<s>[INST] Expectation with square root 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.
 [/INST] 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.
</s>