# Remove Noise from Discrete Signal with given Noise Model

I am interested in finding the true signal $$p \in \mathbb{R}^D$$ of an observed discrete signal $$t \in \mathbb{Z}_{+}^D$$.
I know that each observed $$t_{i}$$ with $$1 \leq i \leq D$$ is the result of a random sample of a Poisson distribution with rate $$\lambda_{i} = p_{i}$$.

For example:

• $$p_{1} = 3.0$$ then $$t_{1} \sim Poisson_{\lambda_{1} = p_{1}}(k) = \frac{\lambda_{1}^k}{k!} e^{-\lambda_{i}} = \frac{3^k}{k!} e^{-3}$$.
• $$p_{2} = 0.2$$ then $$t_{2} \sim Poisson_{\lambda_{2} = p_{2}}(k) = \frac{\lambda_{2}^k}{k!} e^{-\lambda_{i}} = \frac{0.2^k}{k!} e^{-0.2}$$.
• $$p_{3} = 1.1$$ then $$t_{3} \sim Poisson_{\lambda_{3} = p_{3}}(k) =$$ ...

Now I want to find the underlying distribution $$P$$ of $$p$$ under this noise model. Sampling from $$P$$ will generate the true data $$p_i$$ and sampling from a Poisson with each $$p_i$$ as the rate parameter should yield the observed data.
My gut feeling is as follows: observing a high value $$t_i$$ makes it likely that it is drawn from a Poisson with a high $$\lambda$$ value, which should make it likely that the distribution $$P$$ has (some) high values as values.
I think the advantage with respect to the noise removal I have, is that I know the noise model is given by this Poisson process. However, it is unclear to me how I can exploit that.

1. I know that Fourier transformations can be used to remove noise in general and I watched some videos on that (I am from a different background), but cannot get my head around it how to apply it here. Is this even possible?
IIUC, it is usually the other way round, i.e., I have a continuous signal, transform it into a discrete domain, drop the smallest coefficients and convert it back. It is unclear to me if this is applicable here as well.
2. Are there any other approaches or ideas you can think of which are worth trying (keywords I can look up myself are also fine)?

• Hi N8_Coder: so, you have $D$ observatons and each of those observations say, $x_i$ comes from a poisson distribution with rate $\lambda_i$. So, if the rate was constant, the MLE would be the average ( see document in comment ). Since the rate is not constant, it seems like the best you can do is still use the average which is the value of the one observation that you have. Of course, I could still be missing something but it seems like the MLE approach is still valid but with the caveat that the number of observations is equal to one. Commented Jul 27, 2021 at 20:04