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This is my problem: I use data observed with MUSE (which is an astronomical instrument provides cubes, i.e. an image for each wavelength with a certain range, link ) to extract a measure of redshift. Let's call the MUSE cube with $S_{ij}$, where $i$ labels the pixels (i.e. flattening the image, $i \in [0, ..., h\cdot w)$) and $j$ labels the wavelength ($j \in [0, N_\lambda)$). In order to perform such measure, I cross-correlate the spectra (obtained from the MUSE cube extracting the wavelength information corresponding to a pixel, i.e. $S_i$), with a set of templates, which are spectra obtained from known sources, since these templates are redshifted, founding the best cross-matched value (i.e. the maximum of the cross-correlation function for each pixel), I get the redshift (the set of templates is represented with a matrix $T_{jk}$, where $j\in 0,N_\lambda$, and $k\in [0, N_z)$, i.e. a column ($k$) of this matrix is the spectra at $z_k$). This a common and widely-used process and, indeed, it works well.

My question regards the error with which I measure the best redshift (let's call it as $z_{best}$). In fact, MUSE provides also an error for each pixel and for each wavelength, $\sigma_{ij}$ (simplifying the notation, actually it should be written as $\sigma_{S_{ij}}$). I have propagated these error through the cross-correlation relation: $$X_{ik} = \frac{\sum_j S_{ij}T_{jk}}{\sqrt{\sum_jS_{ij}^2\cdot\sum_jT_{jk}^2}} = \frac{\sum_j S_{ij}T_{jk}}{\sqrt{SSS_i\cdot SST_k}} = \frac{\sum_j S_{ij}T_{jk}}{N_{ik}} $$ where $X_{ik}$ is the cross-correlation related to the pixel $i$, to the redshift $z_k$, $SSS_i$ and $SST_k$ are the sum of the squared of the spectra relate to the pixel $i$ (or template related to the redshift $z_k$), $N_{ik}$ is a normalization coefficient.

Using the error-propagation rules, and assuming the templates without error, I have an error on $X_{ik}$: $$\sigma_{X_{ik}}^2 = \sum_j \Big(\frac{\partial X_{ik}}{\partial S_{ij}}\Big)^2\sigma_{ij}^2 = \frac{\sum_j\sigma_{ij}^2T_{jk}^2}{N_{ik}^2} + \frac{X_{ik}^2SST_k^2\sum_jS_{ij}\sigma_{ij}^2}{N_{ik}^4} - 2 \frac{X_{ik}SST_k\sum_j(S_{ij}\sigma_{ij}^2)T_{jk}}{N_{ik}^3} $$ The code has been implemented using python and tensorflow (tf) in order to exploit the GPU:

@tf.function
def make_xcorr_err1(T, S, sigma_S):
    sum_spectra_sq = tf.reduce_sum(tf.square(S), 1) #shape (batch,)
    sum_template_sq = tf.reduce_sum(tf.square(T), 0) #shape (Nz, )
    norm = tf.sqrt(tf.reshape(sum_spectra_sq, (-1,1))*tf.reshape(sum_template_sq, (1,-1))) #shape (batch, Nz)
    xcorr = tf.matmul(S, T, transpose_a = False, transpose_b= False)/norm

    foo1 = tf.matmul(sigma_S**2, T**2, transpose_a = False, transpose_b= False)/norm**2
    foo2 = xcorr**2 * tf.reshape(sum_template_sq**2, (1,-1)) * tf.reshape(tf.reduce_sum((S*sigma_S)**2, 1), (-1,1))/norm**4
    foo3 = - 2 * xcorr * tf.reshape(sum_template_sq, (1,-1)) * tf.matmul(S*(sigma_S)**2, T, transpose_a = False, transpose_b= False)/norm**3

    sigma_xcorr = tf.sqrt(tf.maximum(foo1+foo2+foo3, 0.))

The follow image is the cross-correlation function for a pixel, i.e. $X_{ik}$ for a fixed $i$, $\forall k$, and the area between $\pm3\sigma$ is filled in grey:

xcorr

The red vertical line represents the detected maximum, corresponding to $z\sim0.404$ (which is correct, since I know the true redshift of this source). If a zoom-in around the maximum value:

xcorr_zoom

As you can note, the maximum fluctuates within a range of values, i.e. the error of $X_{ik}$.

Thus my question is: How can I map these fluctations to error on $z_{best}$?

I have thought to some soultions:

  1. For each pixel I can fit the cross-correlation function with a parabola centered around the maximum. Once I have an expression like $a_0 + a_1 z + a_2 z^2$, with a fitted error on $a_n$ and the covariance matrix, I can estimate the maximum and its error. I have discarded this solution since it requires a lot of time, I should fit a parabola for each pixel (my image is about $500\times500$ pixels), then there is the problem of the dimension of the range in $z$.

  2. Trying a boot-strap method: I can generate hundreds of cross-correlation functions, constrained by the estimated errors. It would be as if there were hundreds of functions in the grey area, guassian generated. I can estimate the maximum for each of them, having an idea about the dispersion in $z$, the resulting $z_{best}$ is mean among the $z_{best_n}$. This is clearly faster than the previous method. Can be this approach correct? In this case, is better to use a median, instead of the mean? Is also better use an weighted median, and the error on $z_{best} $ measured as the weighted MAD?

What do you think?

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  • $\begingroup$ Interesting question! In statistics, this is known as inverse prediction, inverse regression or calibration. $\endgroup$ Commented Apr 11, 2020 at 20:26
  • $\begingroup$ @kjetilbhalvorsen Thanks for the editing! Can you suggest me some references (articles or books)? $\endgroup$ Commented Apr 12, 2020 at 9:30

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More of a comment, but too long. This is a variant of calibration, or inverse regression/inverse prediction. One survey paper is this at JSTOR but it does not look explicitly at predicting the peak location. One similar problem is response surface methods, and there is a relevant R package. Some stored google searches that looks promosing: peak detection and confidence intervals and response surface, inverse regression. In a comment whuber points to How to estimate the uncertainty in the zeros of a fitted function? which might be helpful.

Among these I guess you can find something of interest!

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    $\begingroup$ I don't think the comparison with calibration is apt, because this is precisely the kind of curve where the regression function is not invertible. One method of solving the problem is described in my post at stats.stackexchange.com/a/446205/919 concerning finding zeros of the regression function. It's only a small step to finding the critical points of a regression function by finding the zeros of its derivative. $\endgroup$
    – whuber
    Commented Sep 3, 2020 at 18:57

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