# Convergence of sequence of uniforms

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.

• 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 – Henry Jan 18 '19 at 16:24

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.