary digits of a uniform point, let \(R_k=1-2D_k\) . Show that \(\mathbb E[R_jR_k]\) equals \(1\) if \(j=k\) and \(0\) otherwise, and find the law of \(\sum_{k\ge 1}R_k2^{-k}\) .

    Hint

    \(R_k^2=1\) , and for \(j\ne k\) independence gives \(\mathbb E[R_jR_k]=\mathbb E R_j\,\mathbb E R_k=0\) . Since \(\sum_kR_k2^{-k}=1-2\sum_kD_k2^{-k}=1-2U\) , the sum is uniform on \((-1,1)\) by Lemma 2.9.

13. Suppose \(\mathbb E[f(X)g(Y)]=\mathbb E f(X)\,\mathbb E g(Y)\) for all bounded continuous \(f,g:\mathbb R\to\mathbb R\) . Prove that \(X\) and \(Y\) are independent.

    Solution

    Fix real numbers \(x\) and \(y\) . For \(n\ge 1\) let \(f_n\) be the continuous function equal to \(1\) on \((-\infty,x]\) , equal to \(0\) on \([x+1/n,\infty)\) , and linear in between, and define \(g_n\) in the same way with \(y\) in place of \(x\) . Then \(0\le f_n,g_n\le 1\) , and as \(n\to\infty\) , \(f_n(s)\to\mathbf 1_{(-\infty,x]}(s)\) for every \(s\) : if \(s\le x\) the value is always \(1\) , and if \(s>x\) it is \(0\) as soon as \(1/n<s-x\) . The same holds for \(g_n\) . By hypothesis,

    \[ \mathbb E[f_n(X)g_n(Y)]=\mathbb E f_n(X)\,\mathbb E g_n(Y)\qquad\text{for every }n . \]

    All integrands are bounded by \(1\) and converge pointwise, so the Dominated Convergence Theorem applies to each of the three expectations and gives \(\mathbb P(X\le x,Y\le y)=\mathbb P(X\le x)\,\mathbb P(Y\le y)\) . Since \(x\) and \(y\) were arbitrary, Corollary 2.3 shows that \(X\) and \(Y\) are independent. The argument illustrates a general principle: a class of test functions is enough to detect independence as soon as indicators of a generating \(\pi\) -system are bounded pointwise limits of its members.

14. Let \(U_1,\dots,U_n\) be independent and uniform on \((0,1)\) , with minimum \(m\) and maximum \(M\) . Find \(\mathbb P(M\le x)\) and \(\mathbb P(m>x)\) , and decide whether \(m\) and \(M\) are independent.

    Answer

    \(\mathbb P(M\le x)=x^n\) and \(\mathbb P(m>x)=(1-x)^n\) for \(0\le x\le 1\) . They are not independent: \(\mathbb P(m>\tfrac12,M\le\tfrac12)=0\) while both factors are \(2^{-n}>0\) .

15. Let \(U\) be uniform on \((0,1)\) , let \(D_1\) be its first binary digit, and let \(U'=2U-D_1\) . Prove directly from Corollary 2.3 and Theorem 2.2 that \(D_1\) and \(U'\) are independent and that \(U'\) is uniform on \([0,1)\) .

    Solution

    The digit is \(D_1=\mathbf 1_{\{U\ge 1/2\}}\) , and \(U'=2U\) on \(\{D_1=0\}\) , \(U'=2U-1\) on \(\{D_1=1\}\) , so \(0\le U'<1\) . Fix \(d\in\{0,1\}\) and \(0\le x<1\) . Then \(D_1=d\) and \(U'\le x\) hold together exactly when \(d/2\le U\le(d+x)/2\) , an interval of length \(x/2\) . Therefore

    \[ \mathbb P(D_1=d,\,U'\le x)=\frac x2=\mathbb P(D_1=d)\cdot x . \]

    Summing over \(d=0,1\) gives \(\mathbb P(U'\le x)=x\) , so \(U'\) is uniform on \([0,1)\) , and the display becomes \(\mathbb P(D_1=d,U'\le x)=\mathbb P(D_1=d)\,\mathbb P(U'\le x)\) . The classes \(\{\{D_1=0\},\{D_1=1\},\emptyset\}\) and \(\{\{U'\le x\}:x\in\mathbb R\}\) are \(\pi\) -systems generating \(\sigma(D_1)\) and \(\sigma(U')\) , and the product rule holds between them (for \(x\ge 1\) or \(x<0\) it is trivial), so Theorem 2.2 gives independence. Iterating, \(U'\) carries the digits \(D_2,D_3,\dots\) of \(U\) , which is another proof of Proposition 2.8.

