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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?

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  • $\begingroup$ Thank-you! I've reopened it now. $\endgroup$
    – Peter K.
    Aug 26 at 0:28
  • $\begingroup$ 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 $\sigma_2 < \sigma_1$ $\endgroup$
    – Hilmar
    Aug 26 at 6:00

1 Answer 1

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

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).

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

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