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# If two random variables have CDFs that have the same value for all x, can we assume the random variables are equal?

2018-04-20 20:09:27

My text has the following theorem:

Let $X$ have a CDF $F$ and let $Y$ have CDF $G$. If $F(x) = G(x)$ for all $x$, then $\mathbb{P}(X \in A) = \mathbb{P}(Y \in A)$ for all $A$.

I don't see a way that X and Y could assign a different probability to the same event but still have their CDFs be equal at every point. If they disagreed at point j, then F(j) will not equal G(j). Therefore they must be the same?

For elementary probability:

I think it's just saying that if the cdf's are the same then the pdf's are the same. PDF's don't always exist in advanced probability.

Do you seriously prove that? Anyway, it likely comes down to how 'all' $A$ (subsets of $\mathbb R$?) that you can come up with is somehow 'related' to $(-\infty, x]$. For example,

$$A = (a,x] = (-\infty,x] \cap (-\infty,a]^c$$

$$A = (-\infty, x) = (-\infty, x] \cap \{x\}^c$$

$$A = \{x\} = (-\infty, x] \cap (-\infty, x)^c$$

$$A = \{x\} \cup [a,b] = [(-\infty, x] \cap (-\infty, x)^c] \cup [(-\infty,b] \c • For elementary probability: I think it's just saying that if the cdf's are the same then the pdf's are the same. PDF's don't always exist in advanced probability. Do you seriously prove that? Anyway, it likely comes down to how 'all' A (subsets of \mathbb R?) that you can come up with is somehow 'related' to (-\infty, x]. For example,$$A = (a,x] = (-\infty,x] \cap (-\infty,a]^cA = (-\infty, x) = (-\infty, x] \cap \{x\}^cA = \{x\} = (-\infty, x] \cap (-\infty, x)^cA = \{x\} \cup [a,b] = [(-\infty, x] \cap (-\infty, x)^c] \cup [(-\infty,b] \cap (-\infty,a)^c]

Thus, if you can compute $P((-\infty, x])$ for all $x$, then you can compute $P(X \in A)$ for 'all' A. The thing is not all subsets of $\mathbb R$ are 'related' to $(-\infty, x]$. But that's okay because even in advanced probability, we don't mind those subsets. In advanced probability, there may be random variables where that doesn't hold if you try those subsets.