Let $X_n\longrightarrow X$ in distribution and $Y_n\longrightarrow c$ in distribution with $c$ constant. I'm trying to show that $X_n+Y_n\longrightarrow X+c$ in distribution. The correction goes like :

I proves that $$\mathbb P\{X_n+Y_n\leq x\}\leq \mathbb P\{X_n+c\leq x+\varepsilon\}+\mathbb P\{|Y_n-c|>\varepsilon\}$$ and $$\mathbb P\{X_n+c\leq x-\varepsilon\}\leq \mathbb P\{X_n+Y_n\leq x\}+\mathbb P\{|Y_n-c|>\varepsilon\}.$$

And then, they say that since $Y_n\to c$ in probability (I already proves that if $Y_n\to c$ in distribution, then $Y_n\to c$ in probability), the claim follow.

Question : I don't understand why the claim follow. Let $x$ a continuity point of $F_{X+c}$. We have that $$\mathbb P\{X_n+c\leq x-\varepsilon\}-\mathbb P\{X+c\leq x\}-\mathbb P\{|Y_n-c|>\varepsilon\}\leq \mathbb P\{X_n+Y_n\leq x\}-\mathbb P\{X+c\leq x\}\leq \mathbb P\{X_n+c\leq x+\varepsilon\}-\mathbb P\{X+c\leq x\}+\mathbb P\{|Y_n-c|>\varepsilon\}$$

Q1) First, if $X_n\longrightarrow X$ in distribution, I don't see why $X_n+c\longrightarrow X+c$ in distribution. Indeed, if $x$ is a continuity point of $X+c$, we have that $$\mathbb P\{X_n+c\leq x\}=\mathbb P\{X_n\leq x-c\}$$ and since $x-c$ is not a continuity point of $X$, there is no reason for it to converge to $\mathbb P\{X\leq x-c\}$, am I right ?

Q2) So here, there is no reason that $\mathbb P\{X_n+c\leq x\pm\varepsilon\}$ converge to $\mathbb P\{X+c\leq x\pm \varepsilon\}$. So how can I conclude ?

  • $\begingroup$ Sorry but, if $x$ is a continuity point of $X+c$, then $x-c$ is a continuity point of $X$ (those are actually equivalent). Just check the respective CDFs. $\endgroup$
    – Did
    Jan 12 '17 at 18:09

If $x$ is a continuity point of $X+c$, then $x-c$ is a continuity point of $X$. This is evident from a picture. Or you can prove this from the definition of continuity, and the relationship between the CDF of $X$ and the CDF of $X+c$.


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