# Monotonicity of min-entropy

Let $$Z^t = (Y_1,\ldots,Y_t)$$ be a sequence of random variables each taking values in $$Y$$. The random variables are not necessarily i.i.d but we know the joint distributions. i.e for every $$z = (z_1,...,z_t)$$ we know $$P^{Z^t}(z)$$

The min-entropy of a random variable $$X$$ is defined as $$-\log_2(\max P(x) )$$ for x in the value set of $$X$$.

Finally, we define a sequence of values $$H^\infty(t) = -\log_2(\max P^{Z^t}(z))$$ for $$Z^t$$ defined as above.

How can we show that $$H^\infty(t)$$ is monotonically increasing in t? i.e for $$t' \geq t$$ it is the case that $$H^\infty(t') \geq H^\infty(t)$$.

What I am trying without success is to show that $$\max P^{Z^t}(z^t) \geq \max P^{Z^{t'}}(z^{t'})$$ where $$t \leq t'$$.

• If $Y$ is not IID, then your equation for $H^\infty$ doesn't apply. It's not that easy and a common mistake. It will over estimate the entropy in some weird proportion to the auto correlation. Consider, if there's a close but diminishing relationship over many $n$ in $Y_n$, what's $y$ exactly? This equation only applies to IID variables. – Paul Uszak Apr 3 '19 at 20:58
• Sorry the notation is a bit wonky... So in this case, $Y$ is the range of the random variables, the tuple $(Y_1,...,Y_t)$ would be a random variable with range $Y^t$ for which we know the joint distribution. Finally $y$ is an element of $Y^t$. In which case the min-entropy as written should be well defined.. I hope :) – Marc Ilunga Apr 3 '19 at 21:00
• But I am certain that $y \in Y$. I will edit the notation to make it clearer. Thanks and sorry to the terrible formulation :) – Marc Ilunga Apr 3 '19 at 21:17
• I'm saying that in the specific case of non IID data as you suggest, $H^\infty(t) \neq -\log_2(\max P(y))$ as long as $y \in Y$. This equation (as well as $H^{sh}$) only applies to IID variables. The real $H^\infty$ will be lower, perhaps much lower depending on the strength of the auto correlation. It's common to drop the non IID assumption in these situations. Otherwise you end up in a world of Markov chains and pains. – Paul Uszak Apr 3 '19 at 21:33
• I have modified the question to make it clearer, have a look. So I my case i.i.d is not really a concern my random variables are just normal random variables defined over a 'tuple space' so to say. In which case even Shanon entropy applies(checking on Wikipedia). Am I missing somthing? – Marc Ilunga Apr 3 '19 at 21:42

## 2 Answers

Let's make it slightly simpler by proving the case of two variables with no tuple indexing to clutter it up. Fix two random variables $$X$$ and $$Y$$. Is $$H_\infty[X] \leq H_\infty[(X, Y)]$$?

For each $$x$$, we have $$\Pr[X = x] = \sum_y \Pr[X = x, Y = y].$$ Note that since probability masses are always positive, $$\max \Pr[\cdots] \leq \sum \Pr[\cdots]$$; then the sense will get reversed because $$p \mapsto -\log p$$ is a decreasing function:

\begin{align} H_\infty[X] &= -\log \max_x \Pr[X = x] \\ &= -\log \max_x \sum_y \Pr[X = x, Y = y] \\ &\leq -\log \max_x \max_y \Pr[X = x, Y = y] \\ &= -\log \max_{x,y} \Pr[X = x, Y = y] \\ &= H_\infty[(X, Y)]. \end{align}

Then to prove that $$t \mapsto H_\infty[(Z_1, \dots, Z_t)]$$ is increasing, take $$X = (Z_1, \dots, Z_t)$$ and $$Y = Z_{t+1}$$.

• Accepted Answer for simplicity, clarity and didactic value. :) – Marc Ilunga Apr 20 '19 at 6:58

I think I've found the solution.

Let $$z = (z_1,\ldots,z_{t+1})$$ be the value that maximizes $$P^{Z^{t+1}}[\cdot]$$ and the probability is $$Pz$$. Let $$x = (x_1,\ldots,x_t)$$ the value that maximizes $$P^{Z^{t}}[\cdot]$$ and the probability is $$P_x$$; finally let $$Py = \sum_{l \in Y} Pr[(z_1,\ldots, z_t,l)]$$. $$Py$$ is then the probability of $$y = (z_1,\ldots,z_t)$$ i.e the first $$t$$ elements of $$z$$.

It follows then that $$P_z \leq P_y \leq P_x$$.

Which shows that $$H^\infty(t)$$ is monotonically increasing.