Considering this setup, single tone transmission over a channel impaired by AWGN, doppler spread, and inter symbols interference.

We are to compute the BER as a function of the SNR.

We want to find the number of samples that need to be tested and the number of errors that need to be found.

What number of samples/errors need to be simulated to observe a bit-error rate values ($10^{-3}$ for instance) that have 90% confidence level of being within a factor of two of the actual values?


1 Answer 1


For each sample, we note $1$ if error and $0$ otherwise. Then, for bit error rate $q$, the tests are independent Bernoulli random variables $\{X_i\}$ with probability $q$.

The estimate is $\hat{p} = \frac{S}{n}$ where $$S = \sum^n_i X_i \tag{1}$$ and $\mu = \mathbb{E}[X] = np$.

Chernoff bound (see Corollary 5) tells us that for every $0 < t < \mu$, $$\mathbb{P}\left( \mid S-np\mid \leq \frac{t}{\mu}\mu\right) \geq 1 - 2 e^{\frac{-t^2}{3\mu}} \tag{2}$$

Choose $t=\epsilon np$ with $0 < \epsilon < 1$, $$\mathbb{P}\left( \mid S-np\mid \leq \epsilon np\right) \geq 1 - 2 e^{\frac{-\epsilon^2 np}{3}} \tag{3}$$

$$\mathbb{P}\left(p(1-\epsilon) \leq S/n \leq p(1+\epsilon)\right) \geq 1 - 2 e^{\frac{-\epsilon^2 np}{3}} \tag{4}$$

Therefore, $\epsilon$ is relative estimation error and the right hand side of $(4)$ is confidence level which depends on sample size $n$.

For example, if you want estimate a BER within $10^{-3}\times(1 \pm 10\%)$, using sample size $n=8.99 \times 10^5$ will guarantee a confidence level greater than 90%.

Note that choosing another $t$ and/or using another bound may result in a tighter required sample size.


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