Taylor expansions for the moments of functions of random variables

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Template:Short description Template:Multiple issues In probability theory, it is possible to approximate the moments of a function f of a random variable X using Taylor expansions, provided that f is sufficiently differentiable and that the moments of X are finite.


A simulation-based alternative to this approximation is the application of Monte Carlo simulations.

First moment

Given μX and σX2, the mean and the variance of X, respectively,[1] a Taylor expansion of the expected value of f(X) can be found via

E⁡[f(X)]=E⁡[f(μX+(X−μX))]≈E⁡[f(μX)+f′(μX)(X−μX)+12f″(μX)(X−μX)2]=f(μX)+f′(μX)E⁡[X−μX]+12f″(μX)E⁡[(X−μX)2].

Since E[X−μX]=0, the second term vanishes. Also, E[(X−μX)2] is σX2. Therefore,

E⁡[f(X)]≈f(μX)+f″(μX)2σX2.

It is possible to generalize this to functions of more than one variable using multivariate Taylor expansions. For example,[2]

E⁡[XY]≈E⁡[X]E⁡[Y]−cov⁡[X,Y]E⁡[Y]2+E⁡[X]E⁡[Y]3var⁡[Y]

Second moment

Similarly,[1]

var⁡[f(X)]≈(f′(E⁡[X]))2var⁡[X]=(f′(μX))2σX2−14(f″(μX))2σX4

The above is obtained using a second order approximation, following the method used in estimating the first moment. It will be a poor approximation in cases where f(X) is highly non-linear. This is a special case of the delta method.

Indeed, we take E⁡[f(X)]≈f(μX)+f″(μX)2σX2.

With f(X)=g(X)2, we get E⁡[Y2]. The variance is then computed using the formula var⁡[Y]=E⁡[Y2]−μY2.

An example is,[2]

var⁡[XY]≈var⁡[X]E⁡[Y]2−2E⁡[X]E⁡[Y]3cov⁡[X,Y]+E⁡[X]2E⁡[Y]4var⁡[Y].

The second order approximation, when X follows a normal distribution, is:[3]

var⁡[f(X)]≈(f′(E⁡[X]))2var⁡[X]+(f″(E⁡[X]))22(var⁡[X])2=(f′(μX))2σX2+12(f″(μX))2σX4+(f′(μX))(f‴(μX))σX4

First product moment

To find a second-order approximation for the covariance of functions of two random variables (with the same function applied to both), one can proceed as follows. First, note that cov⁡[f(X),f(Y)]=E⁡[f(X)f(Y)]−E⁡[f(X)]E⁡[f(Y)]. Since a second-order expansion for E⁡[f(X)] has already been derived above, it only remains to find E⁡[f(X)f(Y)]. Treating f(X)f(Y) as a two-variable function, the second-order Taylor expansion is as follows:

f(X)f(Y)≈f(μX)f(μY)+(X−μX)f′(μX)f(μY)+(Y−μY)f(μX)f′(μY)+12[(X−μX)2f″(μX)f(μY)+2(X−μX)(Y−μY)f′(μX)f′(μY)+(Y−μY)2f(μX)f″(μY)]

Taking expectation of the above and simplifying—making use of the identities E⁡(X2)=var⁡(X)+[E⁡(X)]2 and E⁡(XY)=cov⁡(X,Y)+[E⁡(X)][E⁡(Y)]—leads to E⁡[f(X)f(Y)]≈f(μX)f(μY)+f′(μX)f′(μY)cov⁡(X,Y)+12f″(μX)f(μY)var⁡(X)+12f(μX)f″(μY)var⁡(Y). Hence,

cov⁡[f(X),f(Y)]≈f(μX)f(μY)+f′(μX)f′(μY)cov⁡(X,Y)+12f″(μX)f(μY)var⁡(X)+12f(μX)f″(μY)var⁡(Y)−[f(μX)+12f″(μX)var⁡(X)][f(μY)+12f″(μY)var⁡(Y)]=f′(μX)f′(μY)cov⁡(X,Y)−14f″(μX)f″(μY)var⁡(X)var⁡(Y)

Random vectors

If X is a random vector, the approximations for the mean and variance of f(X) are given by[4]

E⁡(f(X))=f(μX)+12trace⁡(Hf(μX)ΣX)var⁡(f(X))=∇f(μX)tΣX∇f(μX)+12trace⁡(Hf(μX)ΣXHf(μX)ΣX).

Here ∇f and Hf denote the gradient and the Hessian matrix respectively, and ΣX is the covariance matrix of X.

See also

Notes

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Further reading

  1. ↑ 1.0 1.1 Haym Benaroya, Seon Mi Han, and Mark Nagurka. Probability Models in Engineering and Science. CRC Press, 2005, p166.
  2. ↑ 2.0 2.1 Template:Cite journal
  3. ↑ Template:Cite web
  4. ↑ Template:Cite journal