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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.

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    $\begingroup$ Intuitively, you want to show that $(\partial^2/\partial x\partial y)(\min(x,y)$ is not a function. Start off with $\partial/\partial x \min(x,y) = 1 $ if $x<y$ and $0$ if $x>y$. Now when you take the derivative of this with respect to $y$ you get 0 everywhere except when $x=y$ where you get an infinitely large Dirac-delta like spike. Such things exist, but they are not functions. $\endgroup$ Commented Sep 9, 2018 at 2:45
  • $\begingroup$ @kimchilover Thanks for your comment. I am aware of the result saying that if the partial derivative exists, it is the pdf. However, I am not aware of the converse. If you don't mind, could you point out a reference saying the converse is true (if it is true)? $\endgroup$ Commented Sep 9, 2018 at 2:48
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    $\begingroup$ I offer this as a symptom, not a proof. The way to show there is no joint density is to show that the joint measure has a non-trivial singular component. In your case, to calculate the area of a square's diagonal. It's zero, so the joint measure is singular with respect to $dxdy$, so no density function exists. $\endgroup$ Commented Sep 9, 2018 at 2:53

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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.

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  • $\begingroup$ Another way which is quite similar is to consider the set $A=\{(a,a):a\in [0,1]\}\subset \mathbb{R}^2$. This set is important because it is the set on which "the density would have to infinite in order to exist". Since $(X,Y)$ always lies in this set, we have $$1 =P((X,Y)\in A) =\iint_A f(x,y)dxdy =\int_{x}^{x}\int_0^1 f(x,y)dxdy =\int_{x}^{x}\int_0^1 f(x,y)dxdy=0$$ $\endgroup$
    – toe-pose
    Commented May 17, 2020 at 14:44

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