If $X_n\sim \text{Exp}(a_n)$, then $$X_n\stackrel{\mathcal{L}}{\longrightarrow}X\sim \text{Exp}(a)$$ as $n\to\infty$ and $a_n\to a$, $n\geq$ 1.

I know that,

$\Pr(Y_n\geq a_n+\epsilon)=\Pr(X_i\geq a_n+\epsilon)^{n}= \left(1-F_X(a_n+\epsilon)\right)^{n}$

$$F_n(x)=1-e^{-xa_n}\quad F_n(x)=1-(1-e^{a_n+\epsilon})^{n}.$$ Thus, this converge to $e^{a+\epsilon}$ right?

  • $\begingroup$ The power of $n$ in the second $F_n$ is incorrect. Recall that convergence in distribution is equivalent to convergence of the cdf at all continuity points of $F$. $\endgroup$ – Xi'an Oct 31 '16 at 13:10
  • $\begingroup$ This doesn't make a lot of sense. How did $x$ disappear from the right hand side in your second expression for $F_n(x)$? Also a limit as $n \to \infty$ can't depend on $n$. $\endgroup$ – dsaxton Oct 31 '16 at 13:11

Just use the fact that $e^{-x a_n}$ is a continuous function of $a_n$. This means we can "take the limit inside" like this:

\begin{align} \lim_{n \to \infty} F_n(x) &= \lim_{n \to \infty} \left ( 1 - e^{-x a_n} \right ) \\ &= 1 - e^{- x \lim_{n \to \infty} a_n} \\ &= 1 - e^{- x a} . \end{align}


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