Kyle Sutton

2022-07-10

Let $(\mathrm{\Omega},\mathcal{F},\mathbb{P})$ be a probability space and $X:(\mathrm{\Omega},\mathcal{F})\to (\mathbb{R},{\mathcal{F}}_{E})$ a random variable that admits a density function ${f}_{X}:\mathbb{R}\to \mathbb{R}$.

I have some questions about the information given by the cumulative distribution function of X (${F}_{X}:\mathbb{R}\to [0,1]$).

i) Does the CDF determine ${f}_{X}$ uniquely ?

ii) Does the CDF determine the law of X uniquely ?

iii) Can we find a subset of events that uniquely determine the CDF ?

For i), I would say that as ${F}_{X}(t)={\int}_{-\mathrm{\infty}}^{t}{f}_{X}(x)dx$, the density function is the derivative of the CDF and therefore the CDF can only define one density function. But I feel like maybe we could find a counterexample by considering Lebesgue measure and taking sets of the form (a,b) and [a,b].

For ii), as $\mathrm{\forall}x\in \mathbb{R}$, $\mathbb{P}(X=x)={F}_{X}(x)-{F}_{X}(x-)$ and $\mathbb{P}(X\le x)={F}_{X}(x)$, I am tempted to say that the law of X is indeed by definition determined uniquely by the CDF.

For iii), maybe we could take all events of the form $\{X\in (a,b):a,b\in \mathbb{R}\}$ but I'm not quite sure about that, and particularly regarding the uniqueness.

I have some questions about the information given by the cumulative distribution function of X (${F}_{X}:\mathbb{R}\to [0,1]$).

i) Does the CDF determine ${f}_{X}$ uniquely ?

ii) Does the CDF determine the law of X uniquely ?

iii) Can we find a subset of events that uniquely determine the CDF ?

For i), I would say that as ${F}_{X}(t)={\int}_{-\mathrm{\infty}}^{t}{f}_{X}(x)dx$, the density function is the derivative of the CDF and therefore the CDF can only define one density function. But I feel like maybe we could find a counterexample by considering Lebesgue measure and taking sets of the form (a,b) and [a,b].

For ii), as $\mathrm{\forall}x\in \mathbb{R}$, $\mathbb{P}(X=x)={F}_{X}(x)-{F}_{X}(x-)$ and $\mathbb{P}(X\le x)={F}_{X}(x)$, I am tempted to say that the law of X is indeed by definition determined uniquely by the CDF.

For iii), maybe we could take all events of the form $\{X\in (a,b):a,b\in \mathbb{R}\}$ but I'm not quite sure about that, and particularly regarding the uniqueness.

razdiralem

Beginner2022-07-11Added 15 answers

By subtraction, the CDF $F(x):=P(X\le x)$ determines the probabilities $P(X\in [a,b])$ for every $a,b\in \mathbb{R}$. By an approximation theorem, such as Caratheodory's theorem, this information determines $P(X\in E)$ for every $E\in B(\mathbb{R})$. So (ii) is true.

(i) follows (though, as always, the density is unique only up to almost everywhere equality).

(i) follows (though, as always, the density is unique only up to almost everywhere equality).

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