I have a multivariate function $f:\mathbb{R}^n \to \mathbb{R}$ that takes as input $n$ random variables, each i.i.d. according to a uniform distribution on the interval $(a,b)$. Denote the set of these random variables as $\mathbf{X}.$

I would like to derive an approximate closed-form solution of $E[f(\mathbf{X})]$.

According to https://stats.stackexchange.com/questions/5782/variance-of-a-function-of-one-random-variable, a Taylor series expansion can be done to approximate $E[f(X)]$ for the single variable case:

$$E[f(X)] = f(E[X]) + \frac{f''(E[X])}{2} E[(X- E[X])^2] + R^3$$

where $R^3$ is some remainder term (see link for more details).

Does this extend to the multivariate case as well?

The multivariate taylor for $f(\mathbf{X})$ around $\mathbf{a}$ is

$$f(\mathbf{a})+ (\mathbf{X} - \mathbf{a})^T \nabla f(\mathbf{a}) + (\mathbf{X} - \mathbf{a})^T(H_f(\mathbf{X})^T(\mathbf{X} - \mathbf{a})) + \dots $$

How would the extension for $E[f(\mathbf{X})]$ look? Maybe like this:

$$E[\mathbf{X}] = f(E[\mathbf{X}]) + E[\mathbf{X}- E[\mathbf{X}]]^T\frac{H_f(E[\mathbf{X}])^T}{2} E[\mathbf{X}- E[\mathbf{X}]] + R^3$$

where $E[\mathbf{X}_i] = (a+b)/2, \forall i$.

But doesn't $E[\mathbf{X}- E[\mathbf{X}]]=0$ so the second term is also 0?

How would the extension look and is an extension even possible?

  • $\begingroup$ $\frac{\int\int...\int f(X)dX}{\int\int...\int dX}=\frac{\int\int...\int f(X)dX}{(b-a)^n}$ $\endgroup$ – Djura Marinkov Oct 8 '17 at 14:28
  • 1
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – BGM Oct 8 '17 at 15:24

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