# Noise variance estimation under known channel

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.

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?

• 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 $\sigma_x^2 = 1$. Jul 3 at 17:12
• And it is unbiased. Jul 3 at 17:13

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?

• 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. Jul 3 at 12:14
• Strictly speaking,$\{y_i\}_i$ are realizations, not random variables. Jul 3 at 12:18
• Maybe my wording was not clear. I have edited my question. Jul 3 at 12:58
• @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. Jul 3 at 14:54
• @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! Jul 3 at 15:03

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 $$var(y_{i})$$.

• I have added some argument to make my question clearer in my question and in the MarcusMuller's answer. Jul 3 at 15:37
• 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 $\sigma^2$: $\sigma^2 = \text{Var}(y_i) - \frac{h_i^2}2$ Jul 3 at 15:48
• Let me put all this in an answer. Jul 3 at 15:52