Finding the joint distribution of two independent uniform random variables $X,Y$ given event $E$

Question

If $E = \{\text{either} \ X < 1/3 \ \text{or} \ Y<1/3\}$ for $X\sim Unif(0,1)$ and $Y\sim Unif(0,2)$, does the joint distribution of $X,Y$ given event $E$ exist? I am assuming that $X$ and $Y$ are independent.

Answer 1

Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space, on which you have two independent random variables $X \sim \mathcal U((0,1)), Y \sim \mathcal U((0,2))$. Let $E := \{ \omega \in \Omega : X(\omega) < \frac{1}{3} $ or $Y(\omega) < \frac{1}{3} \}$. You want to find the distribution of $(X,Y)$ but treated as a random variable from $(\Omega,\mathcal F , \mathbb P_E)$ where $\mathbb P_E$ is a probability measure on $(\Omega,\mathcal F)$ such that $\mathbb P_E(A) = \frac{1}{\mathbb P(E)}\mathbb P(E \cap A)$ for any $A \in \mathcal F$. Note that $\mathbb P(E) = 1 - \mathbb P(X > \frac{1}{3})\mathbb P(Y > \frac{1}{3}) = 1 - \frac{2}{3} \frac{5}{6} = \frac{4}{9}$ due to independence

We'll find the CDF, that is $F(t,s) = \mathbb P_E(X \le t, Y \le s)$. For $s$ or $t$ less than $0$ it is obviously $0$. If both of $s,t$ are greater than $\frac{1}{3}$ it is obviously $1$.

Take any $s \in (0, \frac{1}{3}), t>0$

$$ F(t,s) = \frac{9}{4}\mathbb P(X \le s, Y \le t \cap \{ X < \frac{1}{3} \ or \ Y < \frac{1}{3} \}) = \frac{9}{4} \mathbb P(X \le s, Y \le t) = \frac{9}{8}st1_{(0,2)}(t) + \frac{9}{4}s1_{(2,\infty)}(t)$$ since in that case $\mathbb P(X \le s) = s, \mathbb P(Y \le t) = \frac{t}{2}1_{(0, 2)}(t) + 1_{(2,\infty)}(t)$ and we have independence.

Moreover, for $s >0, t \in (0,\frac{1}{3})$ we get: $$ F(t,s) = \frac{9}{4} \mathbb P(X \le s)\mathbb P(Y \le t)= \frac{9}{8}ts1_{(0,1)}(s) + \frac{9}{8}t1_{(1,\infty)}(s)$$

Answer 2

If X and Y are two random variables defined on the same probability space, their joint distribution always exists. If they are independent, then the joint distribution function is given by $$ F_{X,Y}(x,y) = F_X(x)F_Y(y).$$ In your case $(X,Y)$ is uniformly distributed on the rectangle $(0,1)\times (0,2)$. The distribution $(X,Y) | E $ is given by

$$Q_{X,Y}( B|E) = \frac{Q_{X,Y} (B\cap E)}{Q_{X,Y}(E)}.$$

Answer 3

The joint density is the same uniform law but restricted to the desired area

$$f_{XY|E}(x,y)=\frac{9}{8}[\mathbb{1}_{[0;\frac{1}{3})}(x)\mathbb{1}_{[0;2]}(y)+\mathbb{1}_{[\frac{1}{3};1]}(x)\mathbb{1}_{[0;\frac{1}{3}]}(y)]$$