Conditional distribution function when the condition is an inequality How to calculate a conditional distribution when the condition itself is an inequality?
Let me provide a simple example:
$$f(x,y) = 1/4\ , (x,y) \in [-1,0]\times[0,1]$$
$$f(x,y) = 3/4\ , (x,y) \in [0,1]\times[-1/2,1/2]$$
find distribution function of $X$ provided $Y<1/2$.
 A: In general we will have to integrate. But here the geometry is enough. Draw a picture.  (A pcture can also be handy if we need to integrate.)
We have $Y\le 1/2$ if we are in the bottom half of the left-hand square (probability $(1/4)(1/2)$), or if we are in the right-hand square (probability $3/4$), for a total of $7/8$.
Now we can find the cumulative distribution function of $X$, given that $Y\lt 1/2$. Of course this is $0$ if $x\lt -1$, and $1$ if $x \gt 1$. It remains to take care of things when $-1\le x\le 1$. 
For $-1\le x\le 0$, we want to calculate
$$\frac{\Pr((X \le x)\cap (Y\lt 1/2))}{\Pr(Y\lt 1/2)}.$$
The numerator is $(1/4)(1/2)(x-(-1))$. Divide by $7/8$ and simplify. We get 
$\dfrac{x+1}{7}$. 
For $0 \lt x \lt 1$, the calculation is similar. We have 
$$\Pr((X \le x)\cap (Y\lt 1/2))=\frac{1}{8}+\Pr(0\le X \le x)=\frac{1}{8}+\frac{3}{4}x.$$
 Divide by $7/8$. We get $\dfrac{6x+1}{7}$. 
A: 
The density is $\frac17$ on $-1\lt x\lt0$ and $\frac67$ on $0\lt x\lt1$.

To see this, a simple way is to compute $\mathrm P(X\lt x\mid Y\lt\frac12)$ for every $-1\lt x\lt1$ (the computation involves only areas of rectangles). 
Another method is to compute $\mathrm E(u(X)\mid Y\lt\frac12)$ for every measurable bounded function $u$, the result being
proportional to 
$$
\mathrm E(u(X);Y\lt\tfrac12)=\tfrac18\cdot\int_{-1}^0u(x)\mathrm dx+\tfrac34\cdot\int_0^{1}u(x)\mathrm dx.
$$
This proves that the desired density is proportional to $\mathbf 1_{(-1,0)}+6\cdot\mathbf 1_{(0,1)}$. Normalizing to get a total mass $1$ yields the result.
A: The answer should be
f(x,y)=1/7 , x∈[−1,0]
f(x,y)=6/7 , x∈[0,1]
The method is to find the total probability of the area as specified by the condition (y<1/2), ie 1/2*1/4+3/4 =7/8. So for upper half of left square the probability is zero. For lower half it is (1/8)/(7/8)=1/7. For the right square it is (3/4)/(7/8)=6/7.
