Exercise 2.9 - Conditional independence

For question (a), the antecedent $(X\perp W|Z,Y)$ means that $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in \sigma (X)$, $\forall <!--\; FUNCTION\; APPLICATION\; -->w\in \sigma (W)$ and $\forall <!--\; FUNCTION\; APPLICATION\; -->v\in \sigma (Y,Z)$ we have

The antecedent $(X\perp Y|Z)$ can be translated to that $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in \sigma (X)$, $\forall <!--\; FUNCTION\; APPLICATION\; -->y\in \sigma (Y)$ and $\forall <!--\; FUNCTION\; APPLICATION\; -->z\in \sigma (Z)$,

What we desire to obtain is $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in \sigma (X)$, $\forall <!--\; FUNCTION\; APPLICATION\; -->w\in \sigma (W)$ and $\forall <!--\; FUNCTION\; APPLICATION\; -->z\in \sigma (Z)$,

This is correct by having $v$ in the first equation taking values in $\sigma (Z)$ solely. Since $\sigma (Z)\subset \sigma (Y,Z)$.

For question (b), we have the premises: $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in \sigma (X)$, $\forall <!--\; FUNCTION\; APPLICATION\; -->y\in \sigma (Y)$, $\forall <!--\; FUNCTION\; APPLICATION\; -->z\in \sigma (Z)$ and $\forall <!--\; FUNCTION\; APPLICATION\; -->w\in \sigma (W)$:

The desired result if $\forall <!--\; FUNCTION\; APPLICATION\; -->v\in \sigma (Z,W)$,

Let $x$ and $y$ be two disjoint events, $z$ and $w$ be another pair of disjoints, and none of $x\cap z,x\cap w,y\cap z,y\cap w$ is empty w.r.t. the underlying probability measure. Finally, let $v=w\cup z$. One can then check that the equation above does not hold for this setting, hence the deduction in (b) is false.

(b) is intuitively false. A straightforward example is a cryptography example: group signature with three participants. The group signature is a protocol for encryption/verification that ensures a series of security requirements including:

• Anyone participant solely cannot pass the verification.
• Two participants can pass the verification.

For example, let Alice and Bob each hold half of the secret key denoted by $Z$ and $W$ respectively, $Y$ denotes the ciphertext, and $X$ denotes the plaintext. Good group encryption would meet both antecedents in (b) but fails the conclusion. An example of a naive group signature is to use a quadratic function as the secret key:

And provide two different points $z=(t_1,f_1)$, $w=(t_2,f_2)$ from $f$ to Alice and Bob. Both antecedents in (b) are satisfied (by correctly translating the density) since the value of $x$ yields no information about $y$. However, given both $z$ and $w$ then $y$ is a determinstic function of $x$:

and the independence no longer holds. Note that a third participant is necessary to reveal the entire secret key $(x,y,c)$. In practice, the calculation is usually done on an algebraic field/group, e.g. the elliptic curves, to deal with the problem with the density of $x,y,c$ which is usually not uniform in the real number field.