Finding density function of a conditioned function > Let $(X, Y)$ be a point taken uniformly at random on the unit square $[−1, 1]^2$. Let X|E be X conditioned on the event $E$ = ${\\{X + Y \ge 0\\}}$. > > Find the p.d.f. of X|E. * * * Now, the solution is given as follows: ![enter image description here]( As pointed out in the picture, I don't understand why divide by 4 for the numerator and by $\frac{1}{2}$ in the denominator. But here's my reasoning: Divide by 4 because we're taking the area of the shaded region over the entire region which has area 4 as pointed out in the sketch. Would appreciate if someone could explain it and perhaps point me to any good references/links out there! thanks.

For the numerator, it is $2^2$, as you get the probability of $(X\leq x, X+Y\geq 0)$ by the area of the grey triangle divided by the area of the whole square.

For the denumerator, this is the area of the triangle on the right which represents the probability of $X+Y \geq 0$.

![enter image description here](

Solve this simple example: what is the probability that $X>0$ in this figure? It is $0.5$ by intuition, right? Mathematically, it is obtained by divining two areas; the area where $X>0$ (i.e., the right rectangle) and the whole area which represents all possibilities for $X$ (i.e., the whole square).