Let $X_1, X_2, X_3, X_4$ be independent and identically distributed random variables, each assuming the values $1$ and $-1$ with probability $1/2$ each. Then, what is the probability that the following matrix is nonsingular?



closed as off-topic by Jack, Daniel W. Farlow, C. Falcon, Namaste, Juniven Mar 26 '17 at 0:56

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  • $\begingroup$ Maybe do singular instead and later subtract from 1. If the matrices are singular then the diagonal products must be either both ... or both ... $\endgroup$ – quasi Mar 15 '17 at 7:18

The matrix is singular exactly when the determinant is zero, i.e., $X_1X_4 - X_2X_3 = 0$. Now using the fact that $X_2^2 = X_3^2 = 1$ with probability 1, we find that this condition is equivalent to

$$ X_1 X_2 X_3 X_4 = 1. $$

By the symmetry, it easily follows that $\Bbb{P}(X_1 X_2 X_3 X_4 = 1) = \Bbb{P}(X_1 X_2 X_3 X_4 = -1) = \frac{1}{2}$. Therefore the matrix is nonsingular with probability $\frac{1}{2}$.

To be honest, however, you need nothing fancy for this problem. The matrix is evenly distributed over the set of $2\times 2$ matrices having entries $\pm 1$. There are only 16 possibilities, and even a brutal force gives you an answer within a couple of minutes.


Choose $X_4$ last.

Then right before $X_4$ is selected, regardless of the results for $X_1,X_2,X_3$, the matrix has one of the $8$ forms

$$ \begin{bmatrix} \pm 1& \pm 1\\ \pm 1& x\\ \end{bmatrix} $$

where $x$ denotes the yet unselected value of $X_4$.

Then the determinant $D$ has the form $ax + b$ for some $a,b \in \{\pm 1\}$.

Setting $D=0$, we see that the matrix is singular if and only if $x = -\frac{b}{a}$.

But $a,b \in \{\pm 1\}$ implies $-\frac{b}{a}\in \{\pm 1\}$, so the probability that $x = -\frac{b}{a}$ is $\frac{1}{2}$.

Thus, the probability that the matrix is singular is $\frac{1}{2}$, and hence the probability that the matrix is nonsingular is also $\frac{1}{2}$.


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