OK, the final answer is

```
E[sqrt(x^2+y^2)] = s1 * sqrt(2/pi) * EllipticE(1-rho^2)
```

where s1 and rho are defined as in my previous post.

Alternatively, if we put the final answer in the form mentioned in my previous post

```
E[sqrt(x^2+y^2)] = s1*( sqrt(2/pi) + g(rho)*(sqrt(pi/2)-sqrt(2/pi)))
where g(rho) = (EllipticE(1-rho^2) - 1)/(pi/2 - 1)
[ g(rho) goes from 0 to 1 as rho goes from 0 to 1. ]
```

For anyone interested in how I got it, here are the gory details:

We want E[sqrt(x^2+y^2)], which is defined as

```
E[sqrt(x^2+y^2)] = integral[ sqrt(x^2+y^2) * p_x(x) * p_y(y) dx dy, x=-inf..inf, y=-inf..inf]
where
p_x(x) = (1/sqrt(2*pi*s1))*exp(-x^2/(2*s1^2))
p_y(y) = (1/sqrt(2*pi*s2))*exp(-y^2/(2*s2^2))
```

Mathematica and Maple completely barf on this, so we have to re-write in a form that one of them can “digest” it.

One way to re-write this is as

```
E[sqrt(x^2+y^2)] = E[sqrt(s1^2*n1^2 + s2^2*n2^2)]
```

where n1 and n2 are i.i.d. ~ N(0,1)

Then, using polar coordinates

```
R = sqrt(n1^2+n2^2)
theta = atan(n2/n1)
```

R will be Rayleigh with parameter sigma = 1, and theta will be uniform [0,2*pi]

Using the above,

```
n1 = R*cos(theta)
n2 = R*sin(theta)
```

So,

```
E[sqrt(s1^2*n1^2 + s2^2*n2^2)] = integral[ sqrt(s1^2*R^2*cos^2(theta) + s2^2*R^2*sin^2(theta)) * p_R(R) * p_theta(theta) dR dtheta, r=0..inf, theta=0..2*pi]
```

where p_R() is the pdf of a Rayleigh rv parameter sigma=1

and p_theta() is the pdf of a rv that is uniform over [0,2*pi]

The above intergral can be re-written as

```
integral[ R*p_R(R) dR, r=0..inf] * integral[sqrt(s1^2*cos^2(theta) + s2^2*sin^2(theta)), theta=0..2*pi]
```

The first integral is just the mean of a Rayleigh rv with parameter sigma=1, so the mean is sqrt(pi/2)

For the second integral, Mathematica gave me, after some massaging of the answer, (2/pi)*s1*EllipticE(1-rho^2)

So the final answer is

```
E[sqrt(x^2+y^2)] = sqrt(pi/2) * (2/pi)*s1*EllipticE(1-rho^2)
```

which simplifies to

```
E[sqrt(x^2+y^2)] = sqrt(2/pi) * s1 * EllipticE(1-rho^2)
```

For rho = 0, we have EllipticE(1-rho^2) = 1

For rho = 1, we have EllipticE(1-rho^2) = pi/2

So,

if(rho == 1), E[sqrt(x^2+y^2)] = s1*sqrt(pi/2)*

if(rho == 0), E[sqrt(x^2+y^2)] = s1sqrt(2/pi)

which agrees with what was mentioned in my previous post, so it’s a good sanity check.