Here we calculate DM estimation errors given multiple timing perturbations.
Two-frequency timing perturbations
The time of arrival (TOA) at a particular frequency \(\nu\) is given as the infinite-frequency arrival time plus the dispersive delay. We will also include measurement errors \(\epsilon_\nu\) and a chromatic (frequency-dependent) timing perturbation \(t_C\), such that
\[t_\nu = t_\infty + \frac{K\mathrm{DM}}{\nu^2} + t_{C,\nu} + \epsilon_\nu.\]Here, \(K \approx 4.149~\mathrm{ms~GHz^2~pc^{-1}~cm^{3}} = 4.149 \times 10^3~\mu\mathrm{s~GHz^2~pc^{-1}~cm^{3}} = 4.149 \times 10^9~\mu\mathrm{s~MHz^2~pc^{-1}~cm^{3}}\) is the dispersion constant. We estimate the DM by taking TOAs at two frequencies \(\nu\) and \(\nu'\) and calculating
\[\widehat{\mathrm{DM}} = \frac{t_{\nu'} - t_{\nu}}{K(\nu'^{-2} - \nu^{-2})}.\]The estimated infinite-frequency arrival time can then be written in one of two ways as
\[\begin{aligned} \hat{t}_\infty = t_\nu - \frac{K\widehat{\mathrm{DM}}}{\nu^2} &= t_\infty + \frac{K(\mathrm{DM}-\widehat{\mathrm{DM}})}{\nu^2} + t_{C,\nu} + \epsilon_\nu \\ &= t_\infty + \frac{K(\mathrm{DM}-\widehat{\mathrm{DM}})}{\nu'^2} + t_{C,\nu'} + \epsilon_{\nu'}. \end{aligned}\]First we will solve for the DM difference. Substituting the measured TOAs into the equation for \(\widehat{\mathrm{DM}}\) and subtracting from the true DM, we have
\[\begin{aligned} \delta\mathrm{DM} \equiv \mathrm{DM} - \widehat{\mathrm{DM}} &= \mathrm{DM} - \frac{t_{\nu'} - t_{\nu}}{K(\nu'^{-2} - \nu^{-2})} \\ &= \mathrm{DM} - \left[\frac{\left(\cancel{t_\infty} + \frac{K\mathrm{DM}}{\nu'^2} + t_{C,\nu'} + \epsilon_{\nu'}\right) - \left(\cancel{t_\infty} + \frac{K\mathrm{DM}}{\nu^2} + t_{C,\nu} + \epsilon_\nu\right)}{K(\nu'^{-2} - \nu^{-2})}\right] \\ &= \bcancel{\mathrm{DM}} - \left[\frac{\cancel{K}\bcancel{\mathrm{DM}}\cancel{(\nu'^{-2} - \nu^{-2})}}{\cancel{K}\cancel{(\nu'^{-2} - \nu^{-2})}} + \frac{t_{C,\nu'} + \epsilon_{\nu'} - t_{C,\nu} - \epsilon_\nu}{K(\nu'^{-2} - \nu^{-2})}\right] \\ &= -\frac{t_{C,\nu'} - t_{C,\nu} + \epsilon_{\nu'} - \epsilon_\nu}{K(\nu'^{-2} - \nu^{-2})}. \end{aligned}\]The TOA perturbation will be (defining \(r \equiv \nu/\nu'\), with \(\nu > \nu'\)):
\[\begin{aligned} \delta t_\infty \equiv t_\infty - \hat{t}_\infty &= -\frac{K(\mathrm{DM}-\widehat{\mathrm{DM}})}{\nu^2} - t_{C,\nu} - \epsilon_\nu \\ &= -t_{C,\nu} - \epsilon_\nu - \frac{\cancel{K}}{\nu^2}\left[-\frac{t_{C,\nu'} - t_{C,\nu} + \epsilon_{\nu'} - \epsilon_\nu}{\cancel{K}(\nu'^{-2} - \nu^{-2})}\right] \\ &= -t_{C,\nu} - \epsilon_\nu + \left[\frac{t_{C,\nu'} - t_{C,\nu} + \epsilon_{\nu'} - \epsilon_\nu}{r^2 - 1}\right] \\ &= \frac{-(t_{C,\nu} + \epsilon_\nu)(r^2 - 1) + (t_{C,\nu'} - t_{C,\nu} + \epsilon_{\nu'} - \epsilon_\nu)}{r^2 - 1} \\ &= \frac{-r^2 t_{C,\nu} + \cancel{t_{C,\nu}} - r^2 \epsilon_\nu + \cancel{\epsilon_\nu} + t_{C,\nu'} - \cancel{t_{C,\nu}} + \epsilon_{\nu'} - \cancel{\epsilon_\nu}}{r^2 - 1} \\ &= \frac{-r^2 t_{C,\nu} - r^2 \epsilon_\nu + t_{C,\nu'} + \epsilon_{\nu'}}{r^2 - 1}. \end{aligned}\]When the chromatic offsets are zero, we arrive at simply
\[\delta t_\infty = \frac{\epsilon_{\nu'} - r^2 \epsilon_\nu}{r^2 - 1},\]which agrees with Eq. 21 in Cordes, Shannon & Stinebring (2016), assuming the frequency-dependent DM term is zero.
The TOA uncertainty
The result above gives the offset, but one must also consider the TOA uncertainty, \(\sigma_{\delta t_\infty}\), from the variance
\[\sigma_{\delta t_\infty}^2 = \left\langle\delta t_\infty^2\right\rangle - \cancel{\left\langle\delta t_\infty\right\rangle^2} = \left\langle\delta t_\infty^2\right\rangle = \left\langle\left(\frac{\epsilon_{\nu'} - r^2 \epsilon_\nu}{r^2 - 1}\right)^2\right\rangle.\]Let \(X \sim \mathcal{N}(0,\sigma_{\epsilon_{\nu'}}^2)\) and \(Y \sim \mathcal{N}(0,r^4 \sigma_{\epsilon_\nu}^2)\). Then, \(X-Y \sim \mathcal{N}(0,\sigma_{\epsilon_{\nu'}}^2 + r^4 \sigma_{\epsilon_\nu}^2) \equiv \mathcal{N}(0,\sigma_\mathrm{diff}^2)\). If we now define \(Z = (X-Y)^2\), then \(Z \sim \sigma_\mathrm{diff}^2 \chi_1^2\), where \(\chi_1^2\) is the standard chi-squared distribution with one degree of freedom (see chi-squared proof notes). The mean of \(Z\) is \(\langle Z\rangle = \sigma_\mathrm{diff}^2\) and therefore
\[\left\langle\delta t_\infty^2\right\rangle = \frac{\sigma_{\epsilon_{\nu'}}^2 + r^4 \sigma_{\epsilon_\nu}^2}{(r^2-1)^2}.\]If \(\sigma_{\epsilon_{\nu'}} = \sigma_{\epsilon_\nu}\), then we have
\[\sigma_{\delta t_\infty} = \sqrt{\left\langle\delta t_\infty^2\right\rangle} = \sigma_{\epsilon_\nu} \left(\frac{r^4 + 1}{r^4 - 2r^2 + 1}\right)^{1/2}.\]