Category Archives: calculations

No modular iterative way to get joint distribution from covariance matrix

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.

ladder_chain_belt_gold_BT0002

Dirac’s belt trick as a homotopy

Paint one of the “rungs” of the belt.

(When the belt is worn by someone who is standing up, a rung will be a vertical strip.)

Take a video of the performance of the belt trick. Notice how the painted rung is rotating.
The homotopy is $H:X\times [0,1]\to Y$ where $X$ is the interval $[0,1]$ and $Y$ is $\mathrm{SO}(3)$.
$x\mapsto H(x,0)$ is the twisted belt and $x\mapsto H(x,1)$ is the straightened belt.

Each $x$ is a rung of the belt.

$H(x,t)$ is the rotation of rung $x$ at time $t$ in the video.

The rungs do not merely rotate, they are also translated in space during the belt trick. However, we can ignore the translation and focus on the rotation.
For instance, the top rung (where the buckle is) is translated but does not rotate.

hmm

Few paths, fewer words, and the maximum probability of writing 001 in a two-state Hidden Markov Model being $8/27$

The aim of this note is to give the simplest possible non-trivial calculation of the parameters of a HMM that maximize the probability of emitting a certain string.

Let $\{0,1\}$ be our alphabet.
Let $p$ be the probability of emitting 1 in state $s_0$.
Let $q$ be the probability of emitting 1 in state $s_1$.
Let $\epsilon$ be the probability of transitioning from $s_0$ to $s_1$.
Let $\delta$ be the probability of transitioning from $s_1$ to $s_0$.
Let $S(t)$ be the state after transitioning $t$ times, a random variable.
The probability of emitting the string 001 when starting in state $s_0$ is then
$$
f(p,q,\epsilon,\delta)=\Pr(001; S(1)=s_0=S(2))+\Pr(001; S(1)=s_0, S(2)=s_1)$$
$$+\Pr(001; S(1)=s_1, S(2)=s_0)+\Pr(001; S(1)=s_1=S(2))$$
$$=\overline p^2 p \overline\epsilon^2 + \overline p^2q\overline\epsilon\epsilon + \overline p\overline q p\epsilon\delta + \overline p\overline q q \epsilon\overline\delta.
$$
Which choice of parameters $p, q, \epsilon, \delta$ will maximize this probability?
To answer this we first note that
$$\frac{\partial f}{\partial\delta}=0\iff p=1, q=1, \epsilon=0\text{ or }p=q.$$
Going through these possibilities we keep finding values of $f$ bounded above by $\overline p^2p\le 4/27$:

  1. $p=1$ immediately gives $f=0$.
  2. $q=1$ gives $f=\overline p^2 p \overline\epsilon^2 + \overline p^2\overline\epsilon\epsilon=\overline p^2p\overline\epsilon.$
  3. $\epsilon=0$ gives $f=\overline p^2 p.$
  4. $p=q$ gives $f=\overline p^2 p \overline\epsilon^2 + \overline p^2p\overline\epsilon\epsilon + \overline p^2 p\epsilon\delta + \overline p^2 p (\epsilon\overline\delta)=\overline p^2p(\overline\epsilon^2 + \overline\epsilon\epsilon + \epsilon\delta + \epsilon\overline\delta)=\overline p^2p.$

We next consider boundary values for $\delta$.

  1. $\delta=0$. We may assume $p=0$, since there is no use in considering a positive probability of emitting a 1 in state $s_0$ if there is no chance of ever returning to that state. Then upon calculation of $\partial f/\partial q$ we find that $\epsilon=2\overline q$, which gives $f=2q^2\overline q$. This is maximized at $q=2/3$ which corresponds to $\epsilon=2/3$ as well, and gives a value $f=8/27>1/4$.

    This $8/27$ is decomposable as a sum of two disjoint scenarios of probability $4/27$:

    1. One is that after writing the first 0 we stay in state $s_0$, write another 0, and then transition to state $s_1$ to write a 1.
    2. The other is that after writing the first 0 we move to state $s_1$, write the 2nd zero there, and stay there to write the 3rd letter, 1.
  2. $\delta=1$. Then $f=\overline p^2 p \overline\epsilon^2 + \overline p^2q\overline\epsilon\epsilon + \overline p\overline q p\epsilon
    =\overline p(\overline p p \overline\epsilon^2 + \overline pq\overline\epsilon\epsilon +\overline q p\epsilon)$.
    Then $0=\partial f/\partial q = \overline p\epsilon(\overline p\overline\epsilon – p)$ if $p=\frac{\overline\epsilon}{1+\overline\epsilon}$. This gives $f=\frac{\overline\epsilon}{(1+\overline\epsilon)^3}$ (turns out not to depend on $q$) which is maximized for $\epsilon=1/2$ with value $f=4/27$. So we consider boundary values for $q$:

    1. $q=0$. Then $f=\overline pp(\overline p\overline\epsilon^2+\epsilon)\le \frac14\cdot 1$.
    2. $q=1$. Then $f=\overline p^2p\overline\epsilon^2+\overline p^2\overline\epsilon\epsilon$ and $\partial f/\partial\epsilon=\overline p^2(1-2\epsilon-2p\overline\epsilon)=0$ if $\overline p=1/(2\overline\epsilon)$, which gives $f=\overline p/4\le 1/4$.

Now note how much easier this is if we only consider a single path. Then clearly $1/4$ is the maximum possible, via 3 different paths, because of the presence of terms of the form $a\overline a$.
Replacing such occurrences by $1/4$ we upper bound $f$ by
$$\frac14\left(\overline p \overline\epsilon^2 + \overline p^2q + \overline q\epsilon\delta + \overline p \epsilon\overline\delta\right).
$$