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Let $\{ \xi \}^{\infty}_{n=1}$, $\{ \eta \}^{\infty}_{n=1}$, $\{ \zeta \}^{\infty}_{n=1}$ - sequence of random variables.

Proof if

1) $\xi_n \xrightarrow{d} \xi $

2) $\vert \xi_n - \eta_n \vert \leq \zeta_n \vert \xi_n \vert $

3) $\zeta_n \xrightarrow{P} 0 $

then $\eta_n \xrightarrow{d} \xi $

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    This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – 2017-02-15

1 Answers 1

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Use Slutsky's theorem twice.

Prove first that $\zeta_n|\xi_n|\xrightarrow{d} 0$, which implies convergence to zero in probability and convergence $|\xi_n-\eta_n|\xrightarrow{p} 0$.

Then apply a second time Slutsky's theorem to the sum $\eta_n=(\eta_n-\xi_n)+\xi_n$.

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    why $\zeta_n \vert \xi_n \vert \xrightarrow{d} 0$ ? – 2017-02-15
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    @Артем Have you looked at Slutsky's Theorem? – 2017-02-15
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    yes, but in Slutsky's Theorem we proved that $\zeta_n \xi_n \xrightarrow{d} 0$. But, how from this follows your statement? – 2017-02-15
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    @Артем Can you prove that $|\xi_n|\xrightarrow{d}|\xi|$? – 2017-02-15
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    it seems that this is not true – 2017-02-15
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    By the definition of probability convergence.... – 2017-02-15
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    Composition of continuous functions is a continuous function. – 2017-02-15
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    @АртемКуприянов https://en.wikipedia.org/wiki/Continuous_mapping_theorem – 2017-02-15