I would like to estimate the elevation at which species richness is expected to be maximal. The relationships between species richness ($y$) and elevation ($x$) follows a second order polynomial function:
$$y=a*x^2+b*x+c$$
Using R, I obtained an estimation of the parameters $a,\, b$ and $c$; noted $E(a),\, E(b)$ and $E(c)$; as well as an standard deviation around these estimation $SD(a),\, SD(b)$ and $SD(c).$
Now, I can compute the elevation (Xmax=x[y=max(y)]) at which species richness ($y$) is expected to be maximal using:
$X_{max}=-b/(2*a)$ and $E(X_{max})=-E(b)/2*E(A).$
My question is, what is the expected error on $E(X_{max})$ (i.e. $SD(X_{max})$).
I found that $SD(X_{max})$ can be estimate using propagation error estimation:
$$SD(X_{max})=|E(X_{max})|*\sqrt{\frac{V(a)}{E(a)^2}+\frac{V(b)}{E(b)^2}-2\frac{Cov(a,b)}{E(a)*E(b)}}.$$
But what is the covariance $Cov(a,b)$ between the paramerters $a$ and $b$ of second order polynomial function?
If somebody has another idea, I will be happy too!
Best,
Thomas
It is difficult to give good advice without more information about your data. I will give a partial discussion of what I imagine to be the simplest case for your data and best solution for what you want to accomplish.
Suppose you observed the number of species $Y_i$ at $n$ elevations $x_i$ along a particular mountain slope, using the model $$Y_i = \beta_0 + \beta_1 x_i + \beta_2 x_i^2 + e_i,$$ for $i = 1, \dots, n$ and $e_i$ iid $Norm(0, \sigma_e)$.
You will get estimates $\hat \beta_0,\, \hat \beta_1,$ and $\hat \beta_2$ of the coefficients, and $\hat \sigma^2$ of the error variance. Getting confidence intervals for the $\beta_j$ ($j = 0,1,2$) in terms of $\hat \sigma$ and Student's t distribution is straightforward, but depends on the normal distribution of the errors $e_i$ in your model (and the constancy of the variance for different altitudes).
Now it seems you want a confidence interval of the maximum point $\eta = -\beta_1/2\beta_2$ of the parabolic curve of your model. A reasonable point estimate is $\hat\eta = -\hat\beta_1/2\hat\beta_2$ (your $-b/2a$).
However, this estimate does not have a simple distribution. In practice, I suppose there are two difficulties. First, even if your errors were normal, the distribution theory would be messy because, as you note, numerator and denominator are not independent. Second, in practice, the errors are unlikely to be normal and finding the distribution of your point estimate is even further from reach by traditional methods.)
I think your best bet is to bootstrap the residuals from your model. If $$\hat Y_i = \hat \beta_0 + \hat \beta_1 x_i + \hat \beta_2 x_i^2,$$ is the 'predicted' number of species at the $i$th altitude, then the corresponding residual is $r_i = \hat Y_i - Y_i.$ If your model is correct, then the $r_i$ should be a random sample from a distribution useful for determining your desired CI.
Roughly speaking, the bootstrap procedure would sample $n$ values $with\; replacement$ from among these $n$ residuals $r_i$. Call, them $r_i^*$. Then construct re-sampled $Y_i^* = \hat Y_i + r_i^*$. Rerun the regression with these $Y_i^*$ and the original $x_i$, to get $\hat \eta^*.$ Repeat this resampling several thousand times $B$ to get a sequence of $\eta_k^*,$ for $k = 1, \dots, B.$ Their distribution contains the required information on the variability of your point estimate $\hat \eta$ to permit construction of a bootstrap CI.
I will not try to give more details here because I don't know the exact nature of your data. But I think bootstrapping may be the only reliable way to assess the variability of your $\hat \eta$, even if your model and data are not exactly as I imagined.
Note: If you google 'bootstrap regression' you will see quite a variety of resampling approaches for regression. Resampling can be at the level of the data, the residuals (as above), or the distributions of the error and/or parameters.
Addendum: I did a bootstrap on fake data with $n = 56$ elevations evenly spaced from 1000 to 12000, corresponding $Y_i$ (one per elevation) with $\beta_0 = 20,\,\beta_1 = 0.013,\, \beta_2 = 1e-06,$ and normal error with $\sigma_\epsilon = 5$. Hence $\eta = 6500.$ Initial regression had $\hat \eta = 6551.7.$ A 95% Bootstrap CI: $(6362.1, 6736.8).$ I could post R code and a plot on your request.