I don't know how to calculate the expectation when there is some square root in the expression. My problem is this: we have three real random variables $X,Y,Z$, independent and with standard normal distribution $N(0,1)$ and we want to calculate $$E\left(\frac{1}{2} \left(X + Z + \sqrt{X^2 + 4 Y^2 - 2 XZ + Z^2}\right)\right).$$
How can this be done?
Thanks.
On
@jens
The line should read as
$\begin{align} E &= \frac{1}{2}\mathsf{E}\left[\sqrt{4Y^2 + (\sqrt{2}X)^2}\right] \\ \\ &= \frac{\sqrt{2}}{2}\mathsf{E}\left[\sqrt{2Y^2 + X^2}\right] \end{align}$
(I see you have done so.)
Now Mathematica can do the calculation easily and get:
$E=\frac{EllipticE(-1)}{\sqrt{\pi }}\approx 1.0776657899830802$
which agrees with simulations.
I am assuming that $X$, $Y$, and $Z$ are mutually independent, standard Gaussian.
The expression of your expectation can be simplified.
Let us denote your expectation as $E$.
First, notice that $X$ and $Y$ are zero-mean, so $$ E = \frac{1}{2}\mathsf{E}\left[\sqrt{X^2 + 4Y^2 - 2XZ + Z^2}\right] $$
Grouping terms, we have $$ \begin{align} E &= \frac{1}{2}\mathsf{E}\left[\sqrt{4Y^2 + (X-Z)^2}\right] \\ &= \frac{1}{2}\mathsf{E}\left[\sqrt{4Y^2 + (X+Z)^2}\right]. \end{align} $$ The last equality holds because $Z$ and $-Z$ have the same distribution, due to the fact that the distribution of $Z$ is symmetric around $0$. In addition, since $X$ and $Z$ are independent standard Gaussian, their sum is zero-mean Gaussian with variance $2$. That is, the sum $X+Z$ has the same distribution as $\sqrt{2}X$. Hence, $$ \begin{align} E &= \frac{1}{2}\mathsf{E}\left[\sqrt{4Y^2 + (\sqrt{2}X)^2}\right] \\ &= \frac{\sqrt{2}}{2}\mathsf{E}\left[\sqrt{2Y^2 + X^2}\right] \end{align} $$ @bobbym has provided a closed-form expression of this expectation computed with Mathematica in terms of an elliptic integral (see below). The way to derive it is as follows: $$ \begin{align} E &= \frac{\sqrt{2}}{2}\int_0^\infty\int_0^\infty\sqrt{2y^2 + x^2} \cdot \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \cdot \frac{1}{\sqrt{2\pi}}e^{-y^2/2} \ \mathrm{d}x\,\mathrm{d}y \\ &= \frac{\sqrt{2}}{4\pi}\int_0^\infty\int_{-\pi}^{\pi}r^2\sqrt{1+\sin^2\theta} e^{-r^2/2} \ \mathrm{d}r\,\mathrm{d}\theta. \end{align} $$ This is obtained by means of the substitution $$ \begin{align} x &= r\cos(\theta) \\ y &= r\sin(\theta) \end{align} $$ Then you use $$ \int_0^\infty r^2 e^{-r^2/2} \mathrm{d}r = \sqrt{2\pi} $$ and the symmetry of the integral over $\theta$ to show $$ \begin{align} E &= \frac{1}{\sqrt{\pi}}\int_{0}^{\pi}\sqrt{1+\sin^2\theta} \,\mathrm{d}\theta \\ &= \frac{\operatorname{E}(-1)}{\sqrt{\pi}} \end{align} $$ consistently with what @bobbym computed.