Let $X_1, X_2, X_3, X_4$ be random variables, and let $A$ be the following matrix: $$ \left[\begin{matrix} X_1 & X_2\\ X_3 & X_4 \end{matrix}\right]. $$ Assume that $X_1, X_2, X_3, X_4$ are statistically independent, and: $$X_1\sim U([0,a_1])$$ $$X_2\sim U([0,a_2])$$ $$X_3\sim U([0,a_3])$$ $$X_4\sim \operatorname{Exp}(\lambda)$$

Find the probability the the matrix $A$ is invertible.

I was thinking about calculating the determinant and find the probability that that the determinant will be zero. in that case I can find the probability which is $1-P(\det(A)=0)$. I was struggling to find the way to calculate this. I need to know how the random variable $Y=X_1X_4-X_2X_3$ behave. I think that if I will prove that this random variable is continuous then I can say that $P(\det(A)=0)=0$., the problem is that I don't know to do this.

  • $\begingroup$ This is a specific example of a situation in which an event is defined by the zeros of a function. When (a) that function is continuously differentiable; (b) the derivative is nonsingular on the zero set; and (c) for each point in the zero set there is a neighborhood in which at least two of the input random variables are continuous, then the event has zero probability. This is a direct consequence of the Implicit Function Theorem. $\endgroup$ – whuber Nov 19 '18 at 15:43
  • $\begingroup$ are there any restrictions on the $a_{i}$'s? $\endgroup$ – Dale C Dec 25 '19 at 6:15

In the case of a $2 \times 2$ matrix there is a simple formula for the determinant:

$$\det \mathbf{X} = \det \begin{bmatrix} X_1 & X_2 \\ X_3 & X_4 \end{bmatrix} = X_1 \cdot X_4 - X_2 \cdot X_3.$$

So you have:

$$\mathbb{P}(\mathbf{X} \text{ is invertible}) = 1- \mathbb{P}(\det \mathbf{X} = 0) = 1- \mathbb{P}(X_1 \cdot X_4 = X_2 \cdot X_3).$$

In your particular problem all your random variables are continuous, and because the function $(x_1,x_2,x_3,x_4) \mapsto x_1x_4 - x_2x_3$ is differentiable with almost everywhere nonzero derivative, the random variable $X_1 \cdot X_4 - X_2 \cdot X_3$ is also continuous. This means that you have zero probability of $X_1 \cdot X_4 = X_2 \cdot X_3$. Hence, your matrix is invertible with probability one.

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    $\begingroup$ Ben, how you prove that $Y=X_1X_4-X_2X_3$ is continuous? what you have answered is exactly what I have mentioned in the question. does multiplication of continuous random variables is continuous random variable? And does the sum of continuous random variables is random variable? Answering those question will be satisfying. Thank you. $\endgroup$ – Mr.O Nov 19 '18 at 7:57
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    $\begingroup$ That reasoning is (unfortunately) invalid. A counterexample is the function $f(x,y)=(x-x)*(y-y),$ which manifestly is constructed from "basic arithmetic operations." Nevertheless, when $X$ and $Y$ are any two independent RVs, $f(X,Y)$ is discrete. $\endgroup$ – whuber Dec 24 '19 at 16:15
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    $\begingroup$ Good point --- I have edited the answer to avoid the non sequiter. $\endgroup$ – Ben Dec 24 '19 at 21:12
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    $\begingroup$ I edited a small typo out of your post and in the process took the liberty of inserting an explicit justification for the assertion: feel free to remove it if it's inconsistent with your intentions. $\endgroup$ – whuber Dec 24 '19 at 23:46
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    $\begingroup$ @whuber: Please feel free to edit my posts whenever you see a desirable change. I get a notification anyway, so can always reverse in the unlikely event that I don't like it. $\endgroup$ – Ben Dec 25 '19 at 5:46

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