Say I have two zero-mean vectors, with a size of N and the translation between them is k. Say the image signal std is $\sigma_s$ and the noise std is given by $\sigma_n$. What is the variance of the correlation value in case the correlation is estimated by Fourier transform, without zero-padding? Is there a way to find the variance for the Pearson-correlation too?

  • $\begingroup$ I have suggested an answer but I do not feel comfortable about it $\endgroup$ Mar 3 '20 at 14:59

Presented are two vectors of length $l$ with a displacement of $k>0$ between them



where $ω_i\sim N(0,σ_n^2)$ and $a_i\sim N(0,σ_s^2)$. The displacement means that for $k<i<l-k$, $b_{i+k}=a_i$

Our estimation of the circular correlation, without zero-padding, takes the form of

$C_k=\frac{∑_{i=1}^NA_i B_{mod(i-k+m,m)}}{\sqrt{∑_{i=1}^NA_i^2∑_{i=1}^NB_i^2}} \frac{l}{l-k}$

Following derivation

$C_k=\frac{∑_{i=1}^N (a_i+ω_i)(b_{mod(i-k+m,m)}+ω_{mod(i-k+m,m)})}{\sqrt{∑_{i=1}^N(a_i+ω_i)^2∑_{i=1}^N(b_i+ω_i)^2}} \frac{l}{l-k}$

Expected value

The expected value for square of one sample is

$E[(a_i+ω_i )^2 ]=E[a_i^2+a_i ω_i+ω_i^2 ]=σ_s^2+σ_n^2$

Expected value for the circular correlation at the true peak location $E[C_k]=\frac{σ_s^2}{σ_s^2+σ_n^2}$

which means that we use biased estimation of the true peak correlation.

Correlation value not at the true peak location is $E[C_c]=0$


General expression for the variance of multiplication of two samples is given by $var((a_i+ω_i)(b_i+ω_j))=var(a_i b_i+ω_i b_i+a_i ω_i+ω_i ω_j )$

For the correlated samples

$var(a_ib_i+ω_ib_i+a_iω_i+ω_iω_j)=var(2a_iω_i+ω_iω_j)=4σ_s^2 σ_n^2+σ_n^4$

For the uncorrelated samples


Lets neglect the variance of the dynamometer. This leads to the variance of the true peak correlation value

$var(C_k )≈\left(k(σ_s^4+4σ_s^2 σ_n^2+σ_n^4 )+(l-k)(4σ_s^2 σ_n^2+σ_n^4)\right)\left(\frac{l}{l-k}\frac{1}{l(σ_s^2+σ_n^2)}\right)^2$

simplification leads to

$var(C_k)=\frac{kσ_s^4+l(4σ_s^2 σ_n^2+σ_n^4)}{(l-k)^2(σ_s^2+σ_n^2 )^2}$

and for a location which is not the true peak v≠k


here simplification leads to


Therefore, the true peak value

$C_k\sim\left(\frac{σ_s^2}{σ_s^2+σ_n^2},\frac{kσ_s^4+l(4σ_s^2 σ_n^2+σ_n^4)}{(l-k)^2(σ_s^2+σ_n^2 )^2}\right)$

and the false peak value



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