Stochastic Independence and Causal Connection

Abstract

Assumptions of stochastic independence are crucial to statistical models in science. Under what circumstances is it reasonable to suppose that two events are independent? When they are not causally or logically connected, so the standard story goes. But scientific models frequently treat causally dependent events as stochastically independent, raising the question whether there are kinds of causal connection that do not undermine stochastic independence. This paper provides one piece of an answer to this question, treating the simple case of two tossed coins with and without a midair collision.

Appendix: Further Proofs

I will show that if a midair collision between two tossed coins effects a non-deflationary linear transformation of the spin speeds, then the equivalent transformation at the beginning of the tosses will also be non-deflationary, provided that some additional, physically plausible conditions hold. I also show that under the same conditions a deflationary mid-toss collision is equivalent to a beginning-of-toss transformation that is no more deflationary.

Some new notation will simplify the algebra. Define $\hat{t}_{1}$ and $\hat{t}_2$ as the proportions of the total spin time that elapse before and after the collision, so that

$$\begin{aligned} \hat{t}_{1} = t_{1}/T \quad {\text {and}}\quad \hat{t}_2 = t_2/T.\end{aligned}$$

Then the coefficients $d$ and $e$ of the beginning-of-the-toss transformation of $u$ equivalent to the actual interaction can be written

$$\begin{aligned} d&= a\hat{t}_2 + \hat{t}_{1} \\ e&= b\hat{t}_2 \text {.}\end{aligned}$$

The same notation can be used for the coefficients of the beginning-of-the-toss transformation of $v$. If the actual linear transformation of $v$ effected by the collision is:

$$\begin{aligned} v^{\prime} = gv + hu + i \end{aligned}$$

then the beginning-of-toss transformation of $v$ equivalent to the actual interaction is

$$\begin{aligned} (g\hat{t}_2 + \hat{t}_{1})v + h\hat{t}_2 u + i{\hat{t}_2} {\text {.}}\end{aligned}$$

Consider non-deflationary collision dynamics first. The beginning-of-the-toss transformation—the transformation on the $u\,\times\,v$ space—is non-deflationary just in case the absolute value of its determinant is greater than or equal to 1. Let me simplify matters by assuming that the determinant is positive; non-deflation in that case requires that the determinant is greater than or equal to 1 (otherwise, run the following arguments with strategically placed negation operators).

The beginning-of-toss transformation’s determinant is

$$\begin{aligned} \det \left(\begin{array}{cc} a\hat{t}_2 + \hat{t}_{1} & b\hat{t}_2 \\ h\hat{t}_2 & g\hat{t}_2 + \hat{t}_{1} \end{array}\right)&= ag(\hat{t}_2)^{2}+ a\hat{t}_{1} \hat{t}_2 + g\hat{t}_{1} \hat{t}_2 + (\hat{t}_{1})^{2}- bh(\hat{t}_2)^{2}\\&= (ag - bh)(\hat{t}_2)^{2}+ \frac{(a + g)}{2}2\hat{t}_{1} \hat{t}_2 + (\hat{t}_{1})^{2}\text {.}\end{aligned}$$

Since $(\hat{t}_2)^{2}+ 2\hat{t}_{1} \hat{t}_2 + (\hat{t}_{1})^{2}= (\hat{t}_{1} + \hat{t}_2)^{2}= 1$, a sufficient condition for the determinant to be greater than or equal to 1 is that the coefficients of $(\hat{t}_2)^{2}$ and $2\hat{t}_{1} \hat{t}_2$ in the above expression be greater than or equal to 1, that is, that

• 1. $ag - bh \ge 1$, and

• 2. $a + g \ge 2$.

The quantity $ag - bh$ is the determinant of the actual transformation, that is, the transformation actually effected by the midair collision partway through the toss; it is greater than or equal to 1 just in case the original transformation is non-deflationary.

If it is assumed that $a$ and $g$ are positive (i.e., a coin’s pre-interaction spin speed makes a positive rather than a negative contribution to its post-interaction spin speed) and that (because of physical symmetry) the signs of $b$ and $h$ are the same, then (1) entails that $ag \ge 1$, which in turn entails (2). Under these assumptions, then, if the interaction transformation partway through the toss is non-deflationary, its equivalent at the beginning of the toss is also non-deflationary.

What if the mid-toss collision transformation is deflationary? Its degree of deflation is proportional to its determinant $d$, the absolute value of which will be less than one. Assume as before for expository purposes that $d$ is positive. Then it can be shown that under the same conditions assumed in the previous paragraph, the determinant of the equivalent beginning-of-toss transformation is no less than $d$, so that the equivalent beginning-of-toss transformation is no more deflationary than the mid-toss transformation. Proof: A sufficient condition for the beginning-of-toss determinant to be greater than or equal to $d$ can be read off the argument above (using all variable names in the same way):

• 1. $ag - bh \ge d$, and

• 2. $a + g \ge 2d$.

Since $d$ is by definition equal to $ag - bh$, condition (1) is always satisfied. Supposing, as before, that $a$ and $g$ are positive and that $b$ and $h$ have the same sign so that $bh$ is positive, condition (2) is also satisfied (reasoning omitted).

Next consider the case where the coin’s spin speed changes significantly over time. As in the main text, I assume that the speed at any time is a linear function of the initial speed: there exists a function $f(t)$ (not necessarily linear) such that, after time $t$, the speed of a coin with initial speed $u$ will be $f(t) u$. Redefine $\hat{t}_{1}$ and $\hat{t}_2$ so that

$$\begin{aligned} \hat{t}_{1} = F(t_{1}) \big / F(T) \quad {\text {and} }\quad {\hat{t}}_2 = F(t_2) \big / F(T) .\end{aligned}$$

where $F(t) = \int _{0}^{t}f(x) \,dx$. Then the relevant beginning-of-the-toss transformation can be written just as I have written it above.

It can no longer be assumed that $\hat{t}_{1} + \hat{t}_2 = 1$. However, if as I suggested in the main text $f(t)$ is a non-increasing function then as demonstrated below, $\hat{t}_{1} + \hat{t}_2 \ge 1$, which is sufficient for the above reasoning to apply to the more general case, so that if the other conditions stated above apply, then (a) if the actual collision transformation is non-deflationary, the equivalent beginning-of-the-toss transformation is also non-deflationary, and (b) if the actual collision transformation is deflationary, the equivalent beginning-of-the-toss transformation is no more deflationary.

To show that $\hat{t}_{1} + \hat{t}_2 \ge 1$: because $f(t)$ is non-increasing, for $b>0$

$$\begin{aligned} \int _{0}^{a}f(t) \,dt \ge \int _{b}^{b+a}f(t) \,dt \text {.}\end{aligned}$$

It follows from the definition of $F(t)$ that

$$\begin{aligned} \hat{t}_{1} + \hat{t}_2&= \frac{F(t_{1}) + F(t_2)}{F(T)}\\&= \frac{\int _{0}^{t_{1}}f(t) \,dt + \int _{0}^{t_2}f(t) \,dt }{\int _{0}^{T}f(t) \,dt }\\&\ge \frac{\int _{0}^{t_{1}}f(t) \,dt + \int _{t_{1}}^{T}f(t) \,dt }{\int _{0}^{T}f(t) \,dt }\\&\ge 1\text {.}\end{aligned}$$