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Let $X_n$, $Y_n$ and $Y$ be random vectors. If $X_n \xrightarrow[]{d} X$ and $X_n - Y_n \xrightarrow[]{a.s.} 0$. Can we prove that $Y_n \xrightarrow[]{d} X$ ?

Michael Baudin
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    In Van der Vaart, A. W. (2000). Asymptotic statistics (Vol. 3). Cambridge university press, theorem 2.7 (iv) page10 the same result is established but with convergence in probability instead of almost sure convergence. – Michael Baudin Nov 22 '22 at 11:30
  • Doesn't almost sure convergence imply convergence in probability? – whuber Nov 22 '22 at 15:13
  • This is Slutsky's theorem and as whuber told, $\rm a. s. $ convergence implies $\overset{\mathrm P}{\to}.$ – User1865345 Nov 22 '22 at 15:52

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Yes, we can.

First, define $Z_n := X_n-Y_n$

Now, \begin{align} Z_n \overset{a.s}{\to} 0 & \implies Z_n \overset{p}{\to} 0 \quad \text{(van der Vaart, Thm. 2.7. (i))} \\ & \iff \underset{n \to \infty}{\lim}\mathbb{P}(|Z_n-0|>\epsilon)=0, \forall \epsilon>0 \quad \text{(definition of $\overset{p}{\to}$)} \\ & \iff \underset{n \to \infty}{\lim}\mathbb{P}(|(-Z_n)-0|>\epsilon)=0, \forall \epsilon>0 \quad \text{(as $|Z_n-0|=|Z_n|=|-Z_n|=|(-Z_n)-0|$)} \\ & \iff (-Z_n) \overset{p}{\to} 0 \end{align}

where the reference is: Vaart, A. (1998). Asymptotic Statistics (Cambridge Series in Statistical and Probabilistic Mathematics). Cambridge: Cambridge University Press.

Further, we have $Y_n = X_n - (X_n-Y_n) = X_n - Z_n = X_n + (-Z_n)$

So, as $X_n \overset{d}{\to} X$ and $(-Z_n) \overset{p}{\to} 0 $, we have by Slutsky's Lemma (van der Vaart Lemma 2.8 (i)) that $Y_n = X_n + (-Z_n) \overset{d}{\to} X + 0 = X$ i.e. $Y_n \overset{d}{\to} X$

Random_Student
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