questions related to joint distribution and continuous and discrete random variables

Question

(a) If $U$ and $V$ are jointly continuous, show that $P(U =V) = 0$.

(b) Let $X$ be uniformly distributed on $(0,1)$, and let $Y= X$. Then, $X$ and $Y$ are continuous, and $P(X=Y) = 1$. Is there a contradiction?

(a)$P(U =V) = \int_{(u, v) = (u, u)} f(u, v) du dv$, and this integral is equal to zero since it is integrated at one point. Is this correct?

(b) $X$ is uniformly distributed on $(0, 1)$, so we have the distribution function $F(x) = x$ for $x \in (0,1)$. Then, we have an integrable function $f(x) = 1$ such that $\int_0^x 1 dx = x = F(x)$. Thus, $X$ is a continuous random variable. It seems intuitive that $P(X = Y) =1 $ since $X=Y$, but how can we actually compute this? Also, according to the solution, it does not contradict to (a) because $X, Y$ are not jointly continuous (i.e., there exists no integrable function $f$ such that $P(X \le x, Y\le y) = \int_0^x\int_0^y f(u,v) dudv$). How do we know the non-existence of such function?

Answer 1

Hint:

I assume , $Y=X$, you mean distribution of $Y|X=t$ is degenerate in point $t$.

For part a) use the fact that integral over set of measure zero is zero. integral over set of measure zero

For the question "How do we know the non-existence of such function?"

Now for $(x,y)$ a.e

$F_{(X,Y)}(x,y)=P(Y\leq y,X\leq x)=\int_0^1 P(Y\leq y,X\leq x|X=t) f_X(t) dt =\int_0^1 P(t\leq y,t\leq x|X=t) f_X(t) dt =\int_0^{\min(x,y)} dt=\min(x,y)$.

Now $f(x,y)=\frac{d^2}{dx \, dy} F(x,y)=0$ . So for $(x,y)$ a.e $$f(x,y)=0$$. But in this case it's not a distribution function (because it doesn't integrate to unity). Such distribution that the density is not defined called singular distribution.