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I read this conclusion in text book:

If $X$ and $Y$ are i.i.d continuous r.v.s, then $P(X=Y)=0$, but it doesn't give any proof. I am a bit confused here, what is the rational for this conclusion? Is it because for a continuous r.v., the probability of it equaling to any particular value is $0$?

Thanks.

YuiTo Cheng
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wangshuaijie
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  • Sort of related... I find this idea interesting: https://math.stackexchange.com/questions/882784/does-0-chance-mean-impossible/882787 does $P(X=Y)=0$ mean this event is "impossible"? – HJ_beginner Feb 22 '19 at 05:15

4 Answers4

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For a legalistic argument: $P(X=Y)=E( \mathbb 1_{X=Y}) = E( E( \mathbb 1_{X=Y}|Y)) = E( 0) = 0$ since $E(\mathbb 1_{X=Y}|Y) = 0 \text{ a.s.}$. That is, the OP's observation that continuous random variables take on particular values with probability zero, when used with Fubini's theorem (a.k.a. the law of total expectation), does the trick.

kimchi lover
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Try this simple reasoning: Let $Z=X-Y$ (which must also be continuous). Now, \begin{align*}P(X-Y=0)&=P(0 \leq X-Y \leq 0)\\ &= \int_0^0 p(z)dz\\ &=0, \end{align*} where $p$ is the density for $Z$.

Crunchy
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What is ther area of a diagonal in a square?

Here the "honest" calculation:

  • Let $f_X(x),f_Y(y)$ be the continuous probability densities of $X,Y$ and
  • $F_X(x) = P(X \leq x)$, $F_Y(y) = P(Y \leq y)$

Now you have $$P(X=Y) = 1-P(X<Y) - P(X>Y) \stackrel{X,Y\;i.i.d.}{=}1-2P(Y>X)$$ So, we calculate the probability on the RHS: \begin{eqnarray*} P(Y>X) & = & \int_{x =-\infty}^{+\infty}\int_{y =x}^{+\infty}f_Y(y)f_X(x)dydx \\ & = & \int_{x =-\infty}^{+\infty}f_X(x)\int_{y =x}^{+\infty}f_Y(y)dydx \\ & \stackrel{X,Y\; i.i.d.}{=} & \int_{x =-\infty}^{+\infty}f_X(x)(1 -F_X(x))dx\\ & = & 1 - \boxed{\int_{x =-\infty}^{+\infty}f_X(x)F_X(x)dx}\\ \end{eqnarray*}

Now, note that you can show easily by partial integration that $I =\int_{x =-\infty}^{+\infty}f_X(x)F_X(x)dx = \frac{1}{2}$

Indeed we have \begin{eqnarray*} I & = & \int_{x =-\infty}^{+\infty}f_X(x)F_X(x)dx\\ & = & \left. (F_X(x))^2 \right|_{-\infty}^{+\infty}- \int_{x =-\infty}^{+\infty}F_X(x)f_X(x)dx\\ & = & 1-I\\ \end{eqnarray*}

All together $$P(X=Y) = 1-2P(Y>X) = 1-2(1-I) = 1-2\cdot\frac{1}{2} = 0$$

trancelocation
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For every $N$, partition $\mathbb{R}$ to $N$ intervals $I_n^N$ (we allow intersections of probability zero) such that $\mathbb{P}(X\in A_n)=\frac{1}{N}$ and $\mathbb{P}(I_i \cap I_j)=0$. Notice, that this is possible because $X,Y$ are continuous r.v..

So, \begin{align} \mathbb{P}(X=Y) &\leq \mathbb{P}(\bigcup_n (X \in I_n^N,Y \in I_n^N ) )\\ &\leq \sum_{n=1}^N \mathbb{P}(X \in I_n^N,Y \in I_n^N )\\ &\leq \sum_{n=1}^N \mathbb{P}(X \in I_n^N )\mathbb{P}( Y \in I_n^N )\\ &\leq \frac{1}{N} \end{align}

Now, taking $N\to \infty$, we conclude $$\mathbb{P}(X=Y)=0.$$

clark
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