Variance of a derived magnitude

Question

I'm wondering about how to present results on a report (and how to interpret it).

Let $Y = f(\mathbf{X})$ be a random variable. Of course, if we derive its PDF $f_{Y}(y)$, we could present it on the report and the reader would have all the information for that variable. But, suppose we compute its variance approximately (without passing through its PDF) by using the formula \begin{equation*} \text{Var}\left(Y\right) = \sum_{i=1}^{n} \left( \left. \frac{\partial y}{\partial x_i} \right\vert _{\mu_i} \sigma_{X_{i}} \right)^2 \end{equation*} When we present the result as $(\mu_{Y} - \sigma_{Y}, \mu_{Y} + \sigma_{Y})$, we aren't really giving any information about how it's distributed.

When the standard deviation of a variable is given without its PDF, are we supposed to interpret it with some inequality (Chebyshev's, for example) to give us a bound for a confidence interval?

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I'm asking this because I took two laboratory courses reporting magnitudes as mentioned and now, doing a probability course, I've learned that a function of Gaussian distributed variables doesn't follows in general, a Gaussian. So I want to know what's the point on reporting standard deviation for an unknown distribution: using an upper bound for the confidence interval (Chebyshev, the only one that works for any distribution) or there are other reasons.

The question implicitly asks if what I'm saying is correct. If it's not clear what I mean, please leave a comment so I can make an attempt to clarify.

Answer 1

I would say that this approach is used — giving the standard deviations is a useful measure of uncertainty — essentially because lots of quantities encountered in scientific measurements are approximately Gaussian (Central Limit Theorem, blah blah blah ...) and because the new/transformed quantities are still approximately Gaussian under reasonable conditions, so one may still be able to interpret $\pm 2 \sigma$ as an approximate 95% confidence interval, or $\pm 3 \sigma$ as an approximate 99.5% CI.

More specifically, if the input $x$ values are approximately Gaussian, and the function $f$ is not too strongly nonlinear (i.e. $f(x) \approx f(x_0) + (x-x_0) f'(x)$ is reasonably accurate over the range of $x$, where $x_0$ is the mean [or median or whatever] of $x$), then the $y$ will also be approximately Gaussian.

There are lots of other approaches that will give more reliable confidence intervals/expressions of uncertainty (bootstrapping, Bayesian methods, likelihood profiles, "parametric bootstrapping" (i.e. drawing random samples from $\mathbf X$ assuming it is multivariate normal and propagating the error through $f$), calculations based on higher-order Taylor expansions ...), but

• a given researcher may not know how to apply these methods, or may not want to take the trouble (these fancier computations were a much bigger problem in the era before powerful computers and high-level programming languages were universally available);
• in many cases we are stuck knowing just the mean and SD of the original values (e.g. if we are synthesizing results from existing literature), which rules out a lot of the fancier approaches.

Note by the way that your expression assumes that all the $X_i$ are independent, which is not true in all cases (and can cause serious headaches when you forget to include the covariance/cross-terms — ask me how I know!). The more general expression is

$$ \sigma^2_y = \sum_i \left(\frac{\partial y}{\partial x_i}\right)^2 \sigma^2_x + \sum_i \sum_{j \neq i} \frac{\partial y}{\partial x_i} \cdot \frac{\partial y}{\partial x_j} \cdot \textrm{Cov}(x_i,x_j) $$

Answer 2

The answer is not entirely straightforward, for a number of reasons.

First it depends on the field, where you're presenting such a result. In several natural sciences, it is simply customary to provide a result with an uncertainty. How to obtain that result and the uncertainty is another matter, but the general rule is to present it like you have written $$value = \mu_Y \pm \sigma_Y$$ Depending on the actual method, there might be even different $\sigma_Y$ present for the upper and the lower standard deviation. This happens for example, when using numerical Likelihood estimators without a symmetric function for the minimum. The main argument for this kind of expression is that the value obtained by the experiment and the estimation procedure lies with a "probability" of about 68% within the given interval. If one starts to put this more explicitly, then one has to chose the particular brand of probability one chooses to adhere to. Note that the probability is only attributed to the interval, not the value.

Now, the interesting thing here is, that this result holds true for any kind of distribution. You can assume a normal distribution, and by means of whatever (legal) expression transform it to any other distribution. The actual values for the result might change; but the "probability content" doesn't change under these transformations -- it will remain 68% in the given interval. This is the integral of the pdf of this interval; or the cdf value.

This is more convention than anything else. I assume the question is in connection to physics, chemistry or biology. You're of course right, that it might also be of interest to know the underlying distribution. But in a lot of cases this is not accessible, the estimation procedure(s) are still valid, though. So, it's just customary to do it that way.

The mentioning of a confidence interval is somewhat related, both in terminology as in expression. But it's not the same thing as what is mentioned above. It could be an example for a confidence interval. For once, a confidence interval is always associated with a level. The confidence level of the "result expression" from above is 68%.

Even if that is not part of the question, the actual interpretation of what a confidence interval is, is not trivial. See the (debated and debatable) wiki page on confidence intervals