I have a time-varying signal $f(t)$. Couple of probes are trying to measure its values across different time intervals:

\begin{align} X_1 &= f(t = t_0) + e_1 \\ X_2 &= f(t = t_0+h) + e_2 \end{align}

$e_1$ and $e_2$ are two independent random variables with zero mean and variance = $\sigma^2$. Goal is to estimate the rate of change of the signal using following estimator

$$R = (X_2 - X_1)/h$$

How should I calculate $E(R)$ and $\text{Var}(R)$?

  • $\begingroup$ Just to be clear, the noise variables are independent across time and from each other? And $h$ is a known constant? $\endgroup$ Commented Aug 27, 2016 at 12:59
  • $\begingroup$ Yes, noise variables are independent of time. h is a "configurable" parameter like the one in kernel density estimator. $\endgroup$
    – dbose
    Commented Aug 27, 2016 at 14:43

1 Answer 1


Unless I'm missing something, you're overthinking this problem. The expectation operator is linear, so we have: $$ E(R) = \frac{1}{h} \left( E(X_1) - E(X_2) \right) $$

Then all you have to do is plug in some numbers.

Edit: $f$ is a known function and $t_0$ and $h$ are known values, so $E\left(f(t=t_0)\right)=f(t=t_0)$ and $E\left(f(t=t_0+h)\right)=f(t=t_0+h)$. Then by the same linearity used above, we have $$ E\left(f(t=t_0) + e_1\right) = E\left(f(t=t_0)\right) + E\left(e_1\right) $$ and $$ E\left(f(t=t_0 + h) + e_2\right) = E\left(f(t=t_0 + h)\right) + E\left(e_2\right). $$

Since $E(e_1) = E(e_2) = 0$ by assumption, the expectations are just $f(t=t_0)$ and $f(t=t_0+h)$. Then $E(R) = \frac{1}{h}(f(t=t_0) - f(t=t_0+h))$.

  • $\begingroup$ Thanks @ssdecontrol. expectation $\endgroup$
    – dbose
    Commented Aug 29, 2016 at 5:44
  • $\begingroup$ Using Taylor series, this is what I arrived at - $$ E(e_1) = E(e_2) = 0 \\ E(R) = h^{-1}(E[f(t_0+h)] - E[f(t_0)]]) \\ E(R) = h^{-1}E(hf'(t_0)) = f'(t_0) $$ Does this sound good to you? As the $Bias(f'(t),R) = 0$, looks like $R$ is an unbiased estimator for the rate of change. $\endgroup$
    – dbose
    Commented Aug 29, 2016 at 5:52
  • $\begingroup$ @dbose what? you definitely don't need Taylor series for this. $\endgroup$ Commented Aug 29, 2016 at 13:04
  • $\begingroup$ I added the rest of the soultion into the answer, hopefully it clarifies things $\endgroup$ Commented Aug 29, 2016 at 13:10

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