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In the context of reconstruction of climate data from ice cores (see related question at MSE) I came about the following problem. (More background and motivation can be given on demand.)

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be an arbitrary smooth function, and let $p(x)$ be a probability function with $\int_{-\infty}^{\infty} p(x)\text{d}x = 1$. We can write:

$$\int_{-\infty}^{\infty} f(x)\ p(x)\ \text{d}x = f\Big( \int_{-\infty}^{\infty} x \ p(x) \ \text{d}x \Big) + \epsilon$$

My question is for the error $\epsilon$. Can there be a closed form or an estimation of lower and upper bounds in terms of $f$ and $p$? Which mathematical techniques do I have to apply?

There are some extreme cases where an exact solution is easy to obtain:

(1) $f(x) \equiv x_0$. In this case we have

$$\int_{-\infty}^{\infty} x_0\ p(x)\ \text{d}x = x_0 \int_{-\infty}^{\infty}\ p(x) \ \text{d}x = x_0$$

which yields $\epsilon = 0$.

(2) $f(x) = \alpha x$. In this case we have

$$\alpha \int_{-\infty}^{\infty} x \ p(x)\ \text{d}x = \alpha\ \int_{-\infty}^{\infty} x \ p(x) \ \text{d}x + \epsilon$$

which again yields $\epsilon = 0$.

(3) $p(x) = \delta(x)$. In this case we have for the left-hand side

$$\int_{-\infty}^{\infty} f(x)\ \delta(x)\ \text{d}x = f(0)$$

and for the right-hand side

$$f\Big(\int_{-\infty}^{\infty} x\ \delta(x)\ \text{d}x\Big) = f(0)$$

which again yields $\epsilon = 0$.

(4) [omitted]

Instead of the general case, I would also be happy with solutions for the next most simple cases:

(5) $f(x) = x^2$

(6) A normal distribution, essentially

$$p(x) = e^{-x^{2}}$$

Again: Which techniques should be applied to calculate or estimate the error $\epsilon$?

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    $\begingroup$ By Jensens inequality we know $\int_\mathbb{R} f(x) p(x) \, dx \geq f\left( \int_\mathbb{R} x p(x) \, dx \right)$ for all convex functions $f$ and $\int_\mathbb{R} f(x) p(x) \, dx \leq f\left( \int_\mathbb{R} x p(x) \, dx \right)$ for all concave functions $f$. Suppose the support of $p$ is some Interval $I=[a,b]$ with $a \geq b$, then Jensens inequality also tells us, that we have equality if and only if $f$ is linear on $I$. This explains (1), (2) and (3) and might indicate, that bounding the second derivative of $f$ might help. $\endgroup$
    – Ben Deitmar
    Commented Nov 26, 2021 at 13:50
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    $\begingroup$ You can rewrite this with the expectation operator as $\epsilon = E[f(X)]-f(E[X])$. Then case 5 gives $\epsilon$ equal to the variance of $X$, and cases 5 and 6 together is the set-up for the delta method. $\endgroup$
    – user44143
    Commented Nov 26, 2021 at 14:05
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    $\begingroup$ An interesting case is $f(x)=e^{tx}$ giving $\epsilon=M(t)-e^{t\mu}$ where $M(t)$ is the moment generating function and $\mu=E(X)$. If we relax $f(x)$ to be complex valued and set $f(x)=e^{ix\lambda}$ we get $\epsilon=\kappa(\lambda)-e^{i\lambda\mu}$ being $\kappa(\lambda)$ the characteristic function for density $p(x)$. I think these relationships can provide some insight about techniques to deal with this request. $\endgroup$
    – Jorge Zuniga
    Commented Nov 26, 2021 at 16:31
  • $\begingroup$ Note that in this last case if we apply a properly normalized Fourier Inverse Transform we get $\epsilon(x)=f(x)-\delta(x-\mu)$ where $\int_\mathbb{R}\epsilon(x)dx=0$ $\endgroup$
    – Jorge Zuniga
    Commented Nov 26, 2021 at 16:47
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    $\begingroup$ Can there be a closed form or an estimation of lower and upper bounds in terms of $f$ and $p$? Erm... Isn't the definition of $\varepsilon$ a "closed form" in terms of $f$ and $p$? In general, if you want the expressions to be reasonably close, then you need to require that $f$ is nearly linear on almost all of the support of $p$ and that $f$ grows slower that $p$ decays on the tail part. Otherwise it is going to be a rather poor approximation... $\endgroup$
    – fedja
    Commented Nov 30, 2021 at 2:13

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