# What is the Asymptotic Equipartition Property (AEP)?

I am currently studying about Polar Codes in 5G standard and while reading my paper I found something called AEP which is required for channel coding. I surfed the web but didn't found a satisfying answer. Can someone explain what it is, clearly?

When you are sampling a stochastic process $$n$$ times, the larger you make $$n$$, the higher the probability that the series of samples is contained in the so called strongly typical set of outcomes of length $$n$$, where all members have roughly the same probability to be realized. If you let $$n\rightarrow \infty$$ it holds that $$-\frac{1}{n}\log p(X_1,X_2,...,X_n)\rightarrow H(X)$$ with $$H(X)$$ being the entropy rate of the process.
This should be intuitively clear: the longer the sequence, the more possible outcomes, each with possiblity near zero but positive. The number of typical outcomes with $$p\approx\epsilon$$ is growing faster than the number of non typical outcomes as $$n$$ is growing. As $$n\rightarrow \infty$$, the "leftover probability" for non typical outcomes gets smaller and smaller, thus making the above equation valid.
• What is $\frac{1}{n}$, is it the probability of equiprobable random variables $X_{i}$? And also, if $H(X)$ is the entropy, shouldn't there be a summation over here which would be equivalent of the expectation of the information contained? – Himanshu Sharma Apr 9 '19 at 14:29
• I'm sorry, $H(X)$ is the entropy rate, I corrected it in my posting. That's where the $\frac{1}{n}$ is coming from. – Max Apr 10 '19 at 5:35
• "The number of typical outcomes with $p≈ϵ$ is growing faster than the number of non typical outcomes as $n$ is growing." That's not true. For large $n$, the great majority of sequences are not typical. – Matt L. Dec 10 '19 at 14:04