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For $t\in\mathbb{R}$ let $\{X_t\}$ be a family of random variables independent and with the same distribution. Assume the distribution is absolutely continuous with respect to Lebesgue measure. I want to show that the paths of the process $X_t$ are discontinuous almost surely. Does anyone have a hint for this problem?

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1 Answer 1

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Lemma

Let $(Y_n)$ be i.i.d. If $P(Y_n \to Y)>0$ for some random variable $Y$ then $Y$ is degenerate.

Proof of the lemma: $P(Y_{2n+1}-Y_{2n} \to 0)$ is $0$ or $1$ by Kolmogorov's $0-1$ law. If it is $1$ then $Y_{2n+1}-Y_{2n} \to 0$ in distribution. Since this sequence is i.i.d. it follows that $|\phi (t)|=1$ for all $t$ where $\phi$ is the common characteristic function. This implies that $Y_{2n+1}-Y_{2n}$ is degenerate and hence $Y_n$ is degenerate for each $n$. The lemma follows.

Now consider $P(X_{1/n} \to X_0)$. If this probability is $>0$ then the lemma shows that $X_0$ has a degenerate distribution contradicting absolute continuity.

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