Let $X$ be a random variable with cdf $F(\cdot)$ and $$X_n = \frac{n}{n+\sqrt{n}}X$$ for $n\in \Bbb N$. Show that $X_n$ converges to $X$ in distribution.

This is all the information I am given in the problem. We can say a sequence of random variables converges in distribution if $\lim_{n \rightarrow \infty} F_n(x) =F(x)$. Perhaps we could start this problem by looking at the cdf of the nth random variable. $$F_n(x) = P( X_n \le x)= P \left( \frac{n}{n+\sqrt{n}}X \le x \right) = P\left(X \le \frac{n+ \sqrt{n}}{n}x\right)$$ Therefore $$\lim_{n \rightarrow \infty} \ P\left(X \le \frac{n+ \sqrt{n}}{n}x\right) = P(X \le x)=F(x)$$

  • 1
    $\begingroup$ What is the cdf of $X_n$? Do they converge to $F$? $\endgroup$
    – user251257
    Commented Feb 27, 2016 at 8:37
  • $\begingroup$ Yes, your calculation seems fine to me. $\endgroup$
    – Jimmy R.
    Commented Feb 27, 2016 at 11:39
  • 2
    $\begingroup$ Actually, using $<$ instead of $\le$ in the CDF makes the computation wrong since, for every sequence $(x_n)$ with $x_n>x$ and $x_n\to x$, one has $$\lim_{n\to\infty}P(X<x_n)=P(X\le x).$$ $\endgroup$
    – Did
    Commented Feb 27, 2016 at 12:10
  • $\begingroup$ If $X$ is not absolutely continuous then this limit need only hold for points at which $F$ is continuous. $\endgroup$
    – Math1000
    Commented Feb 27, 2016 at 19:37


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