Expectation of cross product of normal distributed variables

Question

Given are 6 independent normal distributed variables $X_i=\mathcal{N}(\mu_i,\sigma^2)_{i=1...6}$ with expectations $\mu_i$ and variance $\sigma^2$. How the expected value of the norm of their cross product can be calculated or approximated?

$$\mathbb{E}\left[ \left \| \begin{pmatrix} X_1\\X_2\\X_3 \end{pmatrix} \times \begin{pmatrix} X_4\\X_5\\X_6 \end{pmatrix} \right\| \right]=\mathbb{E}\left[ \left\|\begin{pmatrix} X_2 X_6-X_3 X_5\\X_3 X_4 - X_1 X_6\\ X_1 X_5-X_2 X_4\end{pmatrix}\right\|\right]=\mathbb{E}\left[\sqrt{(X_2 X_6-X_3 X_5)^2+(X_3 X_4 - X_1 X_6)^2+(X_1 X_5-X_2 X_4)^2}\right]$$

It can be assummed $$\sigma\ll \left\|\begin{pmatrix} \mu_1\\\mu_2\\\mu_3\end{pmatrix}\right\|=\sqrt{\mu_1^2+\mu_2^2+\mu_3^2}$$

$$\sigma\ll \left\|\begin{pmatrix} \mu_4\\\mu_5\\\mu_6\end{pmatrix}\right\|=\sqrt{\mu_4^2+\mu_5^2+\mu_6^2}$$

Answer 1

The approach mentioned by @huber is probably what you need for this particular random variable.

Here is how to perform a 2nd order Taylor series approximation (using Mathematica as your other questions indicate you use this software).

(* Function of interest *)
f = Sqrt[(x[2] x[6] - x[3] x[5])^2 + (x[3] x[4] - x[1] x[6])^2 + (x[1] x[5] - x[2] x[4])^2];

(* Make a conversion to get Mathematica to do the desired 2nd order Taylor series *)
(* Technique from 
https://mathematica.stackexchange.com/questions/15023/multivariable-taylor-expansion-does-not-work-as-expected *)
f = f /. x[i_] -> (z[i] - \[Mu][i]) t + \[Mu][i];

(* Get 2nd order Taylor series *)
taylor = (Series[f, {t, 0, 2}] // Normal) /. t -> 1 // Expand;

(* Take expectation of Taylor series *)
(* Convert z[i]^2 to \[Sigma]^2 + \[Mu][i]^2 *)
mean = taylor //. z[i_]^2 -> \[Sigma]^2 + \[Mu][i]^2;
(* Convert z[i] to \[Mu][i] *)
mean = FullSimplify[mean //. z[i_] -> \[Mu][i]]

$$\frac{\left(\mu_1^2+\mu_2^2+\mu_3^2+\mu_4^2+\mu_5^2+\mu_6^2\right) \sigma^2+2 \left(\left(\mu_5^2+\mu_6^2\right) \mu_1^2-2 \mu_3 \mu_4 \mu_6 \mu_1+\mu_3^2 \left(\mu_4^2+\mu_5^2\right)-2 \mu_2 \mu_5 (\mu_1 \mu_4+\mu_3 \mu_6)+\mu_2^2 \left(\mu_4^2+\mu_6^2\right)\right)}{2 \sqrt{\left(\mu_5^2+\mu_6^2\right) \mu_1^2-2 \mu_3 \mu_4 \mu_6 \mu_1+\mu_3^2 \left(\mu_4^2+\mu_5^2\right)-2 \mu_2 \mu_5 (\mu_1 \mu_4+\mu_3 \mu_6)+\mu_2^2 \left(\mu_4^2+\mu_6^2\right)}}$$

And if we let

$$g=\sqrt{(\mu_2 \mu_6-\mu_3 \mu_5)^2+(\mu_3 \mu_4+\mu_1 \mu_6)^2+(\mu_1 \mu_5-\mu_2 \mu_4)^2}$$

(which is the form of the random variable of interest but with the respective means plugged in), then the approximate mean can be written as

$$g+\frac{\left(\mu_1^2+\mu_2^2+\mu_3^2+\mu_4^2+\mu_5^2+\mu_6^2\right) \sigma^2}{2 g}$$

Now...is this approximation precise enough? Here's an example.

(* Set mean of x[1],...,x[6] *)
xMeans = {1, 2, 3, 3, 7, 5};
\[Mu]Values = Thread[{\[Mu][1], \[Mu][2], \[Mu][3], \[Mu][4], \[Mu][5], \[Mu][6]} -> xMeans]
(* Get a variance satisfying the OP's restriction *)
\[Sigma]X = 0.01 Min[1^2 + 2^2 + 3^2, 3^2 + 7^2 + 5^2]^0.5
(* 0.0374166 *)

(* 2nd order Taylor approximation to mean *)
mean /. \[Mu]Values /. \[Sigma] -> \[Sigma]X
(* 11.7531 *)

(* Estimation of mean by simulations *)
SeedRandom[12345];
n = 10000;
xbar = 0;
Do[xx = RandomVariate[NormalDistribution[0, 1], 6];
 xx = xMeans + \[Sigma]X*xx;
 xbar = xbar + 
   Sqrt[(xx[[2]] xx[[6]] - xx[[3]] xx[[5]])^2 +
        (xx[[3]] xx[[4]] - xx[[1]] xx[[6]])^2 + 
        (xx[[1]] xx[[5]] - xx[[2]] xx[[4]])^2],
 {i, n}]
xbar = xbar/n
(* 11.7543 *)

(* Approximate relative precision *)
(xbar - mean /. \[Mu]Values /. \[Sigma] -> \[Sigma]X)/xbar
(* 0.000103943 *)

If this approximation is not good enough, then you might want to try a higher order Taylor series approximation. Alternatively, finding an approximation for the mean by generating lots of random samples with known values for the parameters is pretty quick, too.