I have this joint distribution
$f_{X,Y}(x,y) = \frac{1}{2}I_A(x,y)$
$A=\{(x,y) \in R^2; -1 < y < 1; 0 < x < 1\} $
What I need is to find the conditional distribution of X for when Y>0
$f_{X|Y}(x|y>0) = ? $
By definition, I know that:
$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_{Y}(y)} $
Soon I have to
$f_{Y}(y) = \int_{0}^{1}\frac{1}{2}dx = \frac{1}{2}, - 1 < y < 1$
$f_{X|Y}(x|y) = \frac{1/2}{1/2} = 1$; $0 < x < 1$
I'm not sure how to proceed from this point. Can anyone show me a way?
On
Hi: The region covering x and y is a rectangle so knowing where $y$ is does not tell you anything where $x$ so this makes them independent. If they region was say say a circle, then this would not be the case.
Notice that, since $x$ and $y$ are independent, the conditional distribution of $x$ is equal to the marginal of $x$ and the marginal of $x = f(x) = I_{[0,1]}(x)$.
On
Can anyone show me a way?
Just use what is essentially the same definition :-
$$\begin{align}f_{X}(x\mid Y\,{>}\,0)~&=~\dfrac{f_{X}(x; Y\,{>}\,0)}{\mathsf P(Y\,{>}\,0)}\\[1ex]&=~\dfrac{\displaystyle\int_{y>0} f_{X,Y}(x,y)\,\mathrm d y}{~\displaystyle\iint_{y>0} f_{X,Y}(x,y)\,\mathrm d y\,\mathrm d x~}\end{align}$$
Suppose I tell you that $(X,Y)$ is a point taken uniformly at random from the rectangle with vertices $(0,-1), (1,-1), (1,1), (0,1)$. Now you randomly sample a large number of these points, but then you take only the ones whose $Y$-coordinate is positive, and you report the $X$-values of those points to me. What do you think the list of numbers I am going to receive will look like?
If you say that such a list will look like it is uniformly distributed on $X \in [0,1]$, you would be correct. So how do we formalize your intuition?
Well, clearly $$\Pr[X \le x \mid Y > 0] = \frac{\Pr[(X \le x) \cap (Y > 0)]}{\Pr[Y > 0]}$$ by the definition of conditional probability. We also know $\Pr[Y > 0] = 1/2$ from your computation. But the numerator is $$\Pr[(X \le x) \cap (Y > 0)] = \int_{v = 0}^1 \int_{u = 0}^x \frac{1}{2} du \, dv = \int_{v=0}^1 \frac{x}{2} \, dv = \frac{x}{2}.$$ And we could also have seen this by reasoning geometrically, since the region satisfying $0 \le X \le x$ and $0 < Y \le 1$ is a rectangle with side lengths $x$ and $1$, thus its area is $x$; but the joint density over this region is $1/2$ so the probability is $x(1/2) = x/2$.
Consequently, $$\Pr[X \le x \mid Y > 0] = \frac{x/2}{1/2} = x,$$ for any $x \in [0,1]$, and this of course is the CDF of a uniform distribution on $[0,1]$.
We can also adapt the above solution to obtain a more general result, e.g., $$\Pr[X \le x \mid Y > y] = \frac{x(1-y)/2}{(1-y)/2} = x.$$ In other words, the conditional distribution of $X$ doesn't depend on $Y$ at all: $X$ and $Y$ are independent random variables, a fact that owes itself to the property that the support of $X$ and $Y$ is a rectangle, and the joint density on that rectangle is uniform. So in fact, we could have written
$$\Pr[(X \le x) \cap (Y > 0)] \overset{\text{ind}}{=} \Pr[X \le x]\Pr[Y > 0]$$
in the first equation, and then the marginal probability $\Pr[Y > 0]$ cancels out, leaving us with $$\Pr[X \le x \mid Y > 0] = \Pr[X \le x].$$