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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 $Y<1/2$.

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    @RodCarvalho But what would be the denominator?2012-09-12

3 Answers 3

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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 $\dfrac{6x+1}{7}$.

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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.

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The answer should be

f(x,y)=2/7 ,(x,y)∈[−1,0]×[0,1/2]

f(x,y)=6/7 ,(x,y)∈[0,1]×[−1/2,1/2]

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 (1/4)/(7/8)=2/8. For the right square it is (3/4)/(7/8)=6/8.