# Detection problem with extra noise observation. Does the extra observations help with detection?

Assuming all “real” number, consider the detection problem of $$x$$ given $$y_1$$ and $$y_2$$ below

$$y_1= x + n_1$$

$$y_2= n_1 + n_2$$

where $$x$$ is either $$+A$$ or $$-A$$ with equal probability, and $$n_1$$ and $$n_2$$ are i.i.d Gaussian random variables with zero mean and variance of $$\sigma^2$$.

Is observation of $$y_2$$ helpful for detection of $$x$$? whether $$x = A$$ or $$x = - A$$?

From $$y_1$$ one can see that it’s a simple BPSK signal detection. However, In my opinion the new observation in $$y_2$$ does not add any new information, since $$n_2$$ and $$n_1$$ are i.i.d Normal gaussian distributions. One idea is for example to write the received signals as $$y=y_1 + \alpha y_2=x+(1+\alpha)n_1+n_2$$. But from my understanding, this linear combination is the same as detecting $$x$$ from $$y_1$$. Is my understanding correct?

• Thank-you! I've reopened it now.
– Peter K.
Aug 26 at 0:28
• You need to specify whether $n_1$ and $n_2$ are uncorrelated and what their standard deviation (in units of $A$) is. You can certainly improve detection for $\sigma_2 < \sigma_1$ Aug 26 at 6:00

$$y=y_1 + \alpha y_2=x+(1+\alpha)n_1+\alpha n_2 \tag{1}$$

As $$n_1$$ and $$n_2$$ are iid $$\mathcal{N}(0,\sigma^2)$$, you have now the new Gausian noise $$\mathcal{N}(0,\nu^2)$$ where $$\nu^2=\left((1+\alpha)^2+\alpha^2\right)\sigma^2$$.

Choose $$\alpha=-\frac{1}{2}$$, then $$\nu^2=\frac{1}{2}\sigma^2$$, better!

The intuition is that by $$y_2$$ you observe another realization of the noise even though $$y_2$$ does not contain $$x$$.

Let's see what happened by analyzing Fisher information (Disclaimer: this is not a rigorous proof as it does not take into account the fact that $$x$$ is BPSK-modulated. This can be done by using the Bayesian interpretation and deriving the posterior distributions from the following likelihoods. Sound hand waving? I do agree).

The sampling distribution $$p(y_1,y_2 \mid x)=p(y_2 \mid y_1, x)p(y_1 \mid x)=\\ C_{12}\times \exp \left( -\frac{1}{2\sigma^2} (y_2 -y_1+x)^2 \right) \exp \left( -\frac{1}{2\sigma^2} (y_1-x)^2 \right)\tag{2}$$ with $$C_{12}$$ some constant. Then the log likelihood function $$L(y_1,y_2;x) = \log p(y_1,y_2 \mid x)$$

Easy to see that $$\mathbb{E}\left[\frac{\partial L(y_1,y_2;x)}{\partial x}\right]=0$$ then Fisher information $$I_{y_1,y_2}(x)=-\mathbb{E}\left[\frac{\partial^2 L(y_1,y_2;x)}{\partial x^2}\right]=\frac{2}{\sigma^2}\tag{3}$$

If only $$y_1=x+n_1$$ is used, with $$C_1$$ some constant, $$p(y_1\mid x)= C_{1}\times \exp \left( -\frac{1}{2\sigma^2} (y_1-x)^2 \right)\tag{4}$$

and the Fisher information

$$I_{y_1}(x)=\frac{1}{\sigma^2}\tag{5}$$

Clearly (3) is better than (5).

• Thanks a lot @AlexTP I've got to refreshen my knowledge. This was very helpful. Aug 27 at 20:55