1
$\begingroup$

I'm just starting estimation theory for my DSP course and I'm a bit confused about the classic example given everywhere which aims to show that two unbiased estimators can have different variances.

enter image description here

enter image description here

I'm just wondering how they did that step that I have a question mark beside. I understand intuiviely that the average of x[n] will be A but I can't mathematically figure that out with the E[] and also what happens to 1/N. If someone could just explain that step in detail.

Another Question: What's the difference between A(hat) and E[A(hat)], I mean shouldn't A(hat) give you A anyways? Then what's the point of the E[A(hat)]?

$\endgroup$
3
$\begingroup$

The expectation of $x[n]$ is

$$E(x[n])=E(A+w[n])=E(A)+E(w[n])=A+0=A\tag{1}$$

because $A$ is deterministic and the noise $w[n]$ is assumed to have a mean of zero. So in that formula you sum up $N$ terms that are equal to $A$, and then you divide that sum by $N$, leaving you with the value $A$.

As for the expectation of the estimated value $\hat{A}$, note that $\hat{A}$ is a random variable because it necessarily depends on the data $x[n]$, which are random, so computing its expected value (and its variance) makes sense.

$\endgroup$
  • $\begingroup$ Ah yes. The additive expected value. Completely forgot. Thank you so much $\endgroup$ – AlfroJang80 Nov 11 '18 at 20:14
  • $\begingroup$ Okay, so a follow-up question for computing the expected value of the A (with upside down hat). Surely, for it's expected value, we no longer have the mean of the entire white noise signal, just the 1st element of it, so we won't be able to say that it's mean is zero now right? So how does he get A for that too? $\endgroup$ – AlfroJang80 Nov 11 '18 at 20:36
  • $\begingroup$ @AlfroJang80: You have $E(w[n])=0$ and that's true for any $n$, also for $n=0$. Note that you consider the zeroth element of all possible realizations of the noise process. $\endgroup$ – Matt L. Nov 12 '18 at 8:15
  • $\begingroup$ Ah I think I get it now. Even though we're considering just one element of the noise, the expected value of it will be 0. Awesome. Thank you so much $\endgroup$ – AlfroJang80 Nov 12 '18 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.