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How can I prove the following: Let $X$ and $Y$ be two random variables. Suppose that their cumulative distribution functions satisfies $F_X(x)=F_Y(x)$ for all $x$. How can I show that $X$ and $Y$ are identically distributed?

I got a hint that this is not easy for beginner and it requires heavy use of $\sigma$-algebras.

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"$X$ and $Y$ are identically distributed" means that $P(X \in B)=P(Y \in B)$ for each Borel set $B$. Knowing that $F_X=F_Y$ means you know $P(X \in (-\infty,a])=P(Y \in (-\infty,a])$ for every real number $a$. So you prove the desired result in three steps:

  1. $P(X \in U)=P(Y \in U)$ for each open $U$.
  2. If $P(X \in U)=P(Y \in U)$ for each open $U$, then $P(X \in F)=P(Y \in F)$ for each closed $F$.
  3. If $P(X \in U)=P(Y \in U)$ for each open $U$ and $P(X \in F)=P(Y \in F)$ for each closed $F$, then $P(X \in B)=P(Y \in B)$ for each Borel $B$.

To prove the first part, start from proving the result for open intervals, and then extend it to general open sets by using the fact that any open set is a countable union of open intervals.

To prove the second part you need only note that a closed set is the complement of an open set, so $P(X \in F)=1-P(X \in F^c)$, and we already have agreement for open sets.

To prove the third part, you show that the set $E=\{ A : P(X \in A)=P(Y \in A) \}$ is a $\sigma$-algebra. This takes some work. First, the above implies that this set contains an algebra (no $\sigma$ here), namely the algebra consisting of open and closed sets. Next, you can check $E$ is closed under increasing unions and decreasing intersections, using continuity of measure. (Here the fact that a probability space has finite measure is required). Then the monotone class theorem implies that $E$ is a $\sigma$-algebra. Now by definition, any $\sigma$-algebra containing all open sets must contain the Borel $\sigma$-algebra.

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    $\begingroup$ In general, $\{A: P(X\in A)=P(Y\in A)\}$ is not a $\sigma$-algebra. $\endgroup$ – user940 Aug 20 '15 at 18:51
  • $\begingroup$ @ByronSchmuland It clearly contains $\emptyset$ and $\Omega$, it should be closed under increasing unions by continuity of measure, and it should be closed under decreasing intersections by continuity of measure (since a probability space has finite measure). Thus it is a monotone class. Now try to apply the monotone class theorem to conclude that it is a $\sigma$-algebra. Where does this argument go wrong? Do you have an example? $\endgroup$ – Ian Aug 20 '15 at 18:51
  • $\begingroup$ @ByronSchmuland I could see a problem if $\{ A : P(X \in A)=P(Y \in A) \}$ did not contain an algebra...but open sets combined with closed sets form an algebra. $\endgroup$ – Ian Aug 20 '15 at 19:01
  • $\begingroup$ Let $P(X=1)=P(X=2)=P(X=3)=1/3$ and $P(Y=3)=2/3, P(Y=4)=1/3$. Then $P(X\in\{1,4\})=P(Y\in\{1,4\})$ and $P(X\in\{2,4\})=P(Y\in\{2,4\})$, but $P(X\in\{4\})\neq P(Y\in\{4\})$. $\endgroup$ – user940 Aug 20 '15 at 19:01
  • $\begingroup$ @ByronSchmuland So your issue is with the general case. I should clarify that I was referring to the same $X,Y$ as before, for which you have equality on an algebra from the preceding argument. $\endgroup$ – Ian Aug 20 '15 at 19:03

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