Finding the distribution of $5X_{1}^2+2X_{1}X_{2}+X_{2}^2$ Suppose $X=[X_{1},X_{2}]$ and $X$~$N_2(μ,Σ)$. I wish to find the distribution of  $5X_{1}^2+2X_{1}X_{2}+X_{2}^2$. Since this is of a quadratic form I do not know a way of solving this. However I kind of feel like its chi-squared distributed with parameter 2. Any answers will be welcomed. Thanks
 A: This looks like the sum of two scaled non-central chi-squares with different non-centrality parameters, which I don't think leads to a single distribution. We have
$$Y = 5X_{1}^2+2X_{1}X_{2}+X_{2}^2  =(2X_1)^2 + (X_1+X_2)^2$$
$$2X_1 \sim N(2\mu_1, 4\sigma^2_1)  = N(m_1, s^2_1) \\ X_1+X_2 \sim N(\mu_1+\mu_2, \sigma^2_1+\sigma^2_2+2\rho\sigma_1\sigma_2) = N(m_2,s_2^2)$$
We then have
$$\tilde Z_1 = \left(\frac {2X_1}{s_1}\right)^2\sim \mathcal \chi^2_{(1)}(\lambda_1=m_1^2/s_1^2)$$ 
$$\tilde Z_2 = \left(\frac {X_1+X_2}{s_2}\right)^2\sim \mathcal \chi^2_{(1)}(\lambda_2=m_2^2/s_2^2)$$
i.e. chi-squares with non-centrality parameters  $\lambda_1$ and $\lambda_2$ respectively.
So your variable is
$$Y = s_1^2\tilde Z_1 + s_2^2\tilde Z_2$$
i.e. the sum of two scaled non-central chi-squares, possibly dependent.
As far as I know, a scaled non-central chi-square does not also follow a non-central chi-square, so you can only proceed further if you look for approximations. If the means of the original variables are zero, and $X_1,\, X_2$ are independent, then matters fully simplify (in such a case $Y$ will follow a Gamma distribution).
A: As Alecos and nivag pointed out, there is no closed form distribution. Only approximations are available. The latest such approximation is by Liu, Tang and Zhang: "A new chi-square approximation to the distribution of non-negative
definite quadratic forms in non-central normal variables", CSDA 2008. See in the references there for other approximations.
A: You could create a numerical estimate of your distribution by a Monte Carlo type approach. i.e. take a large number of random samples from $X_1$ and $X_2$ and calculate the function. As long as you take enough samples the probabilities given by the output should be a good estimate of the true distribution.
As far as I am aware there is no simple, analytical way to calculate the distribution for complicated functions such as this. Although it may be possible in your case as the function is fairly simple (only quadratic).
