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:

$$cov(X,Y) \neq 0$$

  • 2
    $\begingroup$ Are $X$ and $Y$ independent? Are the variances of $X$ and $Y$ known or unknown? $\endgroup$ Aug 15, 2014 at 12:58
  • $\begingroup$ I have calculated the covariance matrix of X and Y. Their covariance is not zero. The variances of X and Y are obtained by calculating with the estimated data provided by the system. $\endgroup$
    – Enthusiast
    Aug 16, 2014 at 4:43

3 Answers 3


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.)


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 $n = 2$, $\alpha_i = v_i$, $\nu_i = 1$, $\nu = 2$, $p = 1$, $\delta_i = \lambda_i$.

  • $\begingroup$ Hi Dougal, Thank you for your solution. Is there any solution in the case that X and Y are not independent? $\endgroup$
    – Enthusiast
    Aug 16, 2014 at 4:46

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!

  • $\begingroup$ Hi, Thank you for your suggestion. Is there any direction if X and Y are not independent? $\endgroup$
    – Enthusiast
    Aug 16, 2014 at 4:48

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