Supposing I have four independent variables $A, B, C$ and $D$, all of which are uniformly distributed from $-1$ to $1$:
- How can I find the probability of both $$ A + D < 0\quad\mbox{and}\quad \frac{\left(A + D\right)^{2}}{4} > AD - BC\quad \mbox{happening}\ ? $$
- I understand that $\operatorname{P}\left(A + D < 0\right) = 1/2$ since it is just the proportion of area of the lower triangle of the $\left[-1, 1\right]\times\left[-1, 1\right]$ square to the whole square, but how do I set up the $\left(~\mbox{quadruple}\,?~\right)$ integral to find the probability of the second event $?$.
- In general for an event depending on four variables, we have $\int\limits_{R} \operatorname{f}\left(a, b, c, d\right)\,{\rm d}R$ where $R$ is the hyper-region we are interested in and $\operatorname{f}\left(a, b, c, d\right)$ is the probability density function, but I'm not sure how to set up each of those.
Would I multiply the two probabilities afterwards $?$
I'm quite green when it comes to probability so apologies if the question is obvious.
This can be done by hand (although I don't know about the qualifier "reasonably").
As done in the other answer (slightly modified), we can let $X = \frac{A-D}{2}$, which since $A$ and $D$ are both $\sim U(-1,1)$ has a triangular distribution centered at $0$ and extending from $-1$ to $1$. If we consider $|X|$, this has a CDF of $P(|X|<x) = 2x-x^2$ for $x \in (0,1)$. Since $|X|$ is positive, we have $P(X^2 < x) = P(|X|<\sqrt{x}) = 2\sqrt{x} - x$. Differentiating gives the PDF $f_{X^2}(x)=\frac{1}{\sqrt{x}} - 1$, for $x \in [0,1]$.
From symmetry, we can see from the joint distribution of $A$ and $D$ that the distribution of $X$ when $A+D>0$ is the same as the distribution of $X$ when $A+D<0$, and as stated in the post, these are equally likely, producing a factor of 1/2.
We can also define $Y=-BC$. If $Y < 0$, then clearly $X^2 > Y$. We have $P(Y<0)=\frac{1}{2}$, again by symmetry.
If $Y\ge 0$ (which occurs half the time), then we want to know if $X^2>|Y|=|B||C|$, which is the product of two independent rv $\sim U(0,1)$, so $|Y|$ has the PDF $f_{|Y|}(y) = -\log(y)$ for $ y \in (0,1)$.
see product distribution of two uniform distribution, what about 3 or more
Since $X^2$ and $|Y|$ are independent, their joint distribution is the product of their distributions, so we want to integrate:
$$\frac{1}{2}\left[ \frac{1}{2} + \frac{1}{2} \int_{y=0}^{y=1} \int_{x=y}^{x=1}\left(\frac{1}{\sqrt{x}} - 1\right)\left(-\log(y)\right)dx\, dy \right]$$
Where the first $\frac{1}{2}$ is from the factor for $A+D>0$, the next $\frac{1}{2}$ is for when $Y<0$, and the next $\frac{1}{2}$ is from the factor for $Y>0$.
For the inner integral: $$\int_{x=y}^{x=1}\left(\frac{1}{\sqrt{x}} - 1\right) dx= (1+y-2\sqrt{y})$$
For the outer integral: $$\int_{y=0}^{y=1} (1+y-2\sqrt{y})\left(-\log(y)\right)dy$$
Letting $u = \sqrt{y}$, so that $dy = 2u \,du$: \begin{align} \int_{u=0}^{u=1} (1+u^2-2u)\left(-2\log(u)\right)2u\, du &= -4 \int_{u=0}^{u=1} (u-2u^2+u^3)\left(\log(u)\right) du \\ &= -4 \left(\int_{u=0}^{u=1} u\log(u) du - 2\int_{u=0}^{u=1} u^2\log(u) du + \int_{u=0}^{u=1} u^3\log(u) du \right) \\ &=4 \left( \frac{1}{4} - \frac{2}{9} + \frac{1}{16} \right) \\ &=1-\frac{8}{9}+\frac{1}{4} \\ &=\frac{13}{36} \\ \end{align}
For the integrals, I looked up in a table that for $m \ge 0$: $$\int x^m \log(x)dx = \frac{x^{m+1}}{m+1} \left( \log(x) - \frac{1}{m+1} \right)$$
Putting this in the main equation gives the probability:
$$\frac{1}{2}\left[ \frac{1}{2} + \frac{1}{2} \frac{13}{36} \right] = \boxed{ \frac{49}{144} } \approx 0.3403$$