Stability estimates on quotients of the form $ \frac{\prod_{j=1}^n a_j}{\prod_{j=1}^n b_j} $

Question

Suppose that $a_j,b_j \in \mathbb C$ are complex numbers, $j=1,\dots,n$, with the property that $|a_j|,|b_j| \geq c > d >0$ where $c,d$ are positive real numbers. I'm interested in the stability of the product $$ \frac{\prod_{j=1}^n a_j}{\prod_{j=1}^n b_j} $$ under noise in the following sense: Let $$ \tilde{a}_j = a_j + \delta_j $$ and $$ \tilde{b}_j = a_j + \gamma_j $$ where $\delta$ and $\gamma$ is $\ell^2$ noise such that $\| \delta \|_{\ell^2},\| \gamma \|_{\ell^2} \leq d$. The assumption on the $\ell^2$ norm to be smaller than $d$ implies that the quotient $$ \frac{\prod_{j=1}^n \tilde a_j}{\prod_{j=1}^n \tilde b_j} $$ is well-defined. The difference $$ \left | \frac{\prod_{j=1}^n \tilde a_j}{\prod_{j=1}^n \tilde b_j} - \frac{\prod_{j=1}^n a_j}{\prod_{j=1}^n b_j} \right | $$ tends to zero as the noise tends to zero (w.r.t the $\ell^2$-norm). I'm searching for a (possibly sharp) estimate on this difference in terms of some norm of the noise and $n$. Is there a way to get such inequalities? This question seems to be quite elementary so maybe there exists some literature on this?

Answer 1

Perhaps the simplest approach is to consider the problem as the calculation of a functional derivative. Let's put $$ (a_1, \ldots, a_n, b_1, \ldots, b_n)\triangleq (\mathbf{a}, \mathbf{b})\in\Bbb C^{2n} $$ and, in an analogous fashion, $$ (\delta_1, \ldots, \delta_n, \gamma_1, \ldots, \gamma_n)\triangleq (\boldsymbol{\delta}, \boldsymbol{\gamma})\in\Bbb C^{2n} $$ Then, defining the complex valued functional $$ \mathfrak{F}\big[(\mathbf{a}, \mathbf{b})\big]= \frac{\prod_{j=1}^n a_j}{\prod_{j=1}^n b_j} $$ it is natural to write $$ \mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]= \frac{\prod_{j=1}^n (a_j+\epsilon\delta_j)}{\prod_{j=1}^n (b_j+\epsilon \gamma_j)}\quad \forall \epsilon >0 $$ and calculating the functional derivative as usual $$ {\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma}) \triangleq \left.\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]\right|_{\epsilon=0} $$ This procedure occasionally shows up in problems of sensitivity analysis in signal processing. This functional derivative is linear respect to the variation $(\boldsymbol{\delta}, \boldsymbol{\gamma})$: if it is a Fréchet derivative, it is possibly relatively easy to obtain a nice estimate in the form $$ \left|\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]-\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})\big]\right|\le c\left|{\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma})\right|\label{1}\tag{☆} $$ with a sharp constant $c$. However, even if it is a Gâteaux derivative (i.e. it depends on the structure of the variation $(\boldsymbol{\delta}, \boldsymbol{\gamma})$) you may able to obtain a sharp estimate.

Addendum. Without going in the details and assuming the (plausible) non degeneracy hypothesis $\left|{\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma})\right| >0$ when $(\boldsymbol{\delta}, \boldsymbol{\gamma})\neq 0$, further fairly elementary steps for possibly arriving at \eqref{1} are the following ones: $$ \begin{split} \left|\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]-\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})\big]\right| &= \int\limits_0^1\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]\mathrm{d}\epsilon \\ & \le \int\limits_0^1\left| \frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big] \right| \mathrm{d}\epsilon \\ & = \left| {\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma})\right|\int\limits_0^1\left| \frac{\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]}{\left.\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]\right|_{\epsilon=0}} \right| \mathrm{d}\epsilon \\ & \le c\left|{\delta \mathfrak{F}}\big[(\mathbf{a}, \mathbf{b})\big](\boldsymbol{\delta}, \boldsymbol{\gamma})\right| \end{split} $$ and therefore the relation $$ \int\limits_0^1\left| \frac{\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]}{\left.\frac{\mathrm{d}}{\mathrm{d}\epsilon}\mathfrak{F}\big[(\mathbf{a}, \mathbf{b})+\epsilon(\boldsymbol{\delta}, \boldsymbol{\gamma})\big]\right|_{\epsilon=0}} \right| \mathrm{d}\epsilon\le c \label{2}\tag{☆☆} $$ may possibly be a first guess for the sought for constant. Note that the integral on the right side of \eqref{2} can be difficult to estimate: however, despite requiring possibly tedious calculations, you can made explicit its form.