# 5 low signal-to-noise measurements of same signal, know noise distribution and signal distribution. Recover?

I have 5 measurements of $x(t)$ at each instance in time, all contaminated with Gaussian noise. My signal-to-noise ratio is bad.

\begin{align} y_1(t)=x(t)+n_1(t)\\ y_2(t)=x(t)+n_2(t)\\ y_3(t)=x(t)+n_3(t)\\ y_4(t)=x(t)+n_4(t)\\ y_5(t)=x(t)+n_5(t)\\ \end{align}

But, I do know the distribution of my noise $n_i$, and the distribution of my signal $x$. What would be the best way to recover my signal even for very bad signal-to-noise (and what would be possible)?

• Is the noise contamination for the $n_i$ independent from $n_j$?
– Sycorax
Commented Sep 1, 2016 at 19:19
• Yes, all noise contaminations n1 to n5 follow the same distribution, but are itself independent from each other Commented Sep 1, 2016 at 19:24
• You write $n_i(t)$. Does this mean that the distribution of noise contamination varies with time? Or, alternatively, at each $t$ is the noise drawn from a normal distribution with fixed mean $\mu_i$ and variance $\sigma_i^2$?
– Sycorax
Commented Sep 1, 2016 at 19:27
• The distribution of the noise is fixed. Sorry for the confusion. At each time t the noise is drawn from a normal distribution with fixed parameters. Please let me know if you have further questions Commented Sep 1, 2016 at 20:23
• When you say you know the distribution of your signal, do you mean the signal is stationary through time and you know its bulk distribution? Or is the distribution time dependent? Commented Sep 1, 2016 at 23:17

If you have $N$ time series, $y_i[t]=x[t]+\epsilon$, for $i=1\ldots N$, then you compute the average signal as $$\bar{y}[t]=\frac{1}{N}\sum_{i=1}^Ny_i[t]$$ If the errors are independent with mean $\langle\epsilon\rangle=0$ and variance $\langle\epsilon^2\rangle=\sigma^2$, then your ensemble-averaged signal will have a reduced error variance given by $$\langle(\bar{y}[t]-x[t])^2\rangle=\tfrac{1}{N}\sigma^2$$
This all holds even if you do not know the noise distribution. In your case, you do know the distribution. Since the errors are Gaussian, the ensemble average will also have Gaussian errors, and is the maximum likelihood estimate of your signal. Because you also know $\sigma$, you can explicitly compute the reduced error variance. From this, you can calculate confidence intervals (bands) on your reconstructed signal.
• Can you give numbers? What is $\sigma$ and what is the amplitude (max - min range) of $x$? Commented Sep 1, 2016 at 23:20