# questions related to joint distribution and continuous and discrete random variables

(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?

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.

• Your argument is good but it's not quite complete, because it's not the case that $f(x,y)=0$ everywhere. However, it is undefined on a set of measure zero, so without changing anything essential you can redefine $f$ to be zero everywhere. But in that case it's not a distribution function (because it doesn't integrate to unity).
– whuber
Apr 22, 2020 at 21:28
• @whuber, Thanks for attention ,I will edit it according your point. Apr 22, 2020 at 21:34