Suppose $A$, and $B$ are events with

$$\Pr(A)=3/12,\quad \Pr(B)=4/12,\quad\Pr(A\cap B)=0$$

Suppose $A’$ and $B’$ are events with

$$\Pr(A’)=3/12,\quad\Pr(B’)=8/12,\quad\Pr(A’\cap B’)=1/12$$

Notice that the covariance matrix $M$ for the Bernoulli random variables $1_A$, $1_B$ is the same as the one for $1_{A’}$, $1_{B’}$.

Now suppose we wanted to take any given joint distribution giving covariance matrix $M$ and extend it to the covariance matrix for $A$, $B$, $C$, where $\Pr(C)=\Pr(C\setminus(A\cup B))=5/12$.

We claim this is impossible if we are given the joint distribution of $A’$ and $B’$. That is, we claim there is no choice of probabilities concerning $C’$ that will give the right covariance matrix.

Note that $\Pr(C’)\in\{5/12, 7/12\}$ is necessary since $\mathrm{Var}(1_C)=\mathrm{Var}(1_{C’})$ is necessary. Moreover

$$0-(3/12)(5/12)=\mathrm{Cov}(A,C)=\mathrm{Cov}(A’,C’)=E(1_{A’}1_{C’})-E(1_{A’})E(1_{C’})=\Pr(A’\cap C’)-(3/12)(5/12\text{ or }7/12)$$

so

$$\Pr(A’\cap C’)=(3/12)(0\text{ or }2/12)=0\text{ or }6/144$$

Similarly for $B’$,

$$0-(4/12)(5/12)=\mathrm{Cov}(B,C)=\mathrm{Cov}(B’,C’)=E(1_{B’}1_{C’})-E(1_{B’})E(1_{C’})=\Pr(B’\cap C’)-(8/12)(5/12\text{ or }7/12)$$

so

$$\Pr(B’\cap C’)=-20/144+(40\text{ or }56)/144= (20\text{ or }36)/144$$

The choice $\Pr(C’)=5/12$. $\Pr(A’\cap C’)=0$, $\Pr(B’\cap C’)=20/144$ gives $\Pr(A’\cup B’\cup C’)=156/144>1$, contradiction.

The other choice $\Pr(C’)=7/12$, $\Pr(A’\cap C’)=6/144$, $\Pr(B’\cap C’)=36/144$ gives $\Pr(C’\setminus(A’\cup B’))\ge 84/144-6/144-36/144=42/144=7/24$, so $\Pr(A’\cup B’\cup C’)\ge 10/12 + 7/24 > 1$, also contradiction.

Q.E.D.