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I have $N$ observations $\{y_i\}_{i=1..N}$ $$y_i = h_i x_i + w_i$$ where $\{x_i\}_i$ are the realizations of uniform distributed QPSK symbol $x$, i.e. $x_i \in \{\pm\frac{1}{\sqrt{2}} \pm j\frac{1}{\sqrt{2}}\}$,

$\{w_i\}_i$ are the realizations of circular complex gaussian noise $w \sim \mathcal{CN}(0,\sigma^2)$,

and $\{h_i\}_i$ are the realizations of random variable $h$, but known (for example via perfect channel estimation of a channel that is static during the duration that spans over $N$ observations).

The random variables $x$ and $w$ are i.i.d. ($h$ is known during the observation duration.)

I want to estimate the noise variance $\sigma^2$.

I tried using $\{\frac{y_i}{h_i}\}_i$ but their variances are scaled by $h_i$ and not be the same anymore.

Any hints?

(well, this is not a homework but maybe it is as simple as a homework for some people. So I keep the homework tag.)


Update: Note that the $N$ values $y_i$ are realizations, not random variables, for the desired $\sigma^2$ estimator.

Let me express my question in another way: I have $N$ values $y_i$ and $N$ values $h_i$, and I am looking for a function $f()$ that $\hat{\sigma}^2 = f(y_1,...,y_N,h_1,...,h_N)$.

So, $\hat{\sigma}^2$ is an estimation of $\sigma^2$, and is a random variable. It would be great that expected value of $\hat{\sigma}^2$ is $\sigma^2$ so that the estimator is unbiased.

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All $x_i$ are identically distributed: discretely uniformly distributed, with the only values being $\{-\sqrt{\frac12}, \sqrt{\frac12}\}$ on both the real and imaginary part.

The $w_i$ are also identically distributed, as per the problem statement.

Therefore, the variance (this being zero-mean) is simply $\sqrt{\frac12}^2=\frac12$ in each of the real and imaginary part.

For the $\mathcal CN$ noise: it has, by definition, indep. real and imaginary part, and their variances are $\frac{\sigma^2}2$ each. Since $h_i$ is known, knowing $\text{Var}(y_i)$ is as good as knowing $\sigma^2$: $\sigma^2 = \text{Var}(y_i) - \frac{h_i^2}2$.

Now, Cath claims that $y_i$ are not IID random variables but realizations of a stoch. process, but such realizations are random variables.

So, $h_ix_i$ and $w_i$ are independent. Meaning that the variance of their sum is just the sum of their variances, $\DeclareMathOperator{\Var}{Var}$

\begin{align} \Var(y_i) &= h_i^2\Var(x_i) + \Var(w_i)\\ &= h_i^2\Var(x_i) + \sigma^2 \quad \|\Var(\Re\{ x_i\}) = \Var(\Im \{x_i\})= \frac 12\,\text{const.}\\ \Var(\Re \{y_i\})&=(\Re \{h_i\})^2\Var(\Re\{ x_i\})+(\Im \{h_i\})^2\Var(\Im \{x_i\}) + \frac{\sigma^2}2\\ &=\frac12\left(\Re \{h_i\})^2+(\Im \{h_i\})^2\right) + \frac{\sigma^2}2\\ &= \frac{\lvert h_i \rvert^2+\sigma^2}{2} \\ &=\Var(\Im \{y_i\}) \quad \text{(symmetry)}\\[1em] \Var{\sum_{i=1}^N \Re\{y_i\}} &= \frac12\left( \sum_{i=1}^N\lvert h_i\rvert^2+ N\sigma^2 \right)\\ \sigma^2&=\frac{2\Var{\sum_{i=1}^N \Re\{y_i\}}-\sum_{i=1}^N\lvert h_i\rvert^2}{N}, \end{align}

same for the imaginary part of $y$. Now, that reads like a very usable approach for an estimator, right?

