Convolution sum vs auto-correlation problem I had an exam one of these days and one of the questions was:

"Knowing that an auto-correlation estimator of a sinal x[n] could be defined by:

$$
R_{xx}[k] = \sum_{n=-\infty}^{+\infty}h[n].x[n+k] , k=0,1,...,n-1
$$
and that the non-periodic convolution:
$$
y[n]=x[n]*h[n] = \sum_{k=-\infty}^{+\infty}h[k].x[n-k]
$$
show how to calculate the autocorrelation with a convolution operation"



And here is my reasoning:

let's say that:
$$
h[n] = x[-n]
$$
Therefore, we would have some $y[n]$ for the following convolution:
$$
y[n] = x[n]*h[n]=x[n]*x[-n] = \sum_{k=-\infty}^{+\infty}x[k].x[k-n]
$$
So it would be nice to invert the signal $y[n]$, finding then a $y[-n]$. However, we will get the same result as before since convolution is a commutative operation:
$$
y[-n]=x[-n]*h[-n]=x[-n]*x[n]=x[n]*x[-n]=\sum_{k=-\infty}^{+\infty}x[-k].x[-k+n]
$$
Therefore, $R_{xx}[n]=y[n]=y[-n]$, since the convolution sum is symmetric with respect to zero.

Question:

Is my reasoning correct by formal mathematics?
The professor gave me a zero in this answer because he says that my assumption $R_{xx}[n]=y[n]=y[-n]$ is wrong because the correct calculation of the autocorrelation has a complex conjugate in one of the terms (so it might not be symmetric with respect to zero). I agree with it even though he has not placed this formalism in the autocorrelation definition. So, following the exam's definitions, is my reasoning correct?

Also, can I say that this will always be true whenever the signal is purely real?
And by consequence of the question above: can some of these be demonstrated for complex signals?
There are two things to consider. First, looking only at the exam question you can say that the convolution of $h[n]$ with a sequence $y[n]$ is given by

$$(h*y)[k]=\sum_nh[n]y[k-n]$$

Now define $y[n]=x[-n]$:

$$(h*y)[k]=\sum_nh[n]x[n-k]$$

Consequently, we have

$$\sum_nh[n]x[n+k]=(h*y)[-k],\quad y[n]=x[-n]$$

So you need to flip the signal $x[n]$, compute the convolution, and flip the result. This is true, given the sum in your exam question. No need to worry about complex conjugates because we just used the definition of the autocorrelation sum in the question.

The other issue is the definition of the autocorrelation for complex sequences, and how it can be implemented by convolution. The usual definition is

$$R_{xx}[k]=\sum_nx^{*}[n]x[n+k]$$

The convolution of $x^*[n]$ with a sequence $h[n]$ is

$$(x^**h)[k]=\sum_nx^*[n]h[k-n]$$

For $h[n]=x[-n]$ we have

$$(x^**h)[k]=\sum_nx^*[n]x[n-k]=R_{xx}[-k]$$

So for complex signals, you need to compute the convolution sum of $x^*[n]$ with $x[-n]$, and flip the result to arrive at the autocorrelation.

Note that $R_{xx}[k]=R_{xx}^*[-k]$, so instead of flipping the result, you can also use its complex conjugate.