16. Let \(X\) and \(Y\) be independent with the same law and finite variance \(\sigma^2\) . Show that \(\mathbb E(X-Y)^2=2\sigma^2\) , and find the variance of the average of \(n\) pairwise uncorrelated variables with common variance \(\sigma^2\) .

    Answer

    \(\mathbb E(X-Y)^2=\operatorname{Var}(X-Y)=\operatorname{Var}X+\operatorname{Var}Y=2\sigma^2\) , since \(\mathbb E(X-Y)=0\) . The average of \(n\) pairwise uncorrelated variables with variance \(\sigma^2\) has variance \(\sigma^2/n\) .

17. A sigma-algebra \(\mathcal T\) is called trivial if \(\mathbb P(A)\in\{0,1\}\) for every \(A\in\mathcal T\) . Prove that a trivial sigma-algebra is independent of every sigma-algebra, and that a sigma-algebra independent of itself is trivial.

    Hint

    If \(\mathbb P(A)\in\{0,1\}\) then \(\mathbb P(A\cap E)=\mathbb P(A)\mathbb P(E)\) for every event \(E\) ; check the two cases. If \(\mathcal T\) is independent of itself then \(\mathbb P(A)=\mathbb P(A\cap A)=\mathbb P(A)^2\) for \(A\in\mathcal T\) .

18. On \(\Omega=\{1,2,3,4\}\) with the uniform probability let \(\mathcal C_1=\{\{1,2\},\{2,3\}\}\) and \(\mathcal C_2=\{\{1,3\}\}\) . Show that \(\mathcal C_1\) and \(\mathcal C_2\) are independent classes while \(\sigma(\mathcal C_1)\) and \(\sigma(\mathcal C_2)\) are not, and identify the hypothesis of Theorem 2.2 that fails.

    Solution

    Write \(A=\{1,2\}\) , \(B=\{2,3\}\) , \(C=\{1,3\}\) . Each has probability \(1/2\) . The classes \(\mathcal C_1=\{A,B\}\) and \(\mathcal C_2=\{C\}\) are independent, because the only products to check are

    \[ \mathbb P(A\cap C)=\mathbb P\{1\}=\tfrac14=\tfrac12\cdot\tfrac12,\qquad \mathbb P(B\cap C)=\mathbb P\{3\}=\tfrac14=\tfrac12\cdot\tfrac12 . \]

    However \(\sigma(\mathcal C_1)\) contains \(A\cap B=\{2\}\) , and \(\mathbb P(\{2\}\cap C)=\mathbb P(\emptyset)=0\) , whereas \(\mathbb P\{2\}\,\mathbb P(C)=\tfrac14\cdot\tfrac12=\tfrac18\) . So \(\sigma(\mathcal C_1)\) and \(\sigma(\mathcal C_2)\) are not independent. In fact \(\sigma(\mathcal C_1)\) is the whole power set, since it contains \(\{2\}\) , \(\{1\}=A\setminus\{2\}\) , and \(\{3\}=B\setminus\{2\}\) , and no event of probability \(1/2\) is independent of every subset.

    The hypothesis that fails is that \(\mathcal C_1\) be a \(\pi\) -system: \(A\cap B\notin\mathcal C_1\) . In the proof of Theorem 2.2 the two measures \(\mu(E)=\mathbb P(E\cap C)\) and \(\nu(E)=\mathbb P(E)\mathbb P(C)\) agree on \(A\) , \(B\) , and \(\Omega\) , and the uniqueness theorem is silent because the class on which they agree is not closed under intersection.