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  • $\begingroup$ OK got it. So the estimator is $\hat{\sigma}^2 = \frac{1}{N} \sum_i \left( \mid y_i \mid^2 - \sigma_x^2 \mid h_i \mid^2\right) = \frac{1}{N} \sum_i \left( \mid y_i \mid^2 - \mid h_i \mid^2\right)$ as $\sigma_x^2 = 1$. $\endgroup$ Jul 3 at 17:12
  • $\begingroup$ And it is unbiased. $\endgroup$ Jul 3 at 17:13
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Any hints?

sure!

hint 1

You can easily estimate the variance of $y_i$.

hint 2 & 2.5

You know the variance of $h_ix_i$ (hint: what does scaling with a scalar do to the variance of $x_i$?).

hint 3

What is the variance of a sum of random variables that are independent?

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  • $\begingroup$ Thanks. However, how can I estimate the variance of $y_i$? Do you mean by the sample variance estimator of $y$? Is it something like $var(y) = \frac{1}{N-1} \sum (y_i - \hat{\mu}_y)^2$ where $\hat{\mu}_y$ is the sample mean of $y$? But $\{y_i\}_i$ are not i.i.d. $\endgroup$ Jul 3 at 12:14
  • $\begingroup$ Strictly speaking,$\{y_i\}_i$ are realizations, not random variables. $\endgroup$ Jul 3 at 12:18
  • $\begingroup$ Maybe my wording was not clear. I have edited my question. $\endgroup$ Jul 3 at 12:58
  • $\begingroup$ @Cath Maillon: Hi. You know the variance of the first term on the RHS. It's $h_t^2 \times$ the variance of a uniform. I think variance of a uniform (a,b) is $(a-b)^2/12$ but look it up. You can estimate the variance of $y_i$ as you described. Then, since the right hand side variances are additive, this means that $\hat\sigma^2 = $ variance of $y_i$ minus the variance of the first term on the right hand side. Note that the $y_i$ are iid because they are the sum of two iid terms. a normal rv and a uniform rv. $\endgroup$
    – mark leeds
    Jul 3 at 14:54
  • $\begingroup$ @markleeds very close! it's a uniform, but not a continuously distributed uniform; the variance as always is $\mathbb E(|x_i - \mathbb E(x_i)|^2)$, and $\mathbb E(x_i)=0$. However, it's easier to consider real and imaginary parts separate and then add them in the end – they're independent! $\endgroup$ Jul 3 at 15:03
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Let me write it here because the comment isn't that all that clear. You have

$Var(y_{i}) = (h_{i}^{2} \times (b-a)^2/12) + \sigma^2$

From the data, you will have, the sample observations, ${y_{i}}$.

You know the mean of the right hand side. Check it but I think it's zero.

So, given that the mean is zero, this implies that

$ \sum var(y_{i}) = \sum_{i=1}^{N} [ h_{i}^2 \times (a-b)^2/12 + \sigma^2 ]$.

So, now you can pull the second one out of the sum and treat it as $N \sigma^2$. Then, you can solve for $\sigma^2$ by bringing the sum of the other terms over to the LHS and then dividing by $N$.

EDIT: This answer is wrong because the LHS is a sum so how to estimate that ? As far as I can tell, the OP is correct and the $h_i$ needs to be eliminated somehow so that the LHS can become $var(y)$ rather than $var(y_{i})$.

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  • $\begingroup$ I have added some argument to make my question clearer in my question and in the MarcusMuller's answer. $\endgroup$ Jul 3 at 15:37
  • $\begingroup$ Mark, you're assuming $x$ is continuously uniformly distributed, but it's discretely uniformly distributed, with the only values being $\{-\sqrt{\frac12}, \sqrt{\frac12}\}$ on both the real and imaginary part. Therefore, the variance (this being zero-mean) is simply $\sqrt{\frac12}^2=\frac12$ in each of the real and imaginary part. Same for the $\mathcal CN$ noise: it has, by definition, indep. real and imaginary part, and their variances are $\frac{\sigma^2}2$ each. Since $h_i$ is known, knowing $\text{Var}(y_i)$ is as good as knowing $\sigma^2$: $\sigma^2 = \text{Var}(y_i) - \frac{h_i^2}2$ $\endgroup$ Jul 3 at 15:48
  • $\begingroup$ Let me put all this in an answer. $\endgroup$ Jul 3 at 15:52

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