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Let $ X_n $ be a family of random variables an $X$ a rv with $X_n \overset{D}\rightarrow X$. Let $ a_n$ a sequence of real numbers with $a_n \overset{n \rightarrow \infty} \rightarrow a $. We say that the Distribution function of $X$ is continuous.

a) Proof that for all $x \in \mathbb{R}$ an every $\epsilon >0$ there exist a $N_{\epsilon,x} \in \mathbb{N}$, thus for all $n \geq N_{\epsilon,x}:$

$P(X+a \leq x - \epsilon) - \epsilon \leq P(X_n + a_n \leq x) \leq P(X+a \leq x + \epsilon) + \epsilon$

b) Show that

$ X_n + a_n \overset{D}\rightarrow X+a$ for $n \rightarrow \infty$

Can anybody give me a hint? I have no idea how to start with the excersice.

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$P(X_n+a_n \leq x) \leq P(X_n \leq x-a+\epsilon)$ for $n$ sufficiently large. Now use the fact that $\lim \sup P(X_n \leq y) \leq P(X\leq y)$ for all $y$. Hence $\lim \sup P(X_n+a_n \leq x-a+\epsilon) \leq P(X \leq x-a+\epsilon)$. It follows that $ P(X_n+a_n \leq x) < P(X+a \leq x+\epsilon)+\epsilon$ for $n$ sufficiently large. The left hand inequality in a) is proved in a similar manner.

b) follows immeditely from a): if $x$ is a continuity point for $X+a$ then the right and left extremes in the inequality in a) both tend to $P(X+a \leq x)$.

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  • $\begingroup$ thanks, i also could show the left inequality in a) $\endgroup$ – Kaya Feb 20 at 15:00

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