How one can show the joint probability density fucntion does not exist?

Question

Let $X \sim \mathcal{U}(0,1)$ be an unifrom random variable and $Y = X$.

I am trying to show that the joint pdf of $X$ and $Y$ does not exist.

Here is what I have gone so far. Suppose there exists a function $f(x,y)$ such that $$ F(x,y) = P(X \le x, Y \le y) = \int_0^x \int_0^y f(t_1,t_2)dt_1dt_2. $$ Note that $1 = f_X(t_1) = \int_0^1 f(t_1,t_2)dt_2$ and $1 = f_X(t_2) = f_Y(t_2) = \int_0^1 f(t_1,t_2)dt_1$.

Since $X = Y$, we have $$ F(x,y) = P(X \le \min\{x,y\}) = \int_0^{\min\{x,y\}} f_X(t_1)dt_1 = \int_0^{\min\{x,y\}} dt_1 = \min\{x,y\}. $$

I was trying to show somehow $f(t_1,t_2) = 1$ based on $f_X(t_1)=1=f_Y(t_2)$ to draw a contradiction. But not sure how to do this.

Any comments/answers will be very appreciated. Thanks.

Answer 1

I found a way to prove this.

Since $Y = X$, we have \begin{align*} P(X \le x, Y \le y) &= P(X \le x, Y \le y, X = Y) \\ &= \int_0^{\min\{x,y\}} \int_{t_2 = t_1} f(t_1,t_2) dt_2 dt_1 = \int_0^{\min\{x,y\}} 0\cdot dt_1 = 0 \end{align*} which is a contradiction.