Given is the $2\times 2$ matrix $$A=\begin{pmatrix}x_1 & y\\ y & x_2\end{pmatrix}$$ What's the probability that this matrix is positive definite for $x_1,x_2,y \in (-1,1)$ uniformly distributed?
I'm using the Eigenvalues to determine the definiteness. $$(x_1-\lambda)(x_2-\lambda)-y^2 \overset{!}{>} 0\Longrightarrow y\in (-1,1)\wedge x_1>\lambda \wedge x_2>\frac{y^2}{x_1-\lambda}+\lambda$$
And now I want to find the CDF of this problem, but I'm not sure how to do it for 3 variables.
Maybe $\int_{-1}^{x_2} \int_{-1}^{x_1} \int_{-1}^y \frac{1}{1-(-1)} du\, dv\, dw$?
Thanks for any tips!
Since the matrix $A$ is symmetric, its eigenvalues $\lambda_1$ and $\lambda_2$ are real. From the properties of trace and determinant we have that $\lambda_1 + \lambda_2 = x_1 + x_2$ and $\lambda_1\lambda_2 = x_1x_2 - y^2$. Therefore, $A$ is positive definite if and only if $x_1 + x_2 > 0$ and $x_1x_2 > y^2$. This means that $x_1>0$ and $x_2>0$ and the set of points $(x_1, x_2, y)\in[-1, 1]^3$ corresponding to positive definite matrices is symmetric in the $y=0$ plane. Therefore, if we denote the volume between the surfaces $y=0$ and $y=\sqrt{x_1x_2}$ over the square $x_1,x_2\in[0,1]^2$ by $V$ then the probability that $A$ is positive definite is $\frac{V}{4}$. Finally, we compute
$$ \begin{align} V=\int_0^1\int_0^1\sqrt{x_1x_2}dx_1dx_2 = \left[\int_0^1\sqrt{x}dx\right]^2 = \frac49 \end{align} $$
and conclude that the probability that $A$ is positive definite is $\frac19$.