I'm having some trouble with this problem:
Let $(X_n)_{n\ge1}$ a sequence of random variables that for all ${n\ge1}$, $X_n$ follows an exponential law with parameter $1/n$. Let $Y_n = X_n – \left\lfloor X_n\right\rfloor$.
$(Y_n)_{n\ge1}$ converges in distribution to which law?
I know for sure it is a Beta distribution, but I don't know the parameters. How should I go about it?
Best Answer
You can try to calculate cumulative distribution function of $Y_n$. Indeed, $Y_n$ takes values $(0,1)$ almost surely. Hence for $t \in (0,1)$ we have: $$ \mathbb P(Y_n \le t) = \mathbb P(X_n \in (k,k+t] ; k \in \mathbb N) = \sum_{k=0}^\infty \int_k^{k+t} \frac{1}{n}\exp(-\frac{1}{n}x)dx = \sum_{k=0}^\infty -\exp(-\frac{1}{n}x) \Big |_{x=k}^{x=t+k} = \sum_{k=0}^\infty \exp(-\frac{k}{n}) - \exp(-\frac{k}{n} - \frac{t}{n}) = \sum_{k=0}^\infty \exp(-\frac{k}{n})(1-\exp(-\frac{t}{n}) = (1-\exp(-\frac{t}{n}))\sum_{k=0}^\infty (\exp(-\frac{1}{n}))^k $$
So that $$F_{Y_n}(t) = \begin{cases} 0 & t \le 0 \\ \frac{1-\exp(-\frac{t}{n})}{1-\exp(-\frac{1}{n})} & t \in (0,1) \\ 1 & t \ge 1 \end{cases}$$
As $n \to \infty$ we have $$F_{Y_n}(t) \to \begin{cases} 0 & t \le 0 \\ t & t \in (0,1) \\ 1 & t \ge 1 \end{cases}$$
So that $Y_n \Rightarrow Y$, where $Y$ is uniformly distributed on $(0,1)$
Calculation of limit: Note that $\exp(x) = 1 + x + o(x)$, hence $$ \frac{1-\exp(-\frac{t}{n})}{1-\exp(-\frac{1}{n})} = \frac{\frac{t}{n} + o(\frac{1}{n})}{\frac{1}{n} + o(\frac{1}{n})} = \frac{t + o(1)}{1 + o(1)} \to t$$