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Let $X\sim\mathrm{ Uniform}(0,1)$. Consider the sequence $X_n = X^n$. I want to study the convergence in law of this sequence.

I did it using the distribution function, I have: $$F_X(x) = x \mathbb{1}_{[0,1](x)} + \mathbb{1}_{[1,\infty](x)}$$

Then:

$$F_{X_n}(x) = x^{\frac{1}{n}}\mathbb{1}_{[0,1]}(x) + \mathbb{1}_{[1,\infty]}(x)$$

As n goes to infinity, I think I get the following function:

$$G(x) = \mathbb{1}_{[0,\infty]}(x)$$ The theory tells me that this is the distribution function of a random $Y$ variable such as $X_n \to Y$ in law. This may sound stupid, but which random variable has such distribution?

To me it feels like it must be $X = 0$ because deriving $G(x)$ gives me just 0 which should be the density.

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  • $\begingroup$ Yes. $F(Y) = \mathbb{1}_{[0,\infty)(y)}$ is the CDF of a constant zero random variable. The pointwise limit of the $X_n$s' CDFs is the similar looking $\mathbb{1}_{(0,\infty)(y)}$ but that is not a CDF as it is not càdlàg $\endgroup$ – Henry Jan 18 at 16:24
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Actually, $F_{X_n}(0)=0$ for all $n$ hence the limiting function is the indicator function of $(0,+\infty)$. This is not a cumulative distribution function, but this is not a problem. Indeed, for the convergence in distribution to a random variable $Y$, we only need the pointwise convergence of the cumulative distribution function of $X_n$ to the cumulative distribution function of $Y$ where this one is continuous. Here $Y=0$ and the only discontinuity point of the c.d.f. of $Y$ is $0$; we have the pointwise convergence at all the other points.

Actually, the convergence $X_n\to 0$ holds almost surely hence in probability hence in distribution.

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