I am running a sensitivity study on the model $y(t) = x(t - \tau)$ where $y(t)$ and $x(t)$ are 2 time signals and $\tau$ a time lag. Basically I want to study the sensitivity of $y$ to a change in $x$. Note that the model is very simple just for the sake of illustrating my question.

What I think about is the derivative of $y$ with respect to $x$, but I cannot see clearly how this could be done. I know the result is 1, but the way I see it is:

\begin{equation} \frac{\partial y}{\partial x(t-\tau)}(t) = \frac{\partial y}{\partial t}(t)\frac{\partial t}{\partial x(t - \tau)} = \frac{\partial x}{\partial t}(t-\tau) \frac{\partial t}{\partial x(t-\tau)} = 1 \end{equation} so that a change in $x$ at $t-\tau$ is seen by the same change at a future time $t$ in $y$.

Is that a rigorous and mathematical way to look at it? I know that I might have messed up the equation and the time arguments, but I would like to learn the rigorous way to write such a problems.


Maybe to show the general case, I am interested in the following analytical models: $y(t) = \sum_i f_i^{p_i}(x(t - \tau_i))$ where $f_i$ is an elementary function. For instance if $f_i(x) = x \ \forall i$, then $y(t) = \sum_i x^{p_i}(t - \tau_i)$. What i am interested in is sensitivities of $y$ wrt to the different $x_i$'s.

  • $\begingroup$ Is there some connection (for example a model) of how $y$ and $x$ are related? I'm curious, for example, if you changed $x$ just a bit, how that would propagate to $y$. Are these measured data? $\endgroup$ Sep 13 at 17:03
  • $\begingroup$ If, on the other hand, all you want to do is explore how $y$ is related to a lagged $x$, you might explore phase relationships in the frequency domain. $\endgroup$ Sep 13 at 18:26
  • $\begingroup$ @Aruralreader the model is that $y(t) = x(t-\tau)$ ... $\endgroup$
    – outlaw
    Sep 13 at 19:50
  • $\begingroup$ So, for two given time series x(t) and y(t) we want to verify the model $y(t)=x(t-\tau)$, where $\tau$ is a given (?) time lag. If the model was perfectly well describing the data then plotting y vs. x data points we'd see they all are on a straight line (y=x). If the model is not perfect then the data points would be forming some kind of cloud, but one can find a fit using least squares fitting, so it will be y = a x + b + $\sigma$, where $\sigma$ is the average error of the fit. These three parameters: a, b, sigma is what describes the sensitivity of y to x, on average, for given data set. $\endgroup$ Sep 14 at 0:49
  • $\begingroup$ If one signal is a time shift of the other, perhaps with an amplitude scaling, this would show up quickly in a (windowed) cross-correlation. Windows and the frequency domain will be your friends. See any signal processing book to get started. $\endgroup$ Sep 14 at 1:10

In essence, what you are asking is the ratio of how $y$ changes at $t$ to how $x$ changes at $t$. That is: $$ \frac{dy}{dx}(t) = \frac{\frac{dy}{dt}(t)}{\frac{dx}{dt}(t)}. $$ Using that $y(t)=x(t-\tau)$, you then get that this is equal to $$ \frac{dy}{dx}(t) = \frac{\frac{dx}{dt}(t-\tau)}{\frac{dx}{dt}(t)}. $$ But this is definitely not equal to one: For example, for $x(t)=t^2$, you would obtain $$ \frac{dy}{dx}(t) = \frac{\frac{dx}{dt}(t-\tau)}{\frac{dx}{dt}(t)} = \frac{t-\tau}{t}. $$ It is true, however, that $$ \lim_{\tau\rightarrow 0}\frac{dy}{dx}(t) = \lim_{\tau\rightarrow 0}\frac{\frac{dx}{dt}(t-\tau)}{\frac{dx}{dt}(t)} = 1. $$

  • $\begingroup$ Thanks for the answer. What I dont see clear is the third equation in the sense that I expected the derivative to be 0, since a change in $x$ at $t$ is only seen in $y$ at $t + \tau$, but not at $t$. $\endgroup$
    – outlaw
    Sep 14 at 6:03
  • $\begingroup$ Can you also please check the updated question $\endgroup$
    – outlaw
    Sep 14 at 6:14
  • 1
    $\begingroup$ @outlaw When you say "I expected the derivative to be zero, since a change in $x$ at $t$ is only seen in $y$ at $t+\tau$ but not $t$", you seem to imply a causal relationship. But that is not the question. The way you formulate it is as a correlation: How are changes at times $t$ and $t+\tau$ correlated, and my derivation shows you how. You would be right, however, if you thought that $x(t)$ is a random function, in which values are completely uncorrelated. In that case, however, you would already have trouble defining time derivatives. $\endgroup$ Sep 14 at 16:39
  • $\begingroup$ I see your point. Can you point out a reference for the definition you used: $\frac{dy}{dx}(t) = \frac{\frac{dy}{dt}(t)}{\frac{dx}{dt}(t)}$ ? I know one can say this is the chain rule, but I dont see the mathematical derivation of it obviously. $\endgroup$
    – outlaw
    Sep 15 at 10:50
  • 1
    $\begingroup$ I don't have a reference, but for sufficiently smooth functions, this makes sense if you consider $dx$, $dy$, $dt$ to be finite changes before making the transition to infinitesimals. $\endgroup$ Sep 15 at 17:33

After more thinking, I am answering my own question. What I was looking for is the following:

Given for example $y(t) = x(t) + z(t-\tau_1)k(t)$, where $x(t), y(t), z(t), k(t)$ time signals, I can write $y(t) = F(t, x(t), z(t-\tau_1), k(t))$ for some function $F$. I want to study the following:

For a fixed time $t$ \begin{align} & \lim_{\epsilon \to 0} \frac{F(t, x(t) + \epsilon, z(t-\tau_1), k(t)) - F(t, x(t),z(t-\tau_1), k(t))}{\epsilon} = \\ &\lim_{\epsilon \to 0} \frac{x(t) + \epsilon + z(t-\tau_1)k(t) - x(t) - z(t-\tau_1)k(t)}{\epsilon} = \lim_{\epsilon\to 0} \frac{\epsilon}{\epsilon} = 1. \end{align} This quantity is what I refer to $\frac{\partial y}{\partial x(t)}(t)$. For example also, \begin{align} \frac{\partial y}{\partial z(t-\tau_1)}(t) := &\lim_{\epsilon \to 0} \frac{F(t, x(t), z(t-\tau_1) + \epsilon, k(t)) - F(t, x(t),z(t-\tau_1), k(t))}{\epsilon} \\ & \lim_{\epsilon \to 0} \frac{(z(t-\tau_1)+\epsilon)k(t) - z(t-\tau_1)k(t)}{\epsilon} = k(t) \end{align}

The partial derivatives notations that I am using might make sense mathematically, but in terms of limits, that is what I was looking for.


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