0
$\begingroup$

The question is the following:

Given two discrete iid random variables $X_1$ and $X_2$ with a probability distribution $p$, Show that: $P(X_1=X_2) \geq \frac{1}{2^{H(p)}}$. Also show that this is an equality iff $p$ is uniform.

Using these lecture notes, I was able to reach this (trivial) step:

$$ P(X_1 = X_2) \geq 2^{-H(p)} $$ $$ P(X_1 = X_2) \geq 2^{\mathbb{E}[\log(p)]} $$

I really don't know where to go from here, How do I get rid of the $\log (p)$? Is this related to Conditional Entropy or Relative Entropy at all? I've seen lots of resources on that online but couldn't relate that information to this question.

I don't even understand this on a basic level. Intuitively, if $p$ is discrete and uniform with $n$ possible values, then $P(X_1=X_2) = \frac{1}{n}$. I see no way in which the formula can be manipulated to transform $2^{\mathbb{E}[\log(p)]}$ into $\frac{1}{n}$.

Feeling like an idiot and I'd love some guidance.

$\endgroup$
  • 1
    $\begingroup$ If $p$ is uniform then $H(p)=\log_2(n)$. $\endgroup$ – Stelios Nov 3 at 23:32
2
$\begingroup$

Below is an expanded version of the proof given in Elements Of Information Theory by T.M Cover (2nd Edition, pg 40 - Lemma 2.10.1) for the inequality in question.

If $X_1$ and $X_2$ are iid (i.e. $X_1 \sim p(x)$ and $X_2 \sim p(x)$ )

\begin{align} P(X_1 = X_2) &= \sum_{x \in \mathcal{X}} P(X_1 = x , X_2 = x) \\ &= \sum_{x \in \mathcal{X}} P(X_1 = x) P(X_2 = x) \\ &= \sum_{x \in \mathcal{X}} p(x) p(x) \\ &= \sum_{x \in \mathcal{X}} p^2(x) \end{align}

Using Jenson's Inequality we know that if a function, $f(y)$, is convex then $\mathbb{E}_{p(x)} \left[ f(y) \right] \geq f\left( \mathbb{E}_{p(x)}[y] \right)$, with equality if and only if $\textbf{y}$ is a constant.

The function $f(y) = 2^{y}$ is convex and so if $z = \mathbb{E}_{p(x)} \left[\phi \right]$, then

$$f(z) = f(\mathbb{E}_{p(x)} \left[\phi \right]) \leq \mathbb{E}_{p(x)} [f (\phi) ]$$

Where

$$\mathbb{E}_{p(x)} [f (\phi) ] = \sum _{x \in \mathcal {X}} p(x) f(\phi)$$

Let $\phi = \log_2 p(x)$. Note that

\begin{align} 2^{-H(X)} &= 2^{\mathbb{E}_{p(x)}[\log _2 p(x)]} \\ &= f(y) |_{y = \mathbb{E}_{p(x)}[\log _2 p(x)]} \\ &= f(y) |_{y = \mathbb{E}_{p(x)}[\phi]} \\ &= f(\mathbb{E}_{p(x)}[\phi]) \\ &\leq \mathbb{E}_{p(x)} [f (\phi) ] \\ &= \mathbb{E}_{p(x)} 2^{\phi} \\ &= \mathbb{E}_{p(x)} 2^{\log _2 p(x)} \\ &= \mathbb{E}_{p(x)} p(x) \\ &= \textstyle{\sum} p(x)p(x)\\ &= \textstyle{\sum} p^2 (x) \\ &= P(X_1 = X_2) \end{align}

What this means is $P(X_1 = X_2) \geq 2^{-H(X)}$, with equality if and only if $\log p(x)$ is a constant. If $\log p(x)$ is constant then $p(x)$ is also constant and so the distribution in that case is uniform. $\square$

$\endgroup$
  • $\begingroup$ Thank you for the reference! I will definitely go through the book! $\endgroup$ – QuantumHoneybees Nov 5 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.