What is the PDF of $[(X-a)^2 + (Y-b)^2]^{1/2}$ where $X$ and $Y$ are two non-standard normal random variables? I have to conduct an experiment getting data from a system. These data are the estimated values, provided by the system, of a true value that we know beforehand. I then compare the estimated values with the true value to analyse how accurate the estimations of the system are.
In this context, we have a set of estimated values and a true value. Each of the estimated values and the true value are represented in terms of $(x,y)$ coordinates. Assume that the coordinate of the true value is $(a,b)$ where $a$ and $b$ are constants. Also assume that the $x$-coordinate and $y$-coordinate of the estimated values are two normally distributed random variables with different means and variances, that is $X \sim \mathcal N(m_1, v_1)$ and $Y \sim \mathcal N(m_2, v_2)$.
My question is: Is there any parametric method to calculate the probability of the random variable 
$$Z= \sqrt{ (X-a)^2 +  (Y-b)^2 }$$
@update: Thank you very much for your help. After analyzing the estimated values, I have learned that the covariance between X and Y is not zero, which is: 
$$cov(X,Y) \neq 0$$
 A: Assume that $X$ and $Y$ are independent with known means/variances.
Unfortunately, I don't think there's a nice standard form for the distribution of $Z$, but I'd be happy to be shown otherwise.
Let $X' = \frac{X - a}{\sqrt{v_1}} \sim \mathcal{N}(\frac{m_1 - a}{\sqrt{v_1}}, 1)$ and $Y' = \frac{Y - b}{\sqrt{v_1}} \sim \mathcal{N}(\frac{m_2 - b}{\sqrt{v_2}}, 1)$.
The square of each is noncentral $\chi^2$ with parameters $k = 1$, $\lambda_1 = \frac{(m_1 - a)^2}{v_1}$ (for $X'$), $\lambda_2 = \frac{(m_2 - b)^2}{v_1}$ (for $Y'$).
We can then write the pdf of their linear combination $Z^2 = v_1 X'^2 + v_2 Y'^2$ with a Laguerre expansion (Castaño-Martínez and López-Blázquez, 2005, (3.2)):
$$
f(z) =
  \frac{1}{2 \beta} e^{-\frac{z}{2\beta}}
  \sum_{k \ge 0}
    \frac{k! c_k}{(1)_k}
    L_k\left( \frac{2 z}{4 \beta \mu_0} \right)
$$
where $\mu_0 > 0$ and $\beta > 0$ are arbitrary parameters,
$L_k$ is the $k$th Laguerre polynomial,
I think $(1)_k$ is the rising factorial so that it's $k!$ and cancels with the $k!$ in the numerator,
and the $c_k$ satisfy these recurrences:
$$
c_0 =
\frac{1}{\mu_0}
\exp\left( - \frac{1}{2} \sum_{i=1}^2 \frac{\lambda_i}{1 + \frac{\beta \mu_0}{v_i (1 - \mu_0)}} \right)
\prod_{i=1}^2 \left( 1 + \tfrac{v_i}{\beta} \left(\tfrac{1}{\mu_0} - 1\right) \right)^{-1/2}
\\
c_k = \frac{1}{k} \sum_{j=0}^{k-1} c_j d_{k-j}
\\
d_j =
- \frac{j \beta}{2 \mu_0}
  \sum_{i=1}^2
    \lambda_i v_i (\beta - v_i)^{j-1}
    \left( \frac{\mu_0}{\beta \mu_0 + v_i(1 - \mu_0)} \right)^{j+1}
+ \sum_{i=1}^2 \tfrac12 \left(\frac{1 - v_i/\beta}{1 + (v_i/\beta) (1/\mu_0 - 1)}\right)^j
$$
In practice, you can truncate the sum after a few values of $k$. The authors show an error bound in (3.9), though according to Bausch (2013) (who gives a more computationally efficient approximation for linear combinations of central chi-squareds) the bound is quite conservative.
If you want the cdf there's a similar expression (3.5):
$$
F(z) =
  \frac{1}{(2 \beta)^2} z e^{-\frac{z}{2\beta}}
  \sum_{k \ge 0}
    \frac{k! m_k}{(2)_k}
    L_k^{(1)} \left( \frac{z}{\beta \mu_0} \right)
$$
where again I think $(2)_k = (k+1)!$, $L_k^{(1)}$ is a generalized Laguerre polynomial, and
$$
m_0 = 8 \exp\left( -\tfrac12 \sum_{i=1}^2 \frac{\lambda_i}{1 + \frac{\beta \mu_0}{v_i (1 - \mu_0)}} \right) \frac{\beta^2}{1-\mu_0} \prod_{i=1}^2 \left( \beta \mu_0 + v_i (1 - \mu_0) \right)^{-1/2}
\\
m_k = \frac{1}{k} \sum_{j=0}^{k-1} m_j d'_{k-j}
\\
d'_j =
  - \frac{j \beta}{2 \mu_0} \sum_{i=1}^2
      \lambda_i v_i (\beta - v_i)^{j-1}
      \left( \frac{\mu_0}{\beta \mu_0 + v_i (1 - \mu_0)} \right)^{j+1}
  + \left( \frac{\mu_0}{\mu_0 - 1} \right)^j
  + \sum_{i=1}^2 \frac{\nu_i}{2} \left( \frac{\mu_0 (\beta - v_i)}{\beta \mu_0 + v_i (1 - \mu_0)} \right)^j
$$
They have truncation error bounds here too (3.12).
Note: I'm not entirely sure why I typed all that out now, but, oh well; I did some simplifications. In the paper's notation of section 3 we have $n = 2$, $\alpha_i = v_i$, $\nu_i = 1$, $\nu = 2$, $p = 1$, $\delta_i = \lambda_i$.
A: Answering separately because adding correlation between $X$ and $Y$ changes the problem a lot, and I think the original answer is useful to keep around.
If $X$ and $Y$ are correlated, really what you're asking is about the distribution of $Z^2 = x^T x$ where $x \sim \mathcal{N}\left( \begin{bmatrix}m_1 - a \\ m_2 - b\end{bmatrix}, \begin{bmatrix} v_1 & c \\ c & v_2 \end{bmatrix} \right)$.
This is a completely different problem. As far as I know, you have to turn to the generalized chi-squared distribution for $Z^2$. I cataloged some papers related to computing things with that distribution in this answer. One option is the approximation of Liu, Tang and Zhang (2009), which finds a good noncentral chi-squared distribution close to $Z^2$.
(I might come back and write up the actual approximation tomorrow, but the paper is quite understandable and available directly from the author.)
A: Set $U = Z^{2}$. Assuming that $X \& Y$ are independently distributed and using the moment generating function method, we get the following:
\begin{align*}
\mathbf{E}(e^{Ut}) = \mathbf{E}(e^{(X-a)^{2}t+(Y-b)^{2}t})\\
= \mathbf{E}(e^{(X-a)^{2}t})\times\mathbf{E}(e^{(Y-b)^{2}t})\\
\end{align*}
From there it should not be difficult...just a couple of integrals to solve, then done!
