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Suppose that we have two continuous independent random variable $X$ and $Y$ with respective densities $f_x$ and $f_y$, how can we prove that $X$ is different than $Y$.

I thought about proving it by contradiction using the definition of the probability of a random variable ( using integral equalities to prove that densities are equals therefore the absurdity ) but I didn't utilize independency there so I felt like it's wrong ( also I'm not sure if two lebesgue integrals are equals in the same interval then the functions integrated are also equals ).

wageeh
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2 Answers2

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In general, if $(X,Y)$ has joint density $f_{X,Y}$, then the probability that $(X,Y)$ lies in some subset $A \subseteq \mathbb{R}^2$ is the two-dimensional integral $$P((X,Y) \in A) = \iint_A f_{X,Y}(x,y) \, dx \, dy.$$ In this case, you do have a joint density: $f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y)$. And in your case, $A = \{(x,y) : x=y\}$ is a diagonal line in the $(x,y)$-plane. The two-dimensional integral of a function over a region with no area (like a line) is zero.

[Update: I noticed you mentioned Lebesgue integrals in your question, so I am guessing you would understand that the integral over a set of measure zero is zero.]

angryavian
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If we have that $X \perp \!\!\! \perp Y$ then we know that $\displaystyle f_{X,Y}(x,y) = f_X(x)\cdot f_Y(y)$.

We can either have $P(X=Y) = 0$ or $f_X \neq f_Y$ as our " measure" of them being different.

For a given $z$ we have

$P(X =z, Y =z) = P(X=z)P(Y=z) = 0 \cdot 0$.

More work is required to get that $P(X=Y)=0$. See @angryavian's answer.

oliverjones
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  • yes, I understand that thank you for answering but is that considered a proof? – wageeh Jan 12 '21 at 20:37
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    @oliverjones By that logic, you could argue that since $P(X=z)=0$ for all $z \in \mathbb{R}$, then $P(X \in \mathbb{R}) = 0$, which isn't true. – angryavian Jan 12 '21 at 20:43
  • @angryavian For a continuous random variable $X$, $\forall x \in \mathbb{R}, P(X=x) =0$. cf https://en.wikipedia.org/wiki/Random_variable#Continuous_random_variable – oliverjones Jan 12 '21 at 20:46
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    @oliverjones Yes, I know, and I stated that in my comment. I am not disputing $P(X=z, Y=z) = 0$; this equality is true. I am saying that this equality does not imply $P(X=Y)=0$. As a counterexample, consider a continuous random variable $X$, and let $Y=X$. Then $P(X=Y)=1$ by construction, but $P(X=z, Y=z) = P(X=z) = 0$. – angryavian Jan 12 '21 at 20:47
  • @angryavian Okay I see the issue there, but your above comment you state it in such a way that it reads as though $P(X =z) = 0 ,\forall z \in \mathbb{R}$ is incorrect. But this is true for a continuous random variable. – oliverjones Jan 12 '21 at 20:50
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    @oliverjones My first comment was trying to explain that the implication "$P(X=z)=0, \forall z$" $\implies$ "$P(X \in \mathbb{R}) = 0$" is incorrect, not that the assumption $P(X=z) = 0, \forall z$ was incorrect. Apologies for the lack of clarity. – angryavian Jan 12 '21 at 20:53
  • I have amended my answer. – oliverjones Jan 12 '21 at 20